TENTH
EDITION
Elementary
Linear Algebra
APPLICATIONS VERSION
/
I
HOWARD ANTON / CHRIS RORRES
STUDENT SOLUTIONS MANUAL
C Trimble & Associates
I
STUDENT SOLUTIONS MANUAL
C Trimble & Associates
ELEMENTARY
LINEAR ALGEBRA
APPLICATIONS VERSION
Tenth Edition
Howard Anton
Professor Emeritus, Drexei University
Chris Rorres
University ofPennsyivania
WILEY
JOHN WILEY & SONS,INC.
Cover art: Norm Christiansen
Copyright © 2011,2005,2000,1994 John Wiley & Sons, Inc. All rights reserved.
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ISBN 978-0-470-45822-8
10 9 8 76 5 4 3
Printed and bound by Bind-Rite/Robbinsville
Contents
Chapter 1
1
Chapter 2
44
Chapter 3
61
Chapter 4
81
Chapter 5
120
Chapter 6
138
Chapter 7
170
Chapter 8
185
Chapter 9
199
Chapter 10
211
Chapter 1
Systems of Linear Equations and Matrices
Section 1.1
Exercise Set 1.1
1. (a),(c), and (f) are linear equations in X], a:2, and X3.
(b) is not linear because of the term X1X3.
(d) is not linear because of the term x~'^.
(e) is not linear because of the term
3. (a) and (d) are linear systems,
(b) is not a linear system because the first and second equations are not linear,
(c) is not a linear system because the second equation is not linear.
5. By inspection,(a) and (d) are both consistent; x^ =3, X2 = 2, X3 = -2, X4 = 1 is a solution of(a) and
X| = 1, X2 = 3, X3 = 2, X4 = 2 is a solution of(d). Note that both systems have infinitely many solutions.
7. (a),(d), and (e) are solutions,
(b) and (c) do not satisfy any of the equations.
9. (a) 7x-5>- = 3
5
3
X=
-y +7
7
Let y = t. The solution is
5
3
-y +-
X=
7
7
y =t
(b) -8x1 + 2x2“5x3 +6x4 = 1
1
X, =-X2
Let X2 = r, X3 =
1
5
3
5
3
1
X3 +-X4
' 4 ^ 8
4 ^ 8
and X4 = t. The solution is
1
Xi = — r — ^ + —f —
' 4
8
4
8
X2=r
X3 =i
X4 =t
2
11. (a)
3
0
2x1
0 0
-4
0
corresponds to
1
=0
3xi -4x2 = 0X2 =1
1
SSM: Elementary Linear Algebra
Chapter 1: Systems of Linear Equations and Matrices
3
(b)
0-2
5
4
-3
7
0
15. If(a, b, c) is a solution of the system, then
aaf +bX]+c = y,, ax\ +bx2+c = y2, and
corresponds to
7
-2
ax^ +bx2,+ c = y3 which simply means that the
points are on the curve.
3a-1
-2x3 =5
7x| +X2 +4x3 =“3-2x2 + -'^3 = ^
(c)
7
J
2
1
-3
5
2 4
0
1
17. The solutions of x^+kx2-c are X| -c-kt,
X2 = t where t is any real number. If these
satisfy x^+1x2 =d,
c- kt + It = d or
corresponds to
c-d ={k- 1) t for all real numbers t. In
particular, if r = 0, then c = d, and if r = 1, then
7xi + 2x2 + -^3“■^■^4 “ 5
X| + 2x2 + 4-^3
=•
k=l.
True/False 1.1
1
0
0
0
0
1
0
0
0
0
1
0
.0
0
0
1
(d)
7
(a)
True; Xj = X2 = ■ • ■ = x„ =0 will be a solution.
(b)
False; only multiplication by nonzero constants
is acceptable.
(c)
True; if k = 6 the system has infinitely many
solutions, while if k^6, the system has no
corresponds to
4
^1
7
^2
-2
^3
3
X4
solution.
4
13. (a) The augmented matrix for -2x| = 6
(d)
True; the equation can be solved for one variable
in terms of the other(s), yielding parametric
equations that give infinitely many solutions.
(e)
False; the system
3x,1 =8
9x, =-3
-2
IS
6
3
9
-3
3x-5y = -7 has the
2x + 9y = 20
6x-10y = -14
solution X = 1, y = 2.
(b) The augmented matrix for
6x1 -X7 +3x3 = 4 is ^
3
*
,0
5X2 “-^3=1
(f)
4
system.
1
5
(g)
True; subtracting one equation from another is
the same as multiplying an equation by -1 and
adding it to another.
(b)
False; the second row corresponds to the
(c) The augmented matrix for
2x2
- 3x4 + X5 = 0
-3x| - X2 + X3
bX] + 2x2 - -'^3 + 2x4 - 3x5 = 6
0
IS
2
-3
6
2
0
-3
1
1
0
0
-1
2
-3
False; multiplying an equation by a nonzero
constant c does not change the solutions of the
equation 0x| +0x2 =
or 0 = -1 which is false.
0
Section 1.2
6
Exercise Set 1.2
(d) The augmented matrix for X| -Xj =7 is
[1
0
0
0
-1
1. (a) The matrix is in both row echelon and
reduced row echelon form.
7].
(b) The matrix is in both row echelon and
reduced row echelon form.
2
Section 1.2
SSM: Elementary Linear Algebra
(c) The matrix is in both row echelon and
(d) The last line of the matrix corresponds to the
reduced row echelon form.
equation OX] +0x2 +0-^3“ which is not
satisfied by any values of X\, X2, and X3,
(d) The matrix is in both row echelon and
so the system is inconsistent.
reduced row echelon form.
(e) The matrix is in both row echelon and
reduced row echelon form.
5. The augmented matrix is
I
I
-I
-2
2
3
3 -7
4
lO
(f) The matrix is in both row echelon and
reduced row echelon form.
Add the first row to the second row and add -3
times the first row to the third row.
(g) The matrix is in row echelon form.
3. (a) The matrix corresponds to the system
X] -3x2
X] = 7+ 3x2“4x3
X2 + 2x3 =2 or X2 = 2-2x3
X3 = 5
X3 = 5
I
I
2
8
0
-I
5
9
0
-lO
-2
-14
Multiply the second row by -1.
2
8
Thus X3 = 5, X2 = 2-2(5)= -8, and
0
1
-5
-9
X| =7+3(-8)-4(5)= -37. The solution is
0
-10
-2
-14
X| =-37, X2 =-8, X3 =5.
Add 10 times the second row to the third row.
2
8
-5
-9
0 0 -52
-104
1
(b) The matrix corresponds to the system
X,
0
+8x3 -5x4 =6
X2 +4x3 -9x4 = 3 or
X3 +X4=2
1
1
Multiply the third row by
X| = 6-8x3 +5x4
X2 = 3-4x3 +9x4 .
X3 = 2- X4
Let X4 = t, then X3 =2-t,
1
1
2
0
1
-5
-9
0
0
1
2
.
52
Add 5 times the third row to the second row and
X2 =3-4(2-0+ 9r = 13/-5, and
-2 times the third row to the first row.
X] =6-8(2-r)+ 5f = 13r -10. The solution
isx|=13f-10, X2=13?-5, X3=-r +2,
X4 =t.
(c) The matrix corresponds to the system
1
1
0
0
1
0
4
1
0 0
1
2
Add -1 times the second row to the first row.
X] +7x2 “2x3
-8x5 =-3
X3 + X4 +6x5 =5 or
X4 +3x5 =9
1
0 0
0
X] --3-1x2+ ^^3 ^As
X3 = 5- X4 -6x5
X4 =9-3x5
3
1
0
1
0 0
1
2
The solution is X| = 3, X2 = 1, X3 = 2.
7. The augmented matrix is
Let X2=s and X5 = t, then X4 = 9-3t,
X3 = 5-(9-30-6f = -3f -4, and
X, =-3-7^+ 2(-3r-4)+8f = -75+ 2r-IL
1
-1
2
1
-1
2
2 -4
0
-1
-1
-2-2 -2
1
1 ‘
The solution is X] =-75+2f-l 1, X2=s,
,3
X3 = -3t-4, X4 = -3t + 9, X5 = t.
Add -2 times the first row to the second row, 1
times the first row to the third row, and -3 times
0 -3 -3.
the first row to the fourth row.
3
SSM:Elementary Linear Algebra
Chapter 1: Systems of Linear Equations and Matrices
1
-1
13. Since the system has more unknowns(4) than
equations (3), it has nontrivial solutions.
2
-1
-1
0
3 -6
0
0
0
1
-2
0
0
,0
3 -6
0
0.
15. Since the system has more unknowns(3)than
equations (2), it has nontrivial solutions.
1
Multiply the second row by -
and then add -1
2
1
17. The augmented matrix is
times the new second row to the third row and
0
add -3 times the new second row to the fourth
1
3 0
2 0 0
1
1
0
row.
Interchange the first and second rows, then add
1
-1
2
0
1
0
0
.0
-1
-1
-2
0
0
0
0
0
0
0
0
-2 times the new first row to the second row.
1
0.
2 0 0
0-3
3 0
0
1
1
0
Add the second row to the first row.
"l 0
0
0-1 -l’
Multiply the second row by —, then add -1
3
1
-2
0
0
0 0
0
0
0
.0 0
0
0
0.
1
2
The corresponding system of equations is
0
1
times the new second row to the third row.
-H’= -l
X
x= w-l
or
=0
y-2z
y = 2z
0 0
Multiply the third row by —, then add the new
y-2s,z-s,w = t.
third row to the second row.
9. In Exercise 5, the following row echelon matrix
1
occurred.
1
2
8
0
1
-5
-9
0 0
1
2
0
2 0
■
Let z = s and w = t. The solution is x = f- 1,
1
0 0
-1
2 0 0
0
1
0
0
0 0
1
0
Add -2 times the second row to the first row.
1
0 0 0
The corresponding system of equations is
0
1
0
0
Xj + X2 + 2x^ — 8
Xi =-X2 -2x3 +8
X2 -5x3 = -9 or X2 = 6x3 -9
X3=2
X3 -2
0 0
1
0
The solution, which can be read from the matrix,
is X] = 0, X2 = 0, X3 = 0.
Since X3 = 2, X2 = 5(2)-9 = 1, and
X] =-1-2(2)+8= 3. The solution is X| =3,
3
19. The augmented matrix is
X2=l, X3=2.
1
1
1
0
,5 -1 1 -1 Oj
1
Multiply the first row by —, then add -5 times
11. From Exercise 7, one row echelon form of the
1
-1
2
-1
-1
0
1
-2
0
0
0
0
0
0
0
.0
0
0
0
0.
augmented matrix is
3
the new first row to the second row.
1
0
I
I
3
3
8
_2
3
3
j »■
f “J
The corresponding system is
x-y + 2z-w = -l
or
3
Multiply the second row by —.
x= y-22 + vv-l
8
y-2z
=0
y = 2z
Let z = s and w = t. Then y = 2s and
fl -L3 -L3 i3 Ol
x=2s-2s + t - 1 = r- 1. The solution is
0
x = t- \,y = 2s, z = s,w = t.
4
1
T
4
1
0
SSM: Elementary Linear Algebra
Section 1.2
1
w
=0
w= y
=0 or x = -y .
-y
Add — times the second row to the first row.
x+y
3
z=0
1 0^0
0
4
z=0
Let y = t. The solution is w = t, x = -t, y = t,
0 1 ^
1 0
4
z = 0.
This corresponds to the system
2
1
=0
+7^3
4
x.
1
^1 = -7-^3
4
1
or
X2 + — Xj+ X4 — 0
5
2
1
4
4
-1
-3
1
5
8
2
1
4
4
10
0
1
2
2
0-1
3
1
1
0
1
3
-2
0
10
4
0
3 4
9
Add -2 times the first row to the second and
-3 0
fourth rows, and add -3 times the first row to the
2
3
-2
1
1
3
0
third row.
-2 0
1
Interchange the first and second rows.
'1 0-1 -3 0'
2
11
Interchange the first and second rows,
'l
0 -2 7 If
3
-2
9
4
2
2
7
1
X3 = 45, X4 = t.
2
0 -2
-3
X2 = -s-t. The solution is x^ -s,X2 =-s-t,
0
3 4
23. The augmented matrix is .^
^2 =-7^3--3^4
Let X3 = 4i and X4 = t. Then Xj =-s and
21. The augmented matrix is
-1
0
4 0
0-2
7
11
-10
-13
-1
7
0-3
7
-16
-25
0
8
-10
-12
1
Multiply the second row by -1, then add 3 times
the new second row to the third row and -1
Add -2 times the first row to the third row and 2
times the new second row to the fourth row.
times the first row to the fourth row.
'1 0
'1 0 -1 -3 0'
0
-2
7
if
1
-7
10
13
0
2
2
4
0
0 0
-14
14
14
0
3
3
7
0
0 0
15
-20
-25
0
1
1
-8
0
1
Multiply the third row by
Multiply the second row by —, then add -3
1
-1 times the new second row to the fourth row.
1
0-1
-3
0
i
l
0
0
0
1
0
0
0
0
-10
0
, then add -15
times the new third row to the fourth row.
2
times the new second row to the third row and
o
14
2 0
0-2
0
1
-7
0
0
1
0 0
7
11
10
13
-1
0 -5
-10
1
Multiply the fourth row by —, then add the
5
Add 10 times the third row to the fourth row,-2
times the third row to the second row, and 3
new fourth row to the third row, add -10 times
the new fourth row to the second row, and add
times the third row to the first row.
-7 times the new fourth row to the first row.
'1 0-1 0 0'
0
1
1
0 0
0
0 0
0
'1 0-2 0 -3'
0 0
1
0
0
0 0
The corresponding system is
1
-7 0-7
0 0
1
0
1
0 0
0
1
2
Add 7 times the third row to the second row and
2 times the third row to the first row.
5
SSM: Elementary Linear Algebra
Chapter 1: Systems of Linear Equations and Matrices
I
0
0 0
-I
0
I
0 0
0
0 0
I
0
0 0 0 1
2
1
1
tbe new first row to the second row.
is /|=-1, /2 = 0, 73 =1, 74 = 2.
3
-3
4
5
2
a
2
2
Add -3 times the first row to the second row and
1
0
0
1
M-b'
3
-4 times the first row to the third row.
9
a I 2b
3
2
-3
4
0
-7
14
-10
9.
2a
The solution is x = —
3
_0 -7 a^-2 a-\4
Add -1 times the second row to the third row.
1
2
-3
4
0
-7
14
-10
0
0
a“-16
a-4
2
9
times the new second row to the first row.
a
1
1
Multiply the second row by —, then add —
^-14 a + 2_
4
I
2
f -^+ b
25. The augmented matrix is
2
a
Multiply the first row by —, then add -3 times
2_
The solution, which can be read from the matrix,
1
I
29. The augmented matrix is 3 6 b '
2b
b
a
9’
3"^ 9 ■
31. Add -2 times the first row to the second row.
'l 3
is in row echelon form. Add -3 times
_0
1.
the second row to the first row.
'1
The third row corresponds to the equation
O'
is another row echelon form for the
0
1
(a“ -16)z = a-4 or (a + 4)(a -4)z = a-4.
matrix.
If a = -4, this equation is Oz = -8, which has no
solution.
33. Let X = sin a, y = cos and z = tan
A:+ 2y + 3z =0
system is 2x +5y +3z = 0.
-x-5y+5z = 0
If a = 4, this equation is Oz = 0, and z is a free
variable. For any other value of a, the solution of
. Thus, if a = 4, the
this equation is z =
a +4
system has infinitely many solutions; if a = -4,
the system has no solution; if a ±4, the system
has exactly one solution.
The augmented matrix is
then the
1
2
3
0
2
5
3
0
-1
-5
5
0
Add -2 times the first row to the second row,
2
27. The augmented matrix is
1
and add the first row to the third row.
2 a'^-5 a-l_'
'l
2
Add -2 times the first row to the second row.
0
1
0
-3
2
0 a^-9 a-3
3 o'
-3 0
8
0
Add 3 times the second row to the third row.
'1 2
The second row corresponds to the equation
0
(a~-9)y = a-3 or (a + 3)ia - 3)y = a - 3.
If a = -3, this equation is Oy = -6, which has no
solution. If a = 3, this equation is Oy = 0, and y is
a free variable. For any other value of a, the
1
3 o'
-3
0
0 0-1
0
Multiply the third row by -1 then add 3 times
the new third row to the second row and -3
times the new third row to the first row.
solution of this equation is y =
a +3
Thus, if a = -3, the system has no solution; if
a = 3, the system has infinitely many solutions;
if a +3, the system has exactly one solution.
1
2
0
0
0
1
0
0
0
0
1
0
Add -2 times the second row to the first row.
6
SSM: Elementary Linear Algebra
1
0
Section 1.2
X= 1 =>x = ±l
0 0 0
1
K =: 3 => y = ±yj3
0 0
0 0
1
0
Z=2^ z=±42
The system has only the trivial solution, which
The solutions are X = ±1, y = ±V3, ^=±^/2.
corresponds to sin or = 0, cosy5 = 0, tan;'= 0.
For 0 < or < Ik, sin or =0 => or = 0, ;r, 2k.
37. (0, 10):r/= 10
{\,lYa + b + c + d = l
(3,-11); 27a +% + 3c + J = -ll
(4, -14); 6Aa+\6b + Ac + d = -14
The system is
For Q< B < 2k, cosy5 = 0=> B-—,—.
2
2
For 0< y< 2k, tan f=0^ 7= 0,
Thus, the original system has
2k.
+b +c+d=l
a
3 ■ 2 • 3 = 18 solutions.
21a +9b +3c+ d =-\l
64a + \6b +4c+ d =-14
35. Let X =x^, Y = y^, and Z = z“, then the
and the augmented
d = 10
system is
1
1
7
27
1
9
3
1
-11
64
16
4
1
-14
0 0
1
10
X+F+Z=6
X-T + 2Z = 2
matrix is
2Z+T-Z=3
0
1
The augmented matrix is
1
1
2
1
1
6
2
2
-1
Add -27 times the first row to the second row
and -64 times the first row to the third row.
3
1
Add -1 times the first row to the second row and
1
1
7
0
-18
-24
-26
-200
0
-48
-60
-63
-462
0
0
0
1
10
-2 times the first row to the third row.
6
0
-2
0
-1
1
-4
-3 -9
Multiply the second row by
, then add 48
18
times the new second row to the third row.
Multiply the second row by -—, then add the
1
1
7
new second row to the third row.
6
0
^ 2
0 0 -| -7
2
0 1 I M
0 0 4 ^
214
0 0
10
0
1
'1
times the new third row to the second row and
1
1
0 1
4
13
100
9
9
1
19
J07
12
6
0
1
10
1 0 4'
1
0
3
0
0 0
1
2
0
3
19
Add
Add -1 times the second row to the first row.
“1 0 0
0
1
0
0
0
1
times the fourth row to the third row,
12
r
13
—^ times the fourth row to the second row,
3 .
9
2
and -1 times the fourth row to the first row.
X
This corresponds to the system
7 ■
1
0 0
0
3
4
2
-1 times the new third row to the first row.
'l
9
Multiply the third row by -.
1
Multiply the third row by —, then add —
7
100
Y
=3.
Z=2
7
SSM: Elementary Linear Algebra
Chapter 1: Systems of Linear Equations and Matrices
1
1
0
1
1
43. (a) There are eight possibilities, p and q are any
0
-3
I 0
10
real numbers.
‘l 0 o'
0 0
1
0
2
0 0
0
1
10
1
p 0
I
0
1 0
0 1
p
q
0
0
0 0
1
0 0
0
0
0 0
q
0
0
1
4
Add — times the third row to the second row
0
3
and -1 times the third row to the first row.
1
1
0 0
0
1
0 0-6
0 0
1
0
0 0 0
1
1
0
1
1
p
0 0
1
0
0 0
0 0
0
0 0 0
0
0 0
0 0
0
p
-5
0 0
2
10
0 0 0
1
0 0 0 , and
0 0 0
0 0 0
0 0 0
Add -1 times the second row to the first row.
'1 0 0 0
0
1
(b) There are 16 possibilities, p, q, r, and s are
any real numbers.
1
0 0-6
1
0
2
0 0 0
1
10
0 0
'1 0 0 0
The coefficients are a = 1,
= -6,c = 2,
d= 10.
Thus the system has exactly one solution.
b
d
1 ^
a
a
c
Q
d
ad-bc
a
1 i
1
0
1
0
0 0 0
1
1 0 0 p
0 1 0 r/
0 0
1
r
0 0 0
0
'l p 0 0
0
0
1
0
0
0
0
1
0
0 0
1
0
0
0 0
0
0 0 0
0
1
0 0
0
0
I 0 p q
1
0
\
r
s
0 0 0
1
0 0
0
0
0 0 0
0
0 0
0
0
1
0, the sequence of matrices is
c
0
0 0
1 0 p 0
q 0
form 0 1 0 £^2 •
0 0 1 J3
a
1
0
39. Since the homogeneous system has only the
trivial solution, using the same steps of GaussJordan elimination will reduce the augmented
matrix of the nonhomogeneous system to the
'l 0 0 J1
41. (a) If a
0
0
\
p
q 0
0
0
1
r
0
0
0
0
0 0
0
0
0
0 0
0
0 0
0
0
0
0 0
0
1 0 p
0 1 q
0
1
p 0 q
0
0
1 ■
0 0 0
1
p 0
0 0
0
0
0 0
0 0
1
0 0 0
0
0 0
0 0
0 0
1
0
I
p
q
0 0 0
1
0
0
0 0
0 0 0 0
0
0
0 0
0 0 0 0
0
0
0 0
a
0
0
1
If a = 0, the sequence of matrices is
1 ^
0 b
c d
c
d
1
0
0
b
0
b
d_
c
c
0
1
0
r
1 ■
0
1
p
0 0 0
0
0 0
0 0 0
0
0 0
0 0 0
0
0 0
p
0
0 0
0 0
p
0
0 0
(b) Since ad- bc^ 0, the lines are neither
identical nor parallel, so the augmented
a
b
k
will reduce to
matrix
c
1
0
k1
0
1
/1
d
I
0 0 0
0
and the system has exactly one
1
0 0 0 0
0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
solution.
8
, and
0 0 0 0
0 0 0 0
SSM: Elementary Linear Algebra
Section 1.3
True/False 1.2
1 +6
5+1
3. (a) D + E = -l +(-l) 0+ 1
(a) True; reduced row echelon form is a row echelon
3+4
form.
(b) False; for instance, adding a nonzero multiple of
the first row to any other row will result in a
7
6
-2
1
3
7
3
7
2+3
1 +2
2+1
4+3
5-1
2-3
5
matrix that is not in row echelon form.
1-6
(c) False; see Exercise 31.
(b) D-E = -l-(-l) 0-1
3-4
(d) True; the sum of the number of leading 1 ’s and
the number of free variables is n.
2-1
-5
4
-1
0
-1
-1
(e) True; a column cannot contain more than one
1-2
4-3
1
leading 1.
(f) False; in row echelon form, there can be nonzero
entries above the leading 1 in a column.
(g) True; since there are n leading 1 ’s, there are no
free variables, so only the trivial solution is
possible.
15
0
(c) 5/1 = -5
10
5
5
(d) -7C =
-7
-28
-14
-21
-7
-35
(h) False; if the system has more equations than
unknowns, a row of zeros does not indicate
(e) 2B - C is not defined since 2B is a 2 x 2
infinitely many solutions.
matrix and C is a 2 x 3 matrix.
(i) False; the system could be inconsistent.
(f) 4E-2D =
Section 1.3
24
4
-4
4
16
4
22
-6
8
-2
4
6
10
0
4
12
2
+
-2
2
4
8
2
6
Exercise Set 1.3
1. (a) BA is undefined.
12
2
10
4
-2
0
2
6
4
8
12
(b) AC+ Z) is 4x2.
(g) -3(D +2E)
/r
(c) AE + 6 is undefined, since A£ is 4 x 4 not
4x5.
= -3
1
5
2
-1
0
1
3
2
4
VL
(d) AB + B is undefined since AB is undefined.
(e) £(A+B)is 5x5.
13
7
= -3 -3
2
5
11 4 10_
(f) £(AC)is 5x2.
(g) E^ A is undefined since E^ is 4x5.
(h) {A^ +E)D is 5x2.
-39
-21
-24
9
-6
-15
-33
-12
-30
0 0
(h) A-A = 0 0
0 0
(i) tr(D)= 1 + 0 + 4 = 5
9
6
SSM: Elementary Linear Algebra
Chapter 1: Systems of Linear Equations and Matrices
/-
1
0) tr(D-3£)= tr
5
18
2
-1 0
1
3 2 4_^ ^ 12
VL
3
9
-3 3 6
3
9
J/
/r
-17
2
-7
2
-3
-5
-9
-1
-5
= tr
Jy
VL
= -17-3-5
= -25
n\
/r
28
(k) 4tr(7^)= 4tr
-7
0
14jy
= 4(28+14)
= 4(42)
= 168
(1) tr(/t) is not defined because A is not a square matrix.
3
0
5. (a) AB =
I
(3-4)+(0-0) -(3- l)+(0-2)
-(l -4)+(2-0) (M)+(2-2)
(l -4)+(L0) -(M)+(l -2)
12 -3‘
-4
5
4
1
(b) BA is not defined since 5 is a 2 x 2 matrix and A is 3 x 2.
18
(c)
5
2
OE)D = -3 3 6 -1 0
3
9
1
1
12 3 9j[_ 3 2 4_
(d) (4B)C =
■(18- l)-(3-l) + (9-3)
(18-5) + (3-0) + (9-2)
(18-2) + (3-l) + (9-4)
-(3-l)-(3- l) + (6-3)
(12-l)-(3- l) + (9-3)
-(3-5) + (3-0) + (6-2)
(12-5) + (3-0) + (9-2)
-(3-2) + (3-1) + (6-4)
(12-2) + (3-l) + (9-4)
42
108
75“
12
-3
21
36
78
63
12
-3
-4
5 * 4 2
4 iJL^ 1 5j
(12-l)-(3-3)
-(4- l) + (5-3)
(4-l) + (l -3)
3
45
9
11
-11
17
7
17
13
(12-4)-(3- l)
-(4-4) + (5-l)
(4-4) + (M)
(12-2)-(3-5)
-(4-2) + (5-5)
(4-2) + (l -5)
10
SSM: Elementary Linear Algebra
3
0
Section 1.3
/r
4
-I
1
4
0
2
3
1
-|\
2
2
(e) A(BC)=
1
3
5J/
0
(4-l)-(l-3) (4-4)-(M) (4-2)-(l-5)
(0-l)+(2-3) (0-4)+(2-l) (0-2)+(2-5)
2
3
0
I
15
3
6
2
10
2
1
(3-l)+(0-6) (3-15)+(0-2) (3-3)+(0-10)
-(M)+(2-6) -(M5)+(2-2) -(1 ■3) + (2-10)
(l- !) + (l-6)
(M5) + (l-2)
(l-3) + (M0)
3
45
9“
11
-11
17
7
17
13
1
(f) cd =
!
4
2
3
I
5
3
4
1
2
5
(1 .1) + (4.4) + (2-2)
(3- l) + (l-4) + (5-2)
21 17'
17
(1-3) + (4-1) + (2-5)
(3-3) + (M) + (5-5)
35
1 05 2]r
3 OlV
1
-1
2
(g) {DAf =
-1
3
2
4
1
1
J/
(l-3)-(5-l) + (2-l)
-(l -3)-(0-l) + (M)
(3-3)-(2-l) + (4-l)
VL.
rv
0
12
(l•0) + (5•2) + (2■l)
-(l-0) + (0-2) + (M)
(3-0) + (2'2) + (4-l) J/
^\T
-2
11
8
vt-
0
-2
J/
I I
12
11
SSM: Elementary Linear Algebra
Chapter 1: Systems of Linear Equations and Matrices
1
3
(h) {,dB)A^ = 4 1 4
2 5 0 ^
3
-1
1
0
2
1
\L
(l-4)+(3-0) -(M)+(3-2) p
(4-4)+(l-0) -(4-l)+(l-2)
0
(2-4)+(5-0) -(2-l)+(5-2) L
-1
1
2
1
0
8
-1
1
2
1
8
(4-3)+(5-0) -(4-l)+(5-2) (4-l)+(5-l)
(16-3)-(20) -(161)-(2-2) (161)-(21)
(8-3)+(8-0) -(8-l)+(8-2) (8-l)+(8-l)
12
6
9
48
-20
14
24
8
16
/r
-\\
1
5
2
1
-1
(i) tr(DD^)= tr -1 0 1 5
3
0 2
3 2 4j[2 1 4
jy
/r
(M)+(5-5)+(2-2) -(M)+(5-0)+(2-l) (l-3)+(5-2)+(2-4)
-(M)+(0-5)+(l-2) (M)+(0-0)+(M) -(L3)+(0-2)+(l-4)
(3-l)+(2-5)+(4-2) -(3-l)+(2-0)+(4-l) (3-3)+(2-2)+(4-4)-jy
= tr
VL.
r'r
30
1
21
1
2
1
21
1
29
= tr
-1/
= 30+2+ 29
= 61
/r
24
0 tr(4£^-D)= tr
-4
4
16
1
5
4 4
-1
0
1
3
2
4
12
8
12
23
-9
14
5
4
3
9
6
8
\L_
rr
2
jy
= tr
\i-
jy
= 23+4+8
= 35
12
SSM: Elementary Linear Algebra
Section 1.3
/r
l
(k) tr(C^A^+2£^)= tr
n\
3
4
1
2
5
I2
3
-1
1
0
2
1
+
VL
-2
8
2
2
2
6
4
6
fv
= tr
12 -2 8
(l-3)+(3-0) -(l l)+(3-2) (M)+(3-l)
2 2
(4-3)+(l -0) -(4-l)+(l -2) (4-l)+(M) + 2
6
4 6
(2-3)
+
(5-0)
-(2-l)
+
(5-2)
(2-l)
+
(5-l)
VI-
/r
3 5 4] ["12 -2 8]'
= tr
12
-2
6
8
5 +
2
2
2
7
6
4
6
-i/
\-
T15
= tr
3 12
14
0
7
12
12
13
J/
= 15+0+13
= 28
(1) ir{{EdfA)= ix
6
1
3
1
-1
1
2
4
4
1
3
2
313^ r 3
0
1
-1
2
5
1
J/
T (6-l)+(l -4)+(3-2) (6-3)+(M)+(3-5)
^\T r
-(l l)+(l-4)+(2-2) -(l-3)+(M)+(2-5)
(4-l)+(l-4)+(3-2) (4-3)+(M)+(3-5) J/
\L
= tr
= tr
nr''
16
34
7
8
-1
3 0
2
14
28
1
1
Vv
n\
3 0
16
7
14
34
8
28
-1
= tr
2
/r
(16-3)-(7-l)+(14- l) (16-0)+(7-2)+(14-l)1^
= tr
(34-3)-(8-l)+(28-l) (34-0)+(8-2)+(28-l)J^
\L
/r-
55
28
= tr
122
VL
44J/
= 55+44
= 99
7. (a) The first row of AB is the first row of A times B.
"6 -2 4"
[ai fi]=[3 -2 7] 0
1 3
[7 7 5
=[18+0+49 -6-2+49 12-6+ 35]
=[67 41 41]
13
3
0
-1
2
SSM: Elementary Linear Algebra
Chapter 1: Systems of Linear Equations and Matrices
(f) The third column of AA is A times the third
(b) The third row of AB is the third row of A
column of A.
times B.
[a,-, B]
6
-2
4
=[0 4 9] 0
I
3
7
[A 33]=
3
-2
7
7
6
5
4
4
0 4 9j[9
21-8+63 ■
7 5_
42+ 20+ 36
=[0+0+63 0+4+63 0+I2+ 45]
0+16+ 81
=[63 67 57]
76
(c) The second column of A5 is A times the
98
second column of B.
97
■3 -2 7] [-2’
[A 63]= 6
5
0
=
4
I
4 9j[ 7
9. (a)
AA =
"-6-2 + 49]
-3
I2
76
48
29
98
24
56
97
-12 + 5 + 28
0 + 4 + 63
41
21
67
-3
3
48
=3 6 +6
5
24
0
4
12
29
(d) The first column of BA is B times the first
-2
3
-2
= -2 6 + 5
56
7
5 +4 4
4
0
9
column of A.
6
[B a|] =
-2
4
3
0
1
3
6
7
7
5
0
76
3
-2
7
98
=7 6 +4
5
+9 4
97
0
4
9
18-12 + 0'
64
14
BB = 21
22
18
77
28
74
0+6+0
(b)
21 + 42 + 0
38
6
6
64
6
63
21
=6 0 +7
3
77
7
5
(e) The third row of AA is the third row of A
fl 4
times A.
22
[83 A]
= [0
4
6
-2
7
9] 6
5
4
0
4
9
= [0 + 24 + 0 0+20 + 36
= [24 56 97]
0+16 + 81]
11. (a)
-2
3
+7
7
5
7
38
6
-2
4
18
=4 0 +3
1
+5 3
74
7
7
5
A=
2
-3
5
•^1
9
-1
1 , x =
jr2 , b = -1
5
4
The equation is
14
4
= -2 0 + 1
28
3
4
7
0
•^3
2-3
5
9
-1
1
^2
5
4
^3
7
0
Section 1.3
SSM: Elementary Linear Algebra
4
0-3
5
(b) A =
1
0
2-5
9
0
1
-1
3
0
2-5
0
3
, x=
7
-1
-1
7
tr(A + B)
•^3
0
^4
2
=(fll 1 + I)+(^22 +^22)
^(^/!« +^nn)
=(<3, 1 +a22+---+ <3„„)+(^|l +/?22
+
•^2
, b=
^1
-8
9
3
= tr(A)+ tr(B)
The equation is
"4 0 -3
1
5
21. The diagonal entries of A + B are a,-,-
^1
3
•^3
0
2
23. (a)
13. (a) The system is
5x| +6x2 -7x3 = 2
-X]-2x2 + ^''‘3 “0
4x2 ~ ^3 = ^
(b) The system is
(b)
Xi + X2 + X3 = 2
=2
2x| + 3x2
5X|-3x2 -6x3 =-9
15. [k
1
1
1
0
1
0
2
0
2
-3
1]
k
0
0
0
0
0
‘^22
0
0
0
0
0
0
«33
0
0
0
0
0
0
0
0
0
0
(344
0
0
«55
0
0
0
0
0
0
«66
a,,
a,2
(3|3
(3|4
^15
«I6
0
(322
^23
<^24
“25
«26
0
0
«33
«34
«35
A36
0
0
0
(344
<^45
(346
0
0
0
0
«55
«56
0
0
0
0
0
«66
ail
0
0
0
0
0
^22
0
0
0
0
0
0
^21
I
k + 2 -1]
^31
«32
«33
0
(341
a42
a43
044
0
0
‘^52
^53
«54
^55
0
.«6I
«62
%3
<^64
%5
%6
an
ai2
0
0
0
0
0
0
(c)
k
=[1'+ 1
a,.
0
1
=[k~+k + k + 2-\]
=[k-+2k + ]]
The only solution of
+ 2A:+1 =0 is ^ =
(d)
a
3
4
d-2c
17.
—1
a+h
d + 2c
-2
gives the
equations
^21
*^22
^23
0
0
«32
^33
«34
0
0
0
0
(343
(344
045
0
0
0
0
<354
<355
«56
0
0
0
0
^65
<366
a =4
d-2c = 3
25. /
d + 2c = -\
1
Xi
1
Xi _\ X\+X2
0 1J[X2
v^2y
^2
a + b = -2
/i \
a = 4=> b = -6
1
(a) /
d-2c = 3,d + 2c = -\ ^2d = 2,d=\
V
d=\=>2c = -2,c = -\
The solution is a = 4, b =-6, c =-I, d = 1.
19. The (y-entry of kA is kajj. If kOjj = 0 for all i,j
then either k = 0 or a^j =0 for all i,j, in which
case A = 0.
15
1y
/
1+n (2
1
y
1
v'y
so
Chapter 1: Systems of Linear Equations and Matrices
(2^
SSM: Elementary Linear Algebra
2+0^ f2^
(b) /
5
(b) The four square roots of
0
V'"/
lo. I 0
are
Vs o] T-Vs o] rVs 0 , and
_ 0 3J’[ 0 3J’[0 -3
1
fi,x)= x
-Vs 0"
2^
1
-3 ■
0
4+3
(c) /V'"/
^
5
0
0 9
True/False 1.3
3 j ^3
(a) True; only square matrices have main diagonals.
y
X
(b) False; an m x n matrix has m row vectors and
n column vectors.
X
+
4
(c) False; matrix multiplication is not commutative.
(2-2'] (0
(d) / f
(d) False; the /th row vector of AB is found by
multiplying the dh row vector of A by B.
-2
I -2
V
(e) True
+
H—V
1
2
1
(f)
-1
-2
Jix)
X
B =
27. LetA = [a::].
v
^ and
False; for example, if A = ^
1
0
0
2
1
, then AB = ^
4
■2
tr(AB) = -1, while tr(A) = 0 and tr(fi) = 3.
fill,
a^2
ai3
X
a,,x + ai2y + ai3Z
021
^22
<^23
3'
a2iX+a22y + a2-iZ
,^31
“32
“33
z
«3l-^ + «32>' + «33^,
1
(g)
x+ y
False; for example, if A = ^
B =
x-y
1
0
0
2
0
A^B^
The matrix equation yields the equations
^ and
-1
, then (ABf = 6* —2^ , while
1
8
3
-2 ■
aj]X + a,2y + ai3Z = a:+y
“21-* + “223' + “23^ = ^“>' •
(h)
“31^ + “32)' + “33^ = 0
True; for a square matrix A the main diagonals
of A and A^ are the same.
For these equations to be true for all values of x,
(i)
y, and z it must be that O] 1 = 1, 012= 1. “21
022 -
one such matrix, A =
29. (a) Both
of
matrix, then A^ is a 4 x 6 matrix and B^ is an
Ujj = 0 for all other ij. There is
1
1
1
1
2
2
2
2 ■
and
'l
1
1
-1
0
0
-1
-1
-1
-1
True; if A is a 6 x 4 matrix and 5 is an m x n
0"
nxm matrix. So B^ A^ is only defined if m = 4,
0 .
and it can only be a 2 x 6 matrix if n = 2.
0
Ci)
True, tr(cA) = cO] 1 + CO22 H
1- co nn
= c(o,i+022 + --- + a„„)
= c tr(A)
are square roots
(k)
True; if A - C = B - C, then o^- -
so it follows that Oy = dy.
16
,
Section 1.4
SSM: Elementary Linear Algebra
I
2
I
(1) False; for example, if A = 3 4 ’
1
0
0
0
, then AC = BC =
2
11. b'' =
3 7 ’
1
and C =
5
B=
4
-3 4 ’
0
1
I
_ I
4
-4
5
8+12 3
2
1
1
20
10
-I
3 0 ■
(m) True; if A is an m x n matrix, then for both AB
and BA to be defined B must be an n x m matrix.
13.
70
45
122
79
ABC =
Then AB will be an m x m matrix and BA will be
an n X n matrix, so m = n in order for the sum to
1
-1
79
-45
(ABC)
be defined.
70-79-122-45L-122
J_r 79 -45]
(n) True; since the yth column vector of AB is
A[/th column vector of B], a column of zeros in
40[-122 70
B will result in a column of zeros in AB.
1
-1
C
1
0
1
(o) False;
’ 0 0
5
1
3 7
70_
5
1
-I
0 0
-1 ^]_ir-i -4'
6(-1)-4(-2)L 2 6j“2L2 6
4
1
3
4
3
B
2.4-(-3)4L^ 2j 2oL-4 2
Section 1.4
1
A-‘
r 2 -1] r 2 -f
3-2-1 -5 -5
3 “ -5
3
Exercise Set 1.4
1
1
1
-I
4
3
-1
40 \L
5. B
2-4-(-3)4L-4 2_
4
/r
1
3
^
2
2
-1
6^2 -5
3
79
-45
40 -122
70
20 -4 2
i
1
5
20
K-i
15. 7A =((7A)-')
i
i
-2
-7
5
10
(-3)(-2)-1-7L-1
-3
3 0
-1
7. D
2-3-0 0 2
-if-"
-1
Ip
0]^ i 0
6 0 2
0 i.
9. Here, a = d =—{e^ +e
and
2
7
1
3
-3
Thus A =
2
1
1
3
7
7
-e ^), so
b =c =
17. /+ 2A =((/+ 2Ar'r'
ad -be =-ie^ +e-^f--(e^ 4
f
5
-2
-1-5-2-4L-4
1 r 5 -2'
13L-4
-1
4
-2j
=-(e^^
+ 2+ e-^^)--(e^^-2
+e
4
4
)
= 1.
2.'
-|-1
Thus,
4(e^+e-^)
17
13
13
A
_L
13
13
4
3
SSM: Elementary Linear Algebra
Chapter 1: Systems of Linear Equations and Matrices
2.
3
19. (a) A’ =
18
13
13
13
1
A
_L
0
13
13
rr3 1
3
1
3
1
2
2
1 J/ 2
1
Thus 2A =
2
13
1
13
and A =
1 1
4
3
1
8
3
2
1
41
15
30
1 1
1
11
Z
6.
13
13
-15
4M 1-15-3oL-30 41
11 -15'
-30
41
(c) A^-2A + 1 =
11
4
6
2
8
3
4
2
1
0
0
1
+
6
2
4
2
(d) p(A)= A-2l =
3
1
2
0
2
1
0
2
2
(e) p{A)= 2A^-A + I
_~22 8]_'3
1
2
1
16 oj
1
0
0
1
+
^'20
7]
14 6
3
(f) piA)= A^-2A + 4I
_'41 15l_r6 2
4
0
0
4
+
30 l lj [4 2
^■39
13]
26 13
3
21. (a)
0
0
3
3
0
0
=
0
0
_0 -3 -
0
3
0
-1
3
0
-3
-1
o]r 0
-8
-6
0 6 -sj
27
0
0
0
26
-18
0
18
26
-1\
3
-3
\l-
90
0
3
0
0
-3
-1
0
0
1
13
13
-1
(b)
9
13
11
-ix
1
VL
0
3
- /
0
18
E
1
01
01
0
0
1
E
01
01
£
1
01
01
1
E
01
01
1-
61
0
0
17
0
0 0
l7
0
+
17
0
I
0
0
0
0
0
1
0
0 0
I
0
0
0
Z- 0
9
9- 0
2-
e+
I
t
0
9
0
0
0
t
9Z
81-
81 0'
9Z
0
0
Zi
PZ-
PZ
Z£
0
31
91-
0
0
17|-
91-
91
l7l-
31-
0
91-
0
0
0
=
5Z_
0
=
LZ_
0
0
=
9I_
0
0
=
81.
I + V- ^VZ = {V)d (3)
0
0
3
0
0
3
0
0
3
9-
I-
0
0
9-
t
9-
t-
0
0
1-
9
9-
3-
0
0
I,
0
= IZ-V =(V}d (p)
9
31
9-
0
0
0
0
0
0
9-
31-
.0
+
3-0
9
0
01
t
01
1
0
01
y snqi • 01
0
lP + VZ-^V = {V)d (J)
I
0 0
17,
9
9-
9
0
0
8-
0
0
930-0
810-0
810-0-
930-0
= I + VZ-^V (3)
6
0
0
LZ
0
01
01
t
0
iz
0
I
oe
Z
0
E
oe
e
0
1
5z
0
z
t
/r
0
01
01
0
0
1
I
6
0
I
0
01
0
E
01
£
0
01
E
0
1
I
-IN
01
E
1
01
E
0
0
0
01
0
1
£
0
0
I- 9 qi
01
9-
I-
0
I-
1 J/
9-
JO 3SJ3AU! aip JBIJ] 3J0iv[ (q)
SI
I
= e(i-'^)=e-'^
£
0
I
VL
01
E
1
0
-
9
I
£
£
1
BjqaSiv JB0un Ajbiu0UJ0|3 :]f^sS
ri uojioas
Chapter 1: Systems of Linear Equations and Matrices
1^22 ■■•rr
27. Since
1
a
nn
0
SSM:Elementary Linear Algebra
^ 0, none of the diagonal entries are zero.
■■ ■
0
II
i
0
Let B =
0
“22
0
0
a,nn
AB = BA = I so A is invertible and B = A *.
29. (a) If A has a row of zeros, then by formula (9) of Section 1.3, AB will have a row of zeros, so A cannot be
invertible.
(b) If B has a column of zeros, then by formula (8) of Section 1.3, AB will have a column of zeros, and B cannot
be invertible.
dB-^A^BAC~^DA~'^B^C~'^=C^
31.
-1
CA"'b"'a”2b(C^) ■dB-'A^BAC~^DA~^B'^C~^ =CA~'b-^A-'^B(C'^)-^c'^
DA~^b'^C~^ = CA~^B~U~^B
DA~^b'^C~^ ■
A^ = CA~'b~'a~^B ■ C^(B^)~' A^
D=
1
I
1
A^
-1
33. (Afir'(AC“')(Z)“'C“'r'D“' =(B“'a"')(AC“‘)(CD)D
1
1
= B“'(A“'A)(C~‘C)(Z)D“')
= B
^11
35. Let X = ^21
^12
^22
^13
^23 •
_^3I
hi
^3.
-1
AX = I
h\+h\
^11 ^21
b2\+h\
hi+hi
^12'*' hi
^22+^32
1
h^+h^
^13"^ ^23 “0
0
hi + ^33
0
0
1
0
0
1
Consider the equations from the first columns of the matrices.
^11 +^3l -1
^11+^21 =0
^21 +^31 =0
1
Thus b^\ =1231 =-^221, so fc,, =-, b2x
1
and b^^ =-.
Working similarly with the other columns of both matrices yields A
20
-1
I
i
_i
2
2
2
1
1
1
2
2
2
i
1
i
2
2
2
Section 1.4
SSM: Elementary Linear Algebra
^1
t>\2 ^13
^22 ^23 •
37. Let X = ^21
.^31 ^32 ^33_
AX=/
^32
^12+^22
^31
b^ ^ +b2\
1
^33
^13+^23
—b^3 +i?23 +^33
.“^11+^21+^! “^12+^22+^32
0
0 0
1
0
0 0
Consider the equations from the first columns of the matrices.
^31=1
b^ ^+b2^ =0
-i'll +^>21 +^31 =0
1
Thus, b2] =-b\\ and -2^7]|+1=0, so b^^ = —, ^21 -“2’
^3'
Working similarly with the other columns of both matrices yields
i 1 _l’
2
2
2
il
A-'
l
2
2
2 •
1
0
0
X,
39. The matrix form of the system is 4^ -2Sj X2
n-1
3
-2
4
5
1
5
3
3
-2
X2
4
5
X,
_ 6
1
X2
4
-3
, so the solution is
-1
3 ■
2
3-5-(-2)4L-4 3
1
5
2
23 -4 2
1
X,
5
23 -4
.■^2
2
-1
3
3
1
J_r-5+6
23
1
4+9
23 13
13
The solution is Xj =—, X2 = —
23
23
41. The matrix form of the system is
6
1
4
-3
-1-1
6
1
^1
4 -3JU2j~
0
-2
-3
6(-3)-1 -4L-4
6
_j_r-3 -f
22
Xi
1
-3
.^2
22
-4
1
6_
-iir o'
6
-2
0+2
22 0-12
1
2
22 -12
1
The solution is x^ = -—, X2
6
11
21
, so the solution is
3-1 r
0
-2 ■
Chapter 1: Systems of Linear Equations and Matrices
51. (a) If A is invertible, then A
-I
SSM: Elementary Linear Algebra
exists.
AB = AC
A“'A5 = A“'aC
IB = IC
B=C
(b) The matrix A in Example 3 is not invertible.
I
-I
I
-I
-I
-I
53. (a) A(A“' + B~')B(A + B)~' =(AA"‘ + AB~‘)B{A + B)
-I
=(/ + Afi“')fi(A + fi)
-1
=(IB + AB~^B){A + B)
-I
={B + A)(A + B)
-I
={A + B){A + B)
=/
I
1
-1
(b) A“'+5“'^(A + S)
55. (/-A)(/ + A + A^+---+ A
yl-l
)=(/-A)/+(/-A)A +(/-A)a2+...+(/-A)A
k-\
= /- A + A - A^ + A^ - A-^ + ■ ■ ■ + A*"'- A^
= /-A
k
= 1-0
=/
True/False 1.4
(a) False; A and B are inverses if and only if AB = BA = I.
(b) False; (A + Bf ={A + B)(A + B)
= A^ + AB + BA + B^
Since AB ^ BA the two terms cannot be combined.
(c) False; (A - B){A + B)= A~ + AB - BA- B-.
Since AB ^ BA, the two terms cannot be combined.
(cl) False; Afi is invertible, but (AS)"' = B~'A~' ^ A-'B-'.
T
7’
(e) False; if A is an m x n matrix and B is and n x p matrix, with m ^ p, then AB and (AB) are defined but A B
not defined.
(f) True; by Theorem 1.4.5.
(g) True;(M + b/ =(kA)'' + B^ = kA'' + B^.
(b) True; by Theorem 1.4.9.
(i) False; p{[)=(aQ+a^+a2+—
which is a matrix, not a scalar.
22
IS
Section 1.5
SSM: Elementary Linear Algebra
(j) True, if the square matrix A has a row of zeros,
the product of A and any matrix B will also have
5. (a) E interchanges the first and second rows of
A.
a row or column of zeros (depending on whether
the product is AB or BA), so AB = I is
impossible. A similar statement can be made if A
0
EA =
1
0
1
-1
-2
5
3
-6
-6
-6
-1
-2
5
-1
has a column of zeros.
(k) False; for example’ 0
1
-1
1 oJ[ 3 -6 -6 -6
0
and
are
(b) E adds -3 times the second row to the third
0
row.
0 0
both invertible, but their sum
.
0
0
2
-1
EA = 0
1
0
1
-3
0
-3
1
2
0
is not
0 0
invertible.
Section 1.5
2
-1
1
-3
0
9
Exercise Set 1.5
1. (a) The matrix results from adding -5 times the
4
0-4-4
5
1
3
3
-4^
5
3
-12
-10
(c) E adds 4 times the third row to the first row.
first row of I2 to the second row; it is an
"1 0 4]ri 4
elementary matrix.
£A= 0
1
0
0 0
13
28
2
5
2
5
3
6
3
6
(b) The matrix is not elementary; more than one
elementary row operation on l2 is required.
7. (a) To obtain B from A, the first and third rows
must be interchanged.
(c) The matrix is not elementary; more than one
elementary row operation on
is required.
0 0 1]
E= 0
1
1
0
0 0
(d) The matrix is not elementary; more than one
elementary row operation on
is required.
(b) To obtain A from B, the first and third rows
must be interchanged.
3. (a) The operation is to add 3 times the second
1
3'
0
1 ■
row to the first row. The matrix is
0 01 01]
£= 0
1
0 0
(b) The operation is to multiply the first row by
(c) To obtain C from A, add -2 times the first
-i 0 o'
1
—. The matrix is
7
row to the third row.
0
1
0 .
0 0
1 0 o'
1
0
1
0
-2
0
1
E=
(c) The operation is to add 5 times the first row
'1 0 o'
to the third row. The matrix is
0
1
5
0
(d) To obtain A from C, add 2 times the first
row to the third row.
0 .
1 0 o'
1
E= 0
1
0
2
0
1
(d) The operation is to interchange the first and
0 0
0
1
1
O'
0 0
1
4
1
0
2
7
0
1
9.
third rows. The matrix is
1
0 0 0'
0 0 0
1
Add -2 times the first row to the second.
23
Chapter 1: Systems of Linear Equations and Matrices
1
4
1
SSM: Elementary Linear Algebra
0
1
_0 -1 -2 1
0
0
Multiply the second row by -1.
'1 4 1
0'
1
1
0
5
-10 0 -2
1
4
-10
1
-3 0
1
5
Add -4 times the second row to the first.
0
3 0
Multiply the second row by —.
0 1 2 -1_
'1 0-7
0
4
1
0
0
1
3 0
1
-2
0
1 1
_0 4 -10
1
-3 0_
5
2-1
-1-1
1
4
-7
4
2
7
2
-1
3
1
5
Add -4 times the second row to the third.
1
-1
0
0
0
0
3
1
0
0
-2 0
1
5
5
1
_7
_4
11.
3 -2 0 1_
0 0-2
Multiply the first row by -1.
'1 -3 -1 O'
1
5
5j
1
Multiply the third row by —.
2
3
-2
0
1
1
Add -3 times the first row to the second.
7
3
3
0
1
0
1
-2
0
1 1
0 0
1
1
JL
2
2
10
5.
0
1 -3 -1 O'
0
0
5
1
5
1
Add -3 times the third row to the first and 2
Multiply the second row by —.
7
1
-3
0
1
-1
times the third row to the second.
0
1
7
0
7.
Add 3 times the second row to the first.
6
1
0 0
1 1
2
10
0
-1
1
1
0 0
1
_1
J.
2
_
2
10
5
1
_ii
-6
2
10
5
3
-1
1
1
2 5^
_1
JZ.
2
2
10
5
'' Off
.0 ' f f
-I-1
3
3
-2
2
1
7
7
13.
1
0
2
5
-1
1
15.
0 0
3 0
1
0
-4 0 0
1
3 4
-1
2
-4 0 0
1
0
1
1
3 0
-10
1
0
5
-10 0 -2
-4
1
4
1
0
2 -9
0 0
1
0
0 0
1
2
4
1
-4
2
-9
0
1
0
0 0
1
Add -2 times the first row to the second and 4
times the first row to the third.
'1
0
1 0'
4
3
2
Multiply the first row by -1.
'1 -3 4 -1 0 0'
0 0
0
-1
-4
Add -3 times the first row to the second and -2
times the first row to the third.
1 0
4
7
Interchange the first and second rows.
'1 0 3 0 1 0'
5
3
1 1
7
3 4
5
1
-3
10-7
0 -10
-3 0
1
4 -1 0 0'
7
1
0
-4 0
2
1
1
Multiply the second row by —.
Interchange the second and third rows.
10
24
SSM: Elementary Linear Algebra
1
-3
0
0
4
0 0
i
_L
10
5
10
7
-4
0
1
-10
Section 1.5
1
3
3
^0 0
0
2
7
6
0
1
0
1
2
7
7
0 0
1
Add 10 times the second row to the third.
1
-3
0
1
0
4 -1
0
Add -2 times the first row to both the second
0 O'
JZ.
1
_L
10
5
10
0 -2
1
and third rows.
0
1
3
3
1
0
1
0
0
1
Since there is a row of zeros on the left side,
'-1 3 -4'
17.
2
4
1
-4
2
-9
1
0
0
1
1
1
i 0 0
-1
1
0
-1
0
1
Add -1 times the second row to the third.
is not invertible.
‘1 3 3 i^
1
1
1
0
0 0
1
0
0 0 0
1
0
1
0
0
0-1
1
1
0
1
1 0 O'
0
1
1
0
1
0
0
1
-1
-1
0
1
0
1
1
0
0
1
0
0 0
0 0-2 -1
-1
1
1
2
1
0
1
1
0
0
1
1
0
1
0
.0 0 M i
i
2J
21.
second rows.
0
0 0
1
0
0 0
1
i
1
1
2
2
2
I
1
1
2
1
0
1
0
1
1
1
1
1
0
0 -1
1
1
0-1
1
1
0
0-1
1
2 6
6
f 0 -3
2
7
6
-1
1
0
2
7
7
0
-1
1
2
-4
0
1
2
12
0
0
2
0
-1
-4
0
Add -1 times the third row to both the first and
1
-1
1 0 0 I 0-3
0
Multiply the third row by —.
19.
1
0 0
0
1
1
2 3 -3'
3 0
1
0
0
Add -3 times the second row to the first.
Add -1 times the second row to the third.
'l 0
1
0-1
Add -3 times the third row to the first.
Add -1 times the first row to the third.
'1 0
0 0
0
1
0 0 0
0 0
0
1
1
0
-5 0 0 0
1
Interchange the first and second rows.
'1
2 12 0 0 1 0 o'
2
1
1
1
2
-4
0
0
")
•>
2
0
0
2
0 0 0
0
-1
-1
-i-l
0
1
1
1
2
2
2
1
1
I
2
2
2
1
1
_1
2
2
2
2
6
6
1
0
0
2
7
6 0
1
0
2
7
7 0 0
1
0 0
0 0
-5
1
0
0 0
1
0
0 0 0
1
Add -2 times the first row to the second.
1
2
12
0
-8
-24
0
0
2
0
-1
0 0
1
0 0
1
-2
0 0
0 0
0
0
^-5 0
1
0
0 0
1
Interchange the second and fourth rows.
'1
2
12 0 0
1 0 O'
1
0-1
Multiply the first row by —.
2
25
^-5 0
0
0
2
0
-8
-24
0 0
0
1
0 0
0
-2
1
1
0
0 0
Chapter 1: Systems of Linear Equations and Matrices
SSM:Elementary Linear Algebra
Multiply the second row by -I.
"l
2
12 0 0
1 0 o'
1
0
0
0
0
4
5
0
2
0 0
0 0
0
0
1
I
I
4
2
0
1
1
4
0
0
1
i
i
1
_1
40
20
10
5
-1
0
0 0
0
_0 -8 -24 0 1 -2 0
0
1
0
-1
9
0 0
0
0 0 0
0
0
9
Add 8 times the second row to the fourth.
"1
2 12
0 0
0
1
4
5
0
0
2
0 0
0
0
8
40
0
1
1 0
o'
0 0
2
1
0
-2
0
-8
0
1
9
-3
0
I
3
0
4
2
-4
0
0
4
I
2
12
0
8
0
2
0
0
0
-1
0
9-1
1
1
0
-1
-4
-5
i
1
I
1
40
20
10
5
Multiply the third row by —.
2
1
0
0
2
1
12
0 0
4
5
0
1
0
0
0
0
-1
-1
0
1
0
1
2
3
-2
6
0
1
0
-1
2 0 0
0
0
0
0
1
0
0
0
0
0 0
4
0
5
0
0
0
0
0
-1
0 0
0
i
0
-2
-4
40
1
9
2
1
2
12
0
0
0
1
4
5
0
0
1
0
0
0
-8
40
0
0
0
i
0
1
i
40
1
20
i
10
0
1
12
4
0
0
1
0
0
0
0
1
0
0
0
0
0
1
I
4
2
0
0
i
_L
__L
__L
40
20
10
1
0-1
3
0
0
1
0
0 0 0
1
6
5
0
1
0 0
0 0
1
0
0 0 0
1
0-1
0 0 0
0 6
2
1
2
0
0
0
0 0
1
0
0 0
I 5 0 0 0 1_
Interchange the second and third rows.
0
'1
i
5_
0 -1 0 -1 0 0 0'
2
0
Add -5 times the fourth row to the second.
2
-2
2 0
0
0
5
Add -2 times the first row to the second.
2
0 0
3
0
1
Multiply the fourth row by
1
Multiply the first row by -1.
'1 0 -1 0 -1 0 0 0'
Add -8 times the third row to the fourth.
1
0 0
23.
0 0 0 ^ 0
0 0 8 40 1 -2 6 -8_
0
0 0 0
0
'1 2 12
0
0
0
3
0
0
0
0
0
1
0
0 6
2
1
0
0
0 0 0
1
1
5
Multiply the second row by -1.
'1 0 -1 0-1 0 0 O'
0
2
_i
5.
Add -12 times the third row to the first and -4
times the third row to the second.
1
0
2
1
0 0
0
0 0
i
-6
1
4
0
0
1
0
0
0
2
9
i
0
1
0
3
-2
0 6
0
0 0
1 5
0 0
2
1
0 0
-1
0
0 0
0
1_
Add -3 times the second row to the third.
O'
'1 0 -1 0-1 0
0
0
0
0 0
6
6
2
1
0 0
I
5
0 0
1
-2 0
0 0
1
0 O'
-1
0
3
0
2
0
0 0
1
40
1
I
20
10
0
2
5_
0 1_
Interchange the third and fourth rows.
'1 0 -1 0-1 0 0 O'
Add -2 times the second row to the first.
0
1
-2 0
0 0
0 0
1
5
0
0 0
6
6
2
0
-1
0
0
1
3 0
Add -6 times the third row to the fourth.
26
Section 1.5
SSM: Elementary Linear Algebra
I
0
-I
0
0
0
0
I
-2
0
0 0-1
0
0
0
5
0 0
0
1
2
3
-6
0 0
0
-I
-24
0
1
i
0 0 0
0
0
Multiply the fourth row by
0
0
-1
0
-1
0
0
0
0
0
-1
0
0 0
1
5
0
0
0
1
0 0
0
1
I
i
24
0
1
-2
0 0
0
1
0 0
-1
1
0
0
0 f
0 0
0
1
0
0
0
0
0 T
1
1
-1
0
0
i
k.
0
0
0
0 f
0
0
0
0
0"
0
0
5
5
5
1
12
24
8
4
i
L
_1
1
12
24
0
0
0
0
n-l
^1
0 ytj 0 0
0 0 /t3 0
0 0 0 /t4
4
0
0
0 0
0
Add -5 times the fourth row to the third.
"1 0 -1 0
0
0
i
-2
i
0
0
0
1
12
0
1
24
1
0
^1
0
0
(b)
4
Add the third row to the first and 2 times the
1
0 0
1
0
0 0
1
0 0 0
1
0 0
0 0
It
1
0 0
1
0
0 0
0
1
0 0 0
1
third row to the second.
1
0 0 0
0
0
1
0 0
0
1
0
0 0 0
■1
2
3
1
24
5
.s
4
2
_5
1
_1
12
24
i
24
n-1
4
.S
12
24
5
5
I
12
4
2
5
i
6
0
0
0 A'2
0
0
0 A'3
0
0
12
0
0
1
0
7
1
5
5
0
0
1
0 0
i
i
12
24
0
0
0 0
0
1
1
and — times the second row to the first.
4
1
1
Add — times the fourth row to the third
k
I
24
0 0
1
4
k
1
4
1
0
i
0
0
0
I -i 0
0
0
0
1
0
0
0
0
1
0
0
1
0
0 i -i
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
1
4
0
0
0
1
it
1
0
0
0
1
0
0
0
1
0
0
k
I
0
0
0
0
1
0 i -k
0
0
Multiply the first row by —, the second
^'l
^'2
0
i
1
row by
1
0 0 1 f
0 0 I
0
k
k
i
i
0
1
1 I
0 0 I
0 0 0
k
k
0
4
6
0
1
12
i
k
4
6
12
Multiply the first and third rows by —.
_i
1
2
k1
25. (a)
12
5
-2
0
0
0
1
I
1
0
1
7
the third row by
k^
-1-1
and the
fourth row by -!-.
27.
k4
c
c
c
1
0
0
1
c
c
1
c
0
1
0
0
0
1
I
I
0
1
I 0
requirement).
27
0
0
1
Multiply the first row by — (so c
c
0
0
1
0 is a
SSM:Elementary Linear Algebra
Chapter 1: Systems of Linear Equations and Matrices
1
1
1
0 0
-3
1
2
2
c
1
c
c
0
1
0
1
1
c
0 0 0
-1-1 ^
1
Add -1 times and the first row to the second and
third rows.
-1-1
-1-1 r
0
1
-1
Vo i
0
1
0
1
-1-1
1
0
-1
1
-i 0
^'1 0 1 1 -4 0 1 0
1
1
1
0
c-1
c-1
I
“ _0 2j[o lj[ 0 lj[l 1_
0 0
c
1
1
0
0
1
Note that other answers are possible.
f
0
0
1
c-1
1
0
-2
31. Use elementary matrices to reduce 0 4
3
0 0
1
c
1
(so
Multiply the second and third rows by
c-1
to 73.
c^\ is a requirement).
1
1
'1 0 2'
i
1
0
0
1
-2
1
0
0 4
3
0
4
3
0 0
1
0 0
1
0
0
1
0 0
1
0
I
1
c(c-l)
0 0
0
1
0
c-1
0
1
1
0 0
0
c
c-1
c(c-l)
1
0
0
0
1
-3
0 0
1
1
0 0
4
3
0 4 0
0 0
1
0
0
Add -1 times tbe third row to the first and
second rows.
1
1
1
0
1
1
0
c-1
0 0
1
0
i «
0
c-1
0 0
i
c(c-l)
1
c-1
0 0
c-1
1
0, 1.
-3
1
-3
1
1
1
2 2_
1 -lir-3 1
0
1
1
0 0
1
1
0
0 0
1
2
2
1
0
0
1
0
2
1
0
1
-3
0
1
0
0
4
3
0 0
1
0
0
0
to
1
0
1
0
0 0
0
-4
1
1
1
0
0 0
1
so
0
1
0 0
0 -2
0 0
0
1
1
0
0 0
» i
h-3
0
Thus
29. Use elementary matrices to reduce
0
1
0 4
0
The matrix is invertible for c
1
0 0
1
c-1
0
0
1
0
1
1
0
-2
0
4
3
0 0
I
0
1
0
0 ijL 1 1
‘ 1 oin o' _
1
I
4
0
-4
n-1 r 1
r
-1 lj[l 1_"
1
0
0
1
1
0
2
1
0
0
= 0
1
0
0
1
-3
0 0
1
0 0
1
'1 0-2
Thus,
= 0
1
0
-1
1
1
0
0
1
-1 0
1
0 1 [0
-1
1
0
-3
1
Vo VL 2
2
1
0
1 0 0
0
1
3
-1-1
0 0
0 i
0
0
1 0 0
0
4 0
0 0 lJ[o 0 lJl_0 0 1_
Note that other answers are possible.
so
28
0
1
SSM: Elementary Linear Algebra
Section 1.6
33. From Exercise 29,
-3
1
2
(c) True; since A and B are row equivalent, there
0
2
-i 0
0
1
I
1
-1
1
1 [0
0
exist elementary matrices £), E2,...,
iJ
that B = Ej^ ---E2E^A. Similarly there exist
elementary matrices E{, E2, ■■■, E'l such that
C = E'r- E'2E[B = Ei-E'2E{E, ■■■E2E,A so a
I
4
i
3
4
and C are row equivalent.
35. From Exercise 31,
1
0
-2
0
4
3
0
0
1
(d)
1
0
1
0
0
0 i
0
0
1
-3
iJL 0
0
1
1
0
0
2
0
1
0
0
0
1
(e)
matrix is invertible.
(f)
2
37.
0
True; interchanging two rows is an elementary
row operation, hence it does not affect whether a
0 74 -74
0
True; a homogeneous system has either exactly
one solution or infinitely many solutions. Since
A is not invertible. Ax = 0 cannot have exactly
one solution.
0
0 0
such
True; adding a multiple of the first row to the
second row is an elementary row operation,
hence it does not affect whether a matrix is
invertible.
1
(g)
Ealse; since the sequence of row operations that
1
2
3
1
4
1
convert an invertible matrix A into /„ is not
2
1
9
unique, the expression of A as a product of
elementary matrices is not unique. For instance,
■-3 1
1 0
1
1
-4 0
1 0
A =
Add -1 times the first row to the second row.
'1
2
0
2
-2
3'
2
1
9
2
1 -2]ri Olfl
0
Add -1 times the first row to the third row.
'1
2
0 2j[o lj[ 0 lj[l 1 _
2
1
2
1
Olfl 5'
0-8
0
1
(This is the matrix from Exercise 29.)
3'
0
2
-2
1
-1
6
Section 1.6
Exercise Set 1.6
Add -1 times the second row to the first row.
'1
0
5'
0
2-2
1
6
1. The matrix form of the system is
'1 iifx, i^r2’
5 6 X2 ~ 9 '
Add the second row to the third row.
1
0
5
0
2
-2
1
1
4
1
1
5
6
= B
6-1
-5
(b)
1
6
6
-1
6-5 -5
-5
1
2
12-9
9
-10 + 9
3
-1
The solution is X\ =3, X2 = -1.
True/False 1.5
(a)
1
False; the product of two elementary matrices is
not necessarily elementary.
3. The matrix form of the system is
True; by Theorem 1.5.2.
“1
3
1
X,
4
2
2
1
x.
-1
2
3
1
3
Find the inverse of the coefficient matrix.
29
SSM: Elementary Linear Algebra
Chapter 1: Systems of Linear Equations and Matrices
3
I
I
0
0
2
2
I
0
I
0
2
3
I
0 0
I
I
3
I
-5
-I
I
0
I
0
0
I
_| -2
-I
-2
1
3
1
0 0
i
i
4
9
i 0
1
0
f 0 i
0
1
0
0
0
0 1
5
1
I
1
5 4 0
0
1
0
1
i 1
5
5
0 0
3
I
1
2 -i 0
2 i
I
0 0
1
I
4
1 0
5
0
I
4
0
1
0
1
1
0
3
0
I
I
1
4
2
4
2
3 -4
0
0 0
0
I
0
i
0 1
1
1
0 0
1
-3
4
I
0
0
-1
1
0 0
1
2
3
-4
0
1
I
1
0 0
0
0
0
0
0 0
2
3
0
-4
4
0
-1
1
2
3
-4
1 +3
8-3-12
The solution is X| = -1, X2 =4,
-1
1
1
X
5
1
1
-4
1
10
-4
1
1
z
0
I
5
5
5
0
5
5
10
3+2
5
0
1-2
-1
3
1
1
5
5
5
I
I
5
5
0
7. The matrix form of the system is
4
3 5
-7
]_ I"i>|]
1 2 x, ^ h.'
= -7.
-1
5. The matrix form of the equation is
1
5
The solution is4: = 1,}'= 5,z = -l.
-4+ 3
3
5
1
1
0
5
1
5
5
0 0
3 0
5
-i 0.
5
0
1
1
1
0 0
0
0 0
-1
0 0-5
1
-2
0 0
I
0
_4
-1
0
0 0
I
-3
-3
I
4
0
0
0
I
5
0
0
0
I
5
0
3
5
1
2
1
-5
6-5 L-1
3
2
-5
-1
3
2b^-5b2
-b\ + 3b2
2 -5] b^
-1
2
b-,
The solution is X| = 2b^ -5b2, X2 = -b^ +3i’2Find the inverse of the coefficient matrix.
1
1
1
1
1
1
-4
0
1
0
systems.
-4
1
1
0
0
1
'1 -5
1
0 0-5
0
5
5
4
0 0
9. Write and reduce the augmented matrix for both
3
2
1
-2'
4
5
0 0
1
0
1
-5
1
-2
0
1
0
17
1
11
30
Section 1.6
SSM: Elementary Linear Algebra
I
-5
I
-2
0
17
1
0
1
0
I
3
b1
13.
1
-2 1 b2
17
22
21
17
17
1
IL
17
17
1
3
1
3
^1
0 7 26,+/72
22
(i) The solution is jc, = —, X2 = —
17
21
(ii) The solution is x,
0 I f6,+l62
17
II
1 0 |6,-|62
-'^2 “77
17
0
11. Write and reduce the augmented matrix for the
four systems.
■4
-7
1
2
4
-7
0
-4
-1
6
3
0
There are no restrictions on b^ and 62.
-5‘
1
1
15.
6
3
1
-4
-1
-5
2
1
0
-15
-4
6
3
1
-28
-13
-9
-2
5
4-5
-3
3
0
2
1
0
8 62
-3 63
-2
5
3 -12
-3
12
-2
1
2
0
1
0
0
I
6
3
4
28
13
3
15
15
15
5.
34
19
1
15
15
5
15
A.
28
13
3
15
15
15
5
7
(i)
5
0
3 -12
0
0
0
1
-2
5
0
1
-4
0
0
0
-46, +62
—
^1
—6, +62 +63
4
The solution is x, =—, Xt = —
'
^1
-46,+62
36,763
15
^
34
15
28
(ii) The solution is x, = —, X2 =■—.
15
15
19
13
1
0
-3
0
1
-4
0
0
0
-f 6, +§62
-|6,+^62
—6, +62 +63
The only restriction is from the third row:
(iii) The solution is x, = —, X2 =
15
-6, +62 +63 =0 or 63 = 6, -63.
15
3
(iv) The solution is x, = —, X2 = —.
5
5
31
SSM: Elementary Linear Algebra
Chapter 1: Systems of Linear Equations and Matrices
1
-1
-2
3
2
5
1
1
bl
17.
-3
2
2
-1
4-3
1
3
1
-1
3
0
b.
-1
11
5
0
-1
11
5
0
1
-11
-5
1
-1
3
2
11
5
0
-1
0
0
0 0
0
0
0 0
1
bi
2b^
3bi+b^
-4fc| +^4
bi
2b\ +b2
bi
—2b^ +i?2 +^4
1
-1
3
2
1
-11
-5
0
0
0
0
0
0
0
0
1
0
0
-8-3
1
-11
-5
0 0
0
0
0 0
0
0
1
0
0 0
i -f '
1
1
0
0
2
_2
1
0
5
5
‘
5
0
0
1
-4
2-5
1
-1
0
5
-2
5
0
1
0
-2
1
-2
0
0
1
-4
1
^1
-2b^ -b2
b\~b2+b^
—2fc| + Z?2 +^4
-1
0
0 0
2-5
3
-1
3
1
0
-2
1
-2
0 0
1
-4
0
0
1
1
2
5
5
2
0
-1
0
2-5
3-1
A" = -2
1
3
2
-1
-2
4
0
-4 2 -5j[3 5
-b,-b2
-2b\ -b2
b^—b2+b'^
—2by + b2 +^>4
7
8
-3 0
5
1
-7
1
11
12
-3
27
26
-6
-8
1
-18
-17
-15
-21
9
-38
-35
2
21. Since Ax = 0 has only the trivial solution, A is
From the bottom two rows, ^1-^2+^ ~ 0
invertible. Thus,
—2/?| +i>2 ^4 — 0.
A^x =0 has only the trivial solution.
Note that (A'^T* =(A"‘)^
Thus bj, = -fc| +^2
^4 2^1 -i»2Expressing b^ and ^>2 terms of b^ and b^
gives b^ =b^+ b^ and b2 = 2Z?3 + b^.
23. Let
is also invertible and
be a fixed solution of Ax = b, and let x be
any other solution. Then
1
1-1 r
-1
1
2
-1
5
7
19. X = 2
3
0
4
0
-3
0
1
0
2
-1
3
5
-7
2
1
1
1
0
0
0 0
1
0
0 0
1
1
-1
2
3
0
2
-1
1
-1
1
1
0
5
-2
-2
0
2
-1
A(x -X])= Ax - Ax, = b-b = 0. Thus
xq = X -X] is a solution to Ax = 0, so x can be
expressed as x = x, +Xo.
Also A(x, +xo)= Ax, +Axo =b +0= b, so
every matrix of the form x,+ Xq is a solution of
8
Ax = b.
True/False 1.6
0 0
1
0
0 0
1
(a) True; if a system of linear equations has more
than one solution, it has infinitely many
solutions.
1
-1
0
1
0
2
1
2
5
-1
0 0
(b) True; if Ax = b has a unique solution, then A is
invertible and Ax = c has the unique solution
f 1 »
0
0
A“'c.
1
32
Section 1.7
SSM: Elementary Linear Algebra
-1
(c) True; if AB = I^,, then B = A
and BA = In
9.
=
1^
0
_
0 (-2)2“
also.
(d) True; elementary row operations do not change
the solution set of a linear system, and row
equivalent matrices can be obtained from one
another by elementary row operations.
(e) True; since (5
1
A-2 =
r2
1
0
0
4
1
0
0 i
0 (-2)-2
-k
1
0
0
1
0
0 (-2)"^
= b, then
0
k
(-2)
= 5'b or A(Sx)= 5b, S\ is a solution
0
of Ay = 5b.
11. ^2 =
(f) True; the system Ax = 4x is equivalent to the
system (A - 4/)x = 0. If the system Ax = 4x has a
unique solution, then so will (A - 4/)x = 0, hence
0
i 0
0 i
0
(i)'
0
0
0
if
0
0
0
0
1
16
A - 4/ is invertible. If A - 4/ is invertible, then
the system (A - 4/)x = 0 has a unique solution
and so does the equivalent system Ax = 4x.
(I)"
0
0
0
(1)"
0
0
0
(I)"
or
0
0
0
or
0
0
0
or
A“2 =
(g) True; if AB were invertible, then both A and B
would have to be invertible.
Section 1.7
4
0
0
9
0
0
0 0
16
Exercise Set 1.7
1. The matrix is a diagonal matrix with nonzero
entries on the diagonal, so it is invertible.
i
0
The inverse is ^
0
5
2^
0
0
0
3*^
0
0
0
4*
3. The matrix is a diagonal matrix with nonzero
entries on the diagonal, so it is invertible.
■-1
0
0
5.
0
0
2
0-1
0
0
2
0
0)2 =-8^0 = 021-
0^0 .
The inverse is
3
13. The matrix is not symmetric since
o'
0
3
15. The matrix is symmetric.
1
6
3
-4
1
4
-1
2
5
4
10
17. The matrix is not symmetric, since
6123 = -6
6 = 032.
19. The matrix is not symmetric, since
7.
5
0
0
-3
2
0
4
-4
0
2
0
1
-5
3
0
3
«I3 =1^^ = «3121. The matrix is not invertible because it is upper
triangular and has a zero on the main diagonal.
0 0 -3j[-6 2 2 2 2
=
■-15
10
0
20
2
-10
6
0
-20'
6
18
-6
-6
-6
-6
23. For A to be symmetric, a^2-‘^2\ or -3 = a + 5,
so o = -8
25. For A to be invertible, the entries on the main
diagonal must be nonzero. Thus
33
I, -2, 4.
Chapter 1: Systems of Linear Equations and Matrices
I
0
0
0
-I
SSM: Elementary Linear Algebra
0
27. By inspection, A = 0
0
T
4=-\{A-AT)
-A)
r
33. If A' A = A, then
=(A'^Af =A‘\A^f =A^A = A, so a is
T
I
Thus
T
symmetric. Since A = A , then A" = A A = A.
is skew-symmetric.
39. From A^ =-A, the entries on the main diagonal
35. (a) A is symmetric since a^j = ciji for all ij.
must be zero and the reflections of entries across
the main diagonal must be opposites.
(b) 4-7“ = -i^ => 24 = 27^ or
/“ = 7^ => i = j, since iJ > 0
0 0 -8]
A= 0
Thus, A is not symmetric unless « = I.
(c) A is symmetric since
0-4
8
= aji for all i,j.
4
0
41. No; if A and B are commuting skew-symmetric
matrices, then
(d) Consider 0^2 and aji-
(ABf =(BA4 =A^b'^= i-A){-B)= AB so
a^2 =2(1)“4-2(24 =18 while
<321
the product of commuting skew-symmetric
matrices is symmetric rather than skewsymmetric.
= 2(24-1-2(14 =10.
A is not symmetric unless n = 1.
43. Let A =
37. (a) If A is invertible and skew-symmetric then
-I
(A-'4 =(A^r' =(-A)
= -A
-I
so A
-I
X
y
0
z
3
a3 = -r
is also skew-symmetric.
. Then
/ 3
3,
yix~+xz+ z-)
, so
=1 and
z'
0
(b) Let A and B be skew-symmetric matrices.
z^ = -8 or X = \,z = -2. Then 3>’ = 30 and
(a^4=(-a4=-a^
(A + b/ = A^ +B'’' =-A~B = -(A + B)
(A-Bf =A^-B'^ =-A + B = -{A-B)
(kAf = L'(A^)= <t(-A)= -kA
v= 10.
t
I
10
-2
True/False 1.7
(c) From the hint, it’s sufficient to prove that
I
1
0
A=
(a) True; since a diagonal matrix must be square and
t
have zeros off the main diagonal, its transpose is
also diagonal.
-(A-I-A ) is symmetric and -(A-A ) is
skew-symmetric.
(b) False; the transpose of an upper triangular matrix
-{A + A^f =-iA'^ +iA^f)=-{A + A'')
2
2
is lower triangular.
2
Thus -^(A-rA^) is symmetric.
1
2
’ 0
1
(c) False; for example
3
0
2
1
+
4
2
2 2 ■
(d) True; the entries above the main diagonal
determine the entries below the main diagonal in
a symmetric matrix.
(e) True; in an upper triangular matrix, the entries
below the main diagonal are all zero.
34
Section 1.8
SSM: Elementary Linear Algebra
(f) False; the inverse of an invertible lower
= 50
X\+X2
triangular matrix is lower triangular.
+ Xj = 30
X3 =-l0
•^l
(g) False; the diagonal entries may be negative, as
long as they are nonzero.
= 10
X2
By inspection, the solution is Xj = 40, X2 =10,
(h) True; adding a diagonal matrix to a lower
triangular matrix will not create nonzero entries
above the main diagonal.
X3 = -10 and the completed network is shown.
50
(i) True; since the entries below the main diagonal
must be zero, so also must be the entries above
the main diagonal.
30’
1
2
(j) False; for example’ 0
1
0
6
2
2 6
5
2
7
60
+
which is symmetric.
40
1
(k) False; for example’ 5
2
1
3
2
5
-5
1
0
2
3. (a) Consider the nodes from left to right on the
top, then the bottom.
+
1
which is upper triangular.
300+ X2 =400+ X3
750+ X3 = 250+ X4
100+X, =400+ X2
200+ x^ = 300+ X|
-,9
1
(I) False; for example
0
1
0
1
0
1 ■
The system is
(in) True; if (M)^ = kA, then
= 100
•^2-^3
^3 -3:4 = -500
0 =(kAf -kA = kA‘'-kA = k{A^ -A) since
k ^ 0, then A = A^.
= 300
X] -X2
+ X4 = 100
-^1
Section 1.8
(b) Reduce the augmented matrix.
“0
1 -1 0
100“
Exercise Set 1.8
0
0
1
-1
-500
1
-1
0
0
300
0
0
1. Label the network with nodes, flow rates, and
flow directions as shown.
50
0
0
60
30
40
Node A: X| +X2 = 50
Node B: x, +X3 = 30
NodeC: X3+50 = 40
Node/): X2+50 = 60
0
0
1
-1
-1
0
0
35
0
300
-1
-500
0
100
100
1
-1
0
0
300
0
0
1
-1
-500
0
-1
0
100
0
0
1
400
300
0
0
0
1
-1
0
100
0
0
1
-1
-500
0-1
0
1
The linear system is
100
400
Chapter 1: Systems of Linear Equations and Matrices
SSM:Elementary Linear Algebra
1
-1
0
0
300
1
1
-1
0
0
1
-1
0
100
0
1
2
4
0
0
1
-1
-500
1
3
0
0 -1
1
500
0-2
1
1
-1
1
-1
0
0
300
0
1
2
4
0
1
-1
0
100
0 0
5
11
0
0
1
-1
-500
0
0
0
0
0
0
From the last row we conclude that
11
/3 = — = 2.2 A, and from back substitution it
1
0 0
1
-100
1
0
-1
-400
0 0
1
-1
-500
0 0 0
0
0
0
Since
then follows that
/ =4-2/3 =4-—=----0.4 A, and
^
-^
5
5
I, = h-I', = — + —=— = 2.6 A. Since /-i is
is a free variable, let X4 = t. The
'
solution to the system is X| = -100+ r,
^
^
5
5
5
^
negative, its direction is opposite to that
indicated in the figure.
X2 = —400+f, x^ = —500+ r, x^ =t.
1. From application of the current law at each of
the four nodes (clockwise from upper left) we
have /| = /2 + /4, 1^=1^ +
I(i — ^3
(c) The flow from A to /? is X4. To keep the
traffic flowing on all roads, each flow rate
must be nonnegative. Thus, from
and /] = I2+1
X3 = -500+ r, the minimum flow from A to
These four equations can be
reduced to the following:
B is 500 vehicles per hour. Thus x^ = 400,
h-h + U’
X2 = 100, x^ =0, X4 = 500.
^4 = ^6
From application of the voltage law to the three
inner loops (left to right) we have
5. From Kirchoffs current law, applied at either of
10= 20/|+ 20/2, 2012 = 20/3, and
the nodes, we have I\+l2 = ^3 ■ From
10+20/3 = 20/5. These equations can be
Kirchoffs voltage law, applied to the left and
simplified to:
right loops, we have 2/( = 2/3 +6 and
2/,+2/2=1, /3=/2, 2/5-2/3=1
2/2+4/3 =8. Thus the currents /|, I2, and
This gives us six equations in the unknown
satisfy the following system of equations:
currents /], ..., /g. But, since 1^ =I2 and
+I2 -/3=0
=6
2/,-2/2
2/2+4/3 =8
/g =14, this can be simplified to a system of
only four equations in the variables /|, I2, I4,
and ly.
Reduce the augmented matrix.
1
1
-1
2-2
0 6
0
4
8
0
2
/1
0
=0
-h-u
=1
2/,+2/2
-2/2
+ 2/5=1
Reduce the augmented matrix.
1
1
-1
1
-1
0
3
1
0
1
2
4
0
1
-1
1
0
2
2
0
0
1
0
-2
0
2
1
1
1
0-2
0
1
-1
0
1
3
2
4
36
-1 -1 0 0'
Section 1.8
SSM: Elementary Linear Algebra
1
-1
0
-1
1
1
0
3x1
8jc1
0
4
2
0
1
0
-2
0
2
1
1
-1
0
-1
-1
homogeneous linear system,
0 0
2X2 -
-2;c4=0
-X4 =0
Reduce the augmented matrix.
"3 0 -1 0 0‘
0 0
1
=0
-^3
0
8
0
0
-2
0
0
2-2
-1
0
0
0
6
-4
1
0
0 -2
4
1
0
1
0 0
3
-1
-1
0
-1
0
0 -2
0
0
6
0
0
8
0
0
0
2
4
1
-4
1
0
0-2 0
-2
-1
1
0
0 0
3
-1
0
1
0
0
0
0
-1
0
0
1
0
-2
i
0
0
8
3
0
2
-2
1
0
i
0
2
■2
-2 0
-1
0
0
0
2
6
1
-4
3
-1
0
1
0
0
0
0
-1
-1
0
0
1
0
-2
0
0
0
0
0
1
0
0
1
-1
0
0
8
3
1
0
0
0
1
1
0
2
1
=
2’
0
1
I
h=l2=U-h-
! A —Oh— —
2
I
— = 0, an d
2
I
^
1
0
3
0
4 0
-2
0
0
I
3
0
0
-1
4 0
1
0
From this we conclude that X4 is a free variable,
and that the general solution of the system is
•
u
3
/6=/4=-- + 2/3=--+1=2’
^
0
2
0
‘
-2
1
-2
From this we conclude that
/i —
3
2
4
0
-1
0
0
1
0
0
8
-1
given by Xa =t, x-,=-xa =-t,
2
4
1
1
2
•^2 =-*3 + 2^4
3
1
4
5
4
= —t,
and
X| =43 =4. The smallest positive integer
9. We seek positive integers X|, X2, X3, and X4
that will balance the chemical equation
solutions are obtained by taking f = 4, in which
X,(C3H8) + X2(02)^X3(C02) + X4(H20)
case we obtain x^ =1, X2 = 5, X3 = 3, and
For each of the atoms in the equation (C, H, and
O), the number of atoms on the left must be
equal to the number of atoms on the right. This
X4 = 4. Thus the balanced equation is
C3Hg +5O2 ^ 3CO2 +4H2O.
leads to the equations 3xi = X3, 8x1 = 2x4, and
2x2 “ 2x3 +X4. These equations form the
37
Chapter 1: Systems of Linear Equations and Matrices
11. We must find positive integers JC], X2, X3, and
X4 that balance the chemical equation
SSM: Elementary Linear Algebra
%
-a
-«3 =-'
+ O')
I
=1
%
a-|(CH3COF)+ X2(H20)
^ X3(CH3COOH)+ X4(HF)
Oq + fli +02 +^3 = 3
Aq +4a|+16^2 +64^3 = — 1
For each of the elements in the equation (C, H,
The augmented matrix of this system is
1
O, and F), the number of atoms on the left must
0
equal the number of atoms on the right. This
leads to the equations x^ = X3,
3X| + 2x2 “4-^3 + -'^4> -’^1 + -^2 “2-^3’
X| = X4. This is a very simple system of
equations having (by inspection) the general
0
I
1
CH3COF+ H2O
1
1
3
-1
4
16
64
0
0
0
solution X| = X2 = X3 = X4 = t.
Thus, taking t= I, the balanced equation is
0
4
16
1
-1
1
3
64
CH3COOH + HF.
13. The graph of the polynomial
0
p{x)= Aq + A|X +A2X“ passes through the points
(1, 1),(2, 2), and (3, 5) if and only if the
0
0
0
1
-1
1
-1
-2
0
1
1
1
2
0
4
16
64
-2
0
0
coefficients Aq, A|, and 02 satisfy the
equations
0
Aq + A| + A2 = 1
Aq + 2a|+4a2 = 2
Aq + 3a| +9a2 = 5
1
0
1
1
0
4
16
64
0
0
1
-1
1
2
Reduce the augmented matrix.
“1 1 1 1“
1
2
4
2
1
3
9
5
0
0
Add -1 times row 1 to rows 2 and 3.
“1
1
1
0
1
3
0
2
8
2
-2
0
0
2
0
0
0
0
20
60
-10
1
0
0
0
1"
4
0
Add -2 times row 2 to row 3.
1
0
2
0
1
3
1
0 0
2
2
2
0
0
1
0
0
0
0
20
60
-10
0
0
0
0
We conclude that A2 = 1, and from back
0
substitution it follows that
-1
= 1 -3a2 = -2, and
1
2
0
1
0
0
0 0
0
60
-10
Aq = 1 - A|-A2 =2. Thus the quadratic
From the first row we see that Aq = 1, and from
polynomial is p(x)= x“-2x +2.
the last two rows we conclude that a, =0 and
15. The graph of the polynomial
A3 =-—. Finally, from back substitution, it
p{x)= Aq + A|X + A2X“ +A3X^ passes through the
13
points (-1, -1),(0, 1),(1, 3), and (4, -1), if and
follov^s that A 1 = 2+ A2 -A3 = — . Thus the
only if the coefficients Aq, A|, A2, and A3
3
satisfy the equations
polynomial is p{x)=--X
6
38
‘
+ — x+1.
6
Section 1.9
SSM: Elementary Linear Algebra
17. (a) The quadratic polynomial
p{x)= flg +a^x + a2X^ passes through the
points (0, 1) and (1,2) if and only if the
coefficients Qq*
-0.25
-0.25
0.90
_i
ro.90 0.25
0.3875 LO-25 0.5oJ’
^^d 02 satisfy the
equations
and so the production needed to provide
=1
%
-1-1
0.50
-I
(b) We have (/-C)
customers $7000 worth of mechanical work
Aq + fll + <32 = 2
and $14,000 worth of body work is given by
The general solution of this system, using
<3| as a parameter, is og = L
x =(/-Cr'd
1
ss
14,000
■$25,290'
$22,581
<k<oo.
3. (a) A consumption matrix for this economy is
(b) Below are graphs of four curves in the
family: y
7000
0.25
0.3875 Lo-25 0.5
ch =\-k. Thus this family of polynomials
is represented by the equation
pix)= \ + kx +{l-k)x^ where
0.9
l + x“ ik = 0),y= 1 +x(k= 1),
"0.1
0.6
0.4'
C= 0.3
0.2
0.3 .
0.4
0.1
0.2
y = \ + 2x-x^ (A: = 2),and y = \+?>x-2x^
(k = 3).
(b) The Leontief equation (/ - C)x = d is
represented by the augmented matrix
■ 0.9 -0.6 -0.4 1930'
-0.3
0.8
-0.3
3860
-0.4
-0.1
0.8
5790
.
The reduced row echelon form of this
matrix is
True/False 1.8
(a)
True; only such networks were considered.
(b)
False; when a current passes through a resistor,
there is a drop in elecU'ical potential in the
0
0
1
0
0
0
0
31,500
26,500
26,300
.
Thus the required production vector is
$31,500'
X =
$26,500 .
$26,300
circuit.
5. The Leontief equation (/ - Qx = d for this
(c)
True
(d)
False; the number of atoms of each element must
economy is
be the same on each side of the equation.
(e)
x = (/-C)“’d
distinct jc-coordinates.
Section 1.9
0.6
0.3
50
0.39 0.5
0.9
60
48
0.39
Exercise Set 1.9
79
123.08
1. (a) A consumption matrix for this economy is
0.50
0.25'
0.25
0.10 ■
0.6
X2
required production vector is
False; this is only true if the n points have
C =
0.9 -0.3]rx,1^[50
60
-0.5
202.56 ■
39
; thu s the
Chapter 1: Systems of Linear Equations and Matrices
7. (a) The Leontief equation for this economy is
i 0 X,1 _ r
0 oJL^2
.^2
SSM:Elementary Linear Algebra
Chapter 1 Supplementary Exercises
. If d^ =2 and c/2 = 0,
1. The corresponding system is
+4x4 = 1
3x|-X2
the system is consistent with general
2xI
solution X, = 4, X2=t {0<t< °°). If
+ 3x3 +3x4 = -1
3 -1 0 4
di=2 and dj = 1 the system is
2
0
3
r
3
-1
inconsistent.
(b) The consumption matrix C =
] -1 0 A
i 0
0
3
2
1
indicates that the entire output of the second
sector is consumed in producing that output;
thus there is nothing left to satisfy any
outside demand. Mathematically, the
1
3
0
3
3
3
-1
I
1
-1 0 I 3
f M -I
0
T 0' IS■ not
Leontief matrix 1-C = 2
0
1
0
invertible.
0
Let X3 = 5 and X4 = t.
(1-Ci i)-C2iCi2 > 0. From this we conclude
9
that the matrix /- C is invertible, and
(/-C)
-1
1
Cl 2
has
_1
2
3
The solution is Xt =-—s
'
9
Xt =
1
l-C = {l'^-df =il-df, wecan
2
^ -
^3=^>
2x| - 4x2 +-'^3=6
+ 3x3 =-l
X2-X3=3
)-' )^ has
-4x1
True/Faise 1.9
False; open sectors are those that do not produce
2
-4
1
6
-4
0
3
-1
0
1
-1
3
> -2
i
3
output.
True
(d)
True; by Theorem 1.9.1.
(e)
True
3
- t—,
2
2
3. The corresponding system is
nonnegative entries.
(c)
3
5
2
conclude that / - C is invertible and that
True
3
2
~ 2^ 2^ 2
has nonnegative entries. Since
(b)
1
-t+-
__3 _3 _J_
is invertible and that
)^ )-'=((/-
4
5
3^ 2
I
(/ - C)-‘ =((/ -
2'^ 2^ 2
1
5—t
T
(a)
4
9
11. If C has row sums less than 1, then C has
column sums less than 1. Thus, from Theorem
'T'
5 _9 _\_ _5
''^1 -“■^2““^4+~
nonnegative entries. This shows that the
economy is productive. Thus the Leontief
equation (/ - Qx = d has a unique solution for
every demand vector d.
(/ -C )
1
2^2 =--2^3--^4-- =
(1-C,,)-C2,Ci2Lc21 1-Cii
1.9.1, it follows that / -
3
5
2
Thus, X3 and X4 are free variables.
9. Since C2iC]2 < 1-Cj|, it follows that
1
i
-I 0 I
■ I i
40
2
-4
0
3
0
1
-1
3
SSM: Elementary Linear Algebra
i
-2
Chapter 1 Supplementary Exercises
3
2
0
-8
5
11
0
1
0
0
1
3
, 3
4
4
5
5
5
3
The solution is x'= — jc + — v, y'=-— x + —y.
1
-2
0
1
1
5
3
2
-1
3
5
11
7. Reduce the augmented matrix.
9
0
1
1
-2
3
1
5
10
44
1
1
1
9
0
4
9
35
2
0
1
-1
3
0
0
-3
35
9
0 1!
4 f
4,
1
-2
i
0
1
-1
3
1
35
3
3
2
0
0
1
0
35
Thus X3 =
3’
26
X2 = X3 +3 =
3
3
6
17
The solution is Xi =-—, x-y =2
3
4
5
4
3
L5
5
2
3’
9.
3
5
3
4
3
0
X
x = —s+—. For x, y, and z to
4
4
a
0
b
2
a
a
4
4
0
a
2
b
Add -1 times the first row to the second.
3
3
5
4
0
35
be positive integers, s must be a positive integer
such that 5s + 1 and -9s + 35 are positive and
divisible by 4. Since -9s + 35 > 0, s < 4, so
s = 1, 2, or 3. The only possibility is s = 3. Thus,
x = 4,y = 2,z = 3.
26
^ ^x
3
5
4_
3
5
4
4
4
5. Reduce the augmented matrix.
1
35
9
35
Xj=-
4
9
y = --.+- and
17
X| =2x2--X3+3=
1
1
4
Thus z is a free variable. Let z = s. Then
, and
1
1
0
a
0
b
2
0
a
4-b
2
0
a
2
b
Add -1 times the second row to the third.
ix
3
2
a
0
b
0
a
4-b
2
0
0
b-2
b-2
-|x+y
(a) The system has a unique solution if the
corresponding matrix can be put in reduced
4x
3
row echelon form without division by zero.
Thus, a ^ 0, b 2 is required,
1
(b) If a ^4 0 and b-2,then X3 is a free variable
and the system has a one-parameter
solution.
(c) If a = 0 and b = 2, the system has a twoparameter solution.
41
SSM: Elementary Linear Algebra
Chapter 1: Systems of Linear Equations and Matrices
(d) If ^
13. (a) Here X must be a 2 X 3 matrix. Let
2 but a = 0, the system has no
solution.
a
b
cl
e / ■
X=
9. Note that K must be a 2 x 2 matrix. Let
c
Then
a
b
c
d
K=
-1
a
b
0
1
c
2 0
0
d
Then
4
a
-2
b
2 0
c
1
d
0
8
6-6
6
1
0
3
I
-4
-2
6
4
2a
b
-b
2c
d
-d
'
1
-2
-1
1
0
0
b+c
a-c
^[ 1 2 0]
-6
6
5
-d + e +3f e +f d-f
0
-4
-2
or
0
1
3
-a + b + 3c
-\
or
-3
e f
“ -3 1 5_ ■
or
Equating entries in the first row leads to the
system
2a +8c
b +4d
-b-4d
-4a +6c
-2b + 3d
2b-3d
2a-4c
b-2d
-b + 2d
-a + b + 3c = \
b +c = 2
6 -6"
6
-4
Thus
Equating entries in the second row yields
the system
-d + e + 3f = -3
e +/ = 1
d
-f=5
I
0
0
2a
+8c
=8
+4d =6
b
-4a
c =0
a
+6c
-2b
These systems can be solved simultaneously
by reducing the following augmented
=6
+3d=-\
matrix.
2a
-4c
b
=-4
-2d=0
-1
1
0
1
2
1
0
0
5
1
0
0
5
0
1
2
Note that we have omitted the 3 equations
obtained by equating elements of the last
-3
3
columns of these matrices because the
information so obtained would be just a repeat of
that gained by equating elements of the second
columns. The augmented matrix of the above
system is
'2 0
0
0
0
-4
0
4
6
0
3
6
1
0
-1
0
6
0
1
1
2
0
0
3
-1 ■
0
-4
0 -2
0_
1
0
The reduced row-echelon form of this matrix is
0
1
'l 0 0 0 o'
0
0
2
0
0
1
0
0
0 0
1
0
0 0 0
1
5
1
2
2
0 -4
0 -2
-1
1
-3
0
5
2
1
1
2
1
1
1 ■
0 0 0 0
0
0 0 0 0
0
0 0-1
1
0
3
0
0
0
1
-1
1
Thus, a =
b = 3, c =
Thus a = 0, b = 2, c = 1, and d= \.
0
6
0
-I
f- 1 and X =
2
The matrix is X =
42
d = 6, e = 0,
3 -r
6 0
1 ’
SSM: Elementary Linear Algebra
Chapter 1 Supplementary Exercises
(b) X must be a 2 X 2 matrix.
15. Since the coordinates of the given points must
satisfy the polynomial, we have
a +b+c=2
p(\)= 2 =>
a -b+c=6
p(-l)=6 =>
M2)= 3 => 4a + 2b +c = 3
X y' . Then
Let X =
z
vv
1 -I 2 _ x + 3y -x 2x+y
^3 0
z + 3w
-z
2z + w
This system has augmented matrix
If we equate matrix entries, this gives us the
equations
x + 3y = -5
z +3w = 6
-x = -l
-z = -3
2z+ w = l
2x+y =0
1
1
2
1
-1
1
6
4
2
1
3
1
0 0
1
Thus X = 1 and z = 3, so that the top two
0
equations give y = -2 and w= \. Since these
0 0
1
which reduces to
0-2
1
3
values are consistent with the bottom two
Thus, a = 1, 6 = -2, and c = 3.
equations, we have that
1
-2'
X=
19. First suppose that A5 '=S 'a. Note that all
3
matrices must be square and of the same size.
Therefore
(c) Again, X must be a 2 x 2 matrix. Let
X=
X
y
z
w
so that BA =
, so that the matrix equation
=
=5“'a.
3x+z
3y + w
-x + 2z -y + 2vv’
x + 2y
4x
Z + 2w
4z
21. Consider the /th row of AB. By the definition of
matrix multiplication, this is
=r^ -2'
4 ■
Q/i + Q/2 +• • •+
This yields the system of equations
=2
2x-2y +z
+
vv
=
-2
-4x + 3y
n
I
1
+z-2w=5
•X
since all entries A(1
2
-2
1
0
2
-4
3
0
1
-2
with matrix
-1
0
1
-2
0
-1
-4
2
1
0 0
^ which
4
1 13
0
37
0
1
0
0
1
0
0
0
0
1
160
0 0
37
reduces to
20
37
46
37
Hence, x =-
113
160
37’
37
, and X =
37
20
, z =-
46
•
n
-y-4z+ 2w =4
vv =
= AB.
A similar argument shows that if AB = BA then
becomes
- 5
or A = B^^AB
37’
1 13
160
37
37
20
46
37
37
43
Chapter 2
Determinants
Section 2.1
4-1
(b) ^23 = 4
Exercise Set 2.1
1 14
4
7
1. Mn =
-1
1
1
2
1
14
il
2
=4
= 28-(-l)= 29
4
6
-(-1)
4 14 ,4 1
4 2^S 1
= 4(2-14)+(8-56)+6(4-4)
q,=(-!)*+* A/,, =29
= -96
2+3
6 -1
M 12 -
C23=(-l)
= 24-3= 21
-3
1+2
Ci2=(-1)
6
M 13 -
4
M,2 =-21
1
= 6-(-21)= 27
4
3
1
4
1
3
-3
4
2+2
C22=(-l)
M 23 -
1
3
7
-1
M 32 -
3+1
C32=(-l)
^33 =
1
-2
6
7
+0
4
6
4
2
-14
4
1
4
3
= 2-21 = -19
M22 = -48
1
6
1
0
14
1
3
2
1
6
3
2
+0
-1
6
1
2
-14
1
1
1
3
C2i=(-1)2+'M2, =-72
= -l-18 = -19
M32 =19
3
5
-2
4
= 12-(-10) = 22
3
5
-2
4
l-l
7
5.
J_ 4 -5
22 2
22
1
i
3
22
= 7-(-12) = 19
7.
C33=(-1)^+3^33=19
0
0
3
3. (a) Mi3 = 4
1
14
2
4
Ci3=(-1)
2+2
= -(2-18)-14(-3-l)
5.
-1
3+2
2
= 72
M3, =-19
3
6
6
-1
= l-6 = -5
M 23 = 5
-2
C3l=(-1)
(d) M2, =
M22=13
-3
M3,=
C22=(-l)
= 4-(-9) = 13
-2
2+3
1
3
= ^8
1
C23=(-l)
2
= -4(2-18)-14(12-4)
= -8-3 = -ll
C2i=(-1)2+'M2, -11
M22 =
6
3
= -4
C,3=(-1)'+3Mi3=27
■2
1
(c) M22 = 4 0 14
7
-3
M2i =
M23 = 96
4
1+3
= 3^4 11
= 3(4-4) = 0
M,3=0
44
-5
7
-7
-2
-5
7
-7
-2
= 10-(-49)=59
-1-1
J_ -2 -7
59
7
-5
1
7
59
59
2
2
59
59
Section 2.1
SSM: Elementary Linear Algebra
a-3
9.
-3
a-2
=(a-3)(a-2)-(5)(-3)
2-5a +6+ 15
=a
^-5a+ 2\
=a
-2
11.
I
4
-2
1
4-2
1
3
5
-7 =
3
5
-7
3
5
1
6
2
1
6
2
1
6
=[(-2)(5)(2)+(1)(-7)(1)+(4)(3)(6)]-[(4)(5)(1)+(-2)(-7)(6)+(1)(3)(2)]
=[-20-7+ 72]-[20+84+6]
= -65
3
13.
0
2
9
0
0
03
0
5 = 2-1
52
-1
-4 1
9
-4
3
1
9
=[(3)(-l)(-4)+(0)(5)(1)+(0)(2)(9)]-[0(-1)(1)+(3)(5)(9)+(0)(2)(-4)]
= 12-135
= -123
15. det(A)=(:L-2)(?i+4)-(-5)
= ?i^+23.-8+5
= X^+2X-3
=(X-])(X + 3)
det(yt) = 0 for X = 1 or -3.
17. det(A)=(A,- 1)(1+ l)-0 =(?t- l)(?i + 1)
det(A)= 0 for ^ = 1 or -1.
3
0
0
'-0^
19. (a) 2 -1
9
-4
I
^0^ ->
-4
1
9
-4
= 3(4-45)
= -123
3
0
0
5_20
(b) 2 -1
9
-4
9
0
0 0
+1
-4
-1
5
-4
= 3(4-45)-2(0-0)+(0-0)
= -123
3
0
0
2
-1
5
0
(c)
-4
1
9
-5^ 0
1
9
-4
=-2(0-0)-(-12-0)-5(27-0)
= 12-135
= -123
45
SSM: Elementary Linear Algebra
Chapter 2: Determinants
3
0
0
9
3
0
I
-4
J.H,
(cl) 2
-4
-9?
»
2 5
= -(-12-0)-9(15-0)
= 12-135
= -123
3
0
0
(e) 2
5 =1
9
-4
0 0_ 3 0
-1
-1 5 ^2 5
=(0-0)-9(15-0)-4(-3-0)
= -135+12
= -123
3
(f)
0
0
0
5=0^
2
9
-4
= -5(27-0)-4(-3-0)
= -135+ 12
= -123
21. Expand along the second column.
det(A)= 5
-3
7
-1
5
= 5[-15-(-7)]= -40
23. Expand along the first column.
det(A)= 1
k
,k k^ , k k^
k
L'
k
L'
k
k~
= {k^ -k^)-{k^ -k^)+(k^ -k^)
=0
25. Expand along the third column, then along the first rows of the 3 x 3 matrices.
3
det(A)= -3 2
2
3
5
3
3
5
2 -2 -3 2 2 -2
10
2
4
0
= 3f3
2 -2_3 2 -2^52 2^ J 2 -2_^2 -2
1^ 10 2 2 2 2 10j l^ l 0 4 0
= -3[3(24)-3(8)+5(16)]-3[3(2)-3(8)+5(-6)]
= -3(128)-3(-48)
= -240
0
27.
0 -1
0
0
0
0 =(1)(-1)(1)= -1
1
46
2
2
4
1
Section 2.1
SSM: Elementary Linear Algebra
0
0 0 0
I
2
0 0
0
4
3 0
I
2
3
I
2
7
-3
0
I
-4
I
0
0
2 2 =(^)(')^2)(3)= 6
0
0
0
29.
31.
=(0)(2)(3)(8)=0
8
3
33. Expand along the third column.
0
sin(0)
cos(^)
0
-cos(6*)
sin(6*)
sm{9)-CQs{6) sin(^)+ cos(6') 1
sin(6>) 005(6*)
-cqs{9) sin(6*)
sin^(6*)-(-cos^(6*))
sin^(6*)+ cos^(6’)
35.
M1
_ I /
1 -
0
= 1 in both matrices, so expanding along the first row or column gives that
= r/i +>L.
True/False 2.1
(a) False; the determinant is ad - be.
1
1
(b) False;
’ 0
0 0
0
0
1
0
0 0
1
I
(c) True; if ;■+y is even then
Similarly, if
= 1 and C^-= (-!)''*"
M= M
then
(-l)'"''-^ =1 so i+j must be even.
a
b
c
(d) True; let A= b
d
e , then C|2
c
e
f
= (-l)^c f^ =-(6/-ce) and C2] =(-l) ^e
arguments show that C23 = C32 and C] 2
I.
(e)
True; the value is det(A) regardless of the row or column chosen.
(f)
True
(g)
False; the determinant would be the product, not the sum.
(h)
False; let A =
1
0
0
1
‘i =-{bf -ce). Similar
/
, then det(A) = 1, but det(cA) =
c
0
0
c
det(cA) = c" A.
47
= c^. In general, if A is an n x n matrix, then
SSM:Elementary Linear Algebra
Chapter 2: Determinants
(i) False; let A =
1
0
0
1
-1
0
and B =
, then
0 -1.
det(A) = det(fi) = 1, but det(A + 6)= 0.
(j) True; let A =
a
^ so that P? =
c
d
a^+bc ab + bd
and
ac+ cd bc+ d^
det(A^)=(a^ +bc)ibc+ d^)-{ab + bd)iac+ cd)
= a^bc+ a^d^+ bcd^ +b^c^-(a^bc +labcd + bcd^)
= a^d^ -2abcd + b^c^
=(ad -bc)^
=(det(A))2
Section 2.2
Exercise Set 2.2
1. det(A)=(-2)(4)-(3)(1)= -8 - 3 = -11
-2 r
A^ =
3
det(
4
)=(-2)(4)-(1)(3)=-8-3 = -11
3. det(A)=[(2)(2)(6)+(-1)(4)(5)+(3)(l)(-3)] -[(3)(2)(5)+(2)(4)(-3)+(-1)(1)(6)]
=[24-20-9]-[30-24-6]
= -5
det(A^)=[(2)(2)(6)+(l)(-3)(3)+(5)(-l)(4)]-[(5)(2)(3)+(2)(-3)(4)+(1)(-1)(6)]
=[24-9-20]-[30-24-6]
= -5
5. The matrix is
1
0
0 0
0
1
0 0
with the third row multiplied by -5.
= -5
0 0-5 0
0 0
0
1
7. The matrix is
1
0 0 0
0 0
0
with the second and third rows interchanged.
1
1
0
0 0
0 0 0
1
9. The matrix is
1
0
0
1
0
with -9 times the fourth row added to the second row.
0
0-9
=1
0 0
1
0
0 0 0
1
48
0
I
I
I-
I
I
S
t
3-
9-
8-
I
0
I
I
I-
0
I
0
0
I
0
0
I
9-3-
s
e
9-
8-
0
I
1
1
I
I
9
£
\
z- =
0 0
0 0
0 0
I
t
0
I
0 0
0 0
0 0
1
£
0
1
££ =
0
3
I
31
9
8
0
I
1
1
I
I
0 0
0 0
0 0
1
£
0
(I !-)£- =
3-
£
I
e
0
11-
0
0
0
0
I
I
I
I
0 0
3 0
P-O
Z
0
I
I
0
0
0
0
3-
£
\
9
£
17
1-0
£
I
£
5
I
L£
1
3-
1
P
1
I
I
I-
1
3-
0 0 0
9
3
£
£
I
0
I
9
0
L
3- £ = 3-
3-
6
1
1
L
0
3-
•ei
£
9=
(9-)- =
1
3
I
I
1
0
3
£
1
0
I
£
0
0I
0
I-
3
0
£
I
0-
1
1
I
1
£
3-
0
0
I
1
3 0
1
0
I
I
I
3
£
I
I- 0
3-
3 0
01
3
£
17
3
£
SI
3
0
9-
0
1
e- 0 0
3
0
1
3
0 0
I
I
1
3 0
3
£
0£ =
3I
Z-
0 0
I
0
0 0
00
I
0
0 0
3
0
1
0
1
0
I
1
0
1
I
Z£
I
£
I
I-
I
1
I-
0
£
I
0
0 £- =
•LI
S
\
0
0 £- =
I
0 0
3 0
I
3
17
9
£
9
9=
1
I
I
3
1
I
I-
I
I
1
I
I
0
I
I
3
SZ uojiods
1
3
£
I
0I
3
I
I
£
I
IT
0
Bjq06|v JB0un Ajb}U0uj0|3 :i/\iss
SSM: Elementary Linear Algebra
Chapter 2: Determinants
19. Exercise 14: First add multiples of the first row
0
to the remaining rows, then expand by cofactors
1
1
1 0 1 0
2 i
i
M i 0
3 I 0 0
along the first column. Repeat these steps with
the 3 X 3 determinant, then evaluate the 2 x 2
I I i 0
1 f 0 0
3
determinant.
1
0
1
I
1
1
-2
3
1
1
-2
3
1
5
-9
6
3
0
1
-9
-2
i 0 ^
-I
2
-6
-2
0
0
-3
-1
0
12
0
-1
i i
=(-l) 2
3
3
3
1
-9
-2
0
-3
-I
t 0 .
12
0
-I
2
i
I
3
3
3
6
2
2
2
1
I
0
3
3
I
1
-9
-2
0
-3
-1
-f » -f
0
108
23
r t^
-3
108
3
i
I
2
2
5
_2
3
3
23
5^
6j
=(-3)(23)-(-l)(108)
3
= 39
3
Exercise 15: Add -2 times the second row to the
first row, then expand by cofactors along the
6
first column. For the 3 x 3 determinant, add
Exercise 17: Add 2 times the first row to the
multiples of the first row to the remaining row,
then expand by cofactors of the first column.
Finally, evaluate the 2 x 2 determinant.
second row, then expand by cofactors along the
2
1
3
0
0
2
0
1
1
0
1
1
determinant, add -2 times the first row to the
-1
1^1 0 1
2
first column two times. For the 3x3
second row, then expand by cofactors along the
first column. Finally, evaluate the 2 x 2
1
0
0
2
1
0
2
0
1
2
3
determinant.
3
5
3
-7
0-4
2
3
1
5
3
0-1
1
2
6
8
0
1
0
1
0
2
1
1
0
0
0
1
1
-1
2
6
8
I
-1
-2
=(-1)2
1
0
0
0
1
0
1
2
3
0
0
2
0
0
0 0
1
1
-1
0
2
0
4
I
1
0
^0 1 0 1
2
0
2
1
1
4
0 0
1
f
1
0
1
=(-1)2
I
I
0
1
=-(^-2)
=6
Exercise 16: Add — times the first row to the
1
1
0
1
second row, then expand by cofactors along the
0
1
-1
fourth column. For the 3x3 determinant, add -2
times the second row to the third row, then
expand by cofactors along the second column.
Finally, evaluate the 2 x 2 determinant.
0
2
I
= -[!-(-!)]
= -2
50
1
Section 2.3
SSM: Elementary Linear Algebra
21. Interchange the second and third rows, then the
1
first and second rows to arrive at the determinant
whose value is given.
d e f
d e f
a b
c
8 h
i
a b
c
8 h
i
a
b
d
8
e f
h
i
29.
1
a
b
c
2 ^2
a
c
1
1
b-a
c-a
=-6
=0
0 b^-a^
23. Remove the common factors of 3 from the first
b-a
row,-1 from the second row, and 4 from the
c
value is given.
3a
3b
a
3c
b
c
a
b
c
e
4h
f
4i
a
b
-a
=(b- a)(c+ a){c -a)-(c- a){b + a){b -a)
={b- a){c -d)[c + a-{b + a)]
={b -a)(c-a)(c-b)
-e -f =3 -d -e -f
4h
4i
4g 4h
4i
d
4g
c-a
^b^-a^ -2 -2
=(b-a)(c^ -a~)-{c-a){b^ -a^)
third row to arrive at the determinant whose
-d
4g
-2 -a-2
c
True/False 2.2
(a) True; interchanging two rows of a matrix causes
c
the determinant to be multiplied by -1, so two
such interchanges result in a multiplication by
= -12 d
e f
i
8 h
= -12(-6)
(b) True; consider the transposes of the matrices so
= 72
that the column operations performed on A to get
B are row operations on aJ.
25. Add -1 times the third row to the first row to
arrive at the matrix whose value is given.
a+g
b + h c+ i
a
e
f =d
d
h
8
i
(c) False; adding a multiple of one row to another
b c
e / =-6
h
8
does not change the determinant. In this case,
det(fi) = det(A).
i
(d) False; multiplying each row by its row number
27. Remove the common factor of -3 from the first
causes the determinant to be multiplied by the
row number, so det(6) = n\ det(A).
row, then add 4 times the second row to the third
to arrive at the matrix whose value is given.
■3b
-3a
d
g-4d
-3
(e) True; since the columns are identical, they are
-3c
proportional, so det(A)= 0 by Theorem 2.2.5.
f
i-4f
e
h-4e
(f)
True; -1 times the second and fourth rows can
a
b
c
be added to the sixth row without changing
d
e
f
i-4f
det(A). Since this introduces a row of zeros,
det(A) = 0.
g-4d
h-4e
a
b
c
= -3 d
e
f
8
h
/
Section 2.3
Exercise Set 2.3
= -3(-6)
1. det(A) = -4-6 = -10
d et(2A) =
51
-2
4
6
8
= -16-24 = -40 = 22(-10)
Chapter 2: Determinants
SSM: Elementary Linear Algebra
13. Expand by cofactors of the first row.
3. det(A)=[20- 1 + 36]-[6 + 8 - 15] = 56
^
2
1
-6
det(A)= 2
det(-2A)=-6 ^
-2
-2
8
-10
0
3 6
-2(6)= 12
Since det(A) 0,A is invertible.
=[-160+8-288]-[-48-64+120]
15. det(A)=(A:-3)(jt-2)-4
448
= lc^-5k +6-4
= k^-5k +2
=(-2)^56)
9-1
8
5. det(Afi)=31
1
17
10
0
2
Use the quadratic formula to solve
k^-5k +2= 0.
-(-5)±V(-5)^-4(l)(2) 5±Vl7
=[18-170-0]-[80+0-62]
-1
-3
det(BA)= 17
2
2(1)
= -170
6
A is invertible for k ^
5±^fn
2
11 4
10
5
2
17. det(A)=[2+12;(+36]-[41:+18+12]
=[-22-120+510]-[660-20-102]
= 8A:+8
= -170
A is invertible for 1:5^ -1.
det(A)= 2
2
1
3
4
= 2(8-3)= 10
19. A is the same matrix as in Exercise 7, which was
shown to be invertible (det(A)= -1). The
det(B)=[l-10+0]-[15+0-7]=-17
3 0
cofactors of A are Cu = -3, C|2 = -(-3)= 3,
3
det(A +5)=10 5 2
5
0
Ci3=-4+2= -2, C2, =-(15-20)= 5,
3
C22=6-10 =^, C23 =-(8-10)= 2,
=5^5 33
C3,=0+5 = 5, C32 =-(0+5)= -5, and
C33 =-2+5 = 3.
=5(9-15)
= -30
The matrix of cofactors is
?idet(A)+ det(fi)
7. Add multiples of the second row to the first and
third row to simplify evaluating the determinant.
0
3
2
adj(A)= —
det(A)
= 9-10 = -l
2
-2
5
-4
2
5
-5
3
5
5
1
3 -4 -5
-2
3 5
0
3
-3
A-‘
5
det(A)=-l -1 0=-(-l)
-3
2
3
" 3 -5 -5'
3
3
= -3
Since det(A)^ 0, A is invertible.
9. Expand by cofactors of the first column.
4
5
2 -2
-3
21. A is the same matrix as in Exercise 9, which was
shown to be invertible (det(A)= 4). The
^ “^=2(2)=4
det(A)= 2;0
cofactors of A are Cj
|
= 2, C|2 = 0, C13 = 0,
C21 = —(—6)= 6, C22 = 4, C23 = 0,
Since det(A) 0, A is invertible.
j =9 — 5 = 4, 632 = —(—6)= 6, and 6*33 = 2.
11. The third column of A is twice the first column,
so det(A)= 0 and A is not invertible.
52
SSM: Elementary Linear Algebra
Section 2.3
2 0 0
The matrix of cofactors is
2
A
1
1
4
0
C23-(-l)l 3 9=0
4
6
2
1
6
3
2
4
1
3
acij(A)=- 0 4 6
C24 - 1
3
8 =0
"^[o 0 2
1
3
2
3
1
1
-1
3
6
det(A)
2
0
^
ii
2
C3i = 5 2 2 =0
1 f
3
2
2
0 0 i
1
23. Use row operations and cofactor expansion to
simplify the determinant.
1
det(A)=
3
1
1
1
0-1
0 0
0
0
7
8
0
0
1
1
-1
0 0
1
3
0
1
=(-l)
1
7
8
1
1
2
1
3
1
2
3
1
C34 =(-1) 2 5 2 =-(-l)= l
8
1
2
C33 = 2 5 2
1
0 7
1
C32 =(-1) 2 2 2 =0
1
3
2
3
1
1
C4|=(-1)5 2 2 =-(l)=-l
3
8
9
=-(7-8)
=1
Since det(A) 0, A is invertible. The cofactors of
1
1
C42 - 2
2
2 =0
1
8
9
A are
5
2
2
3
8
9 =-4
3
2
2
1
C,
2
Ci3 -
2
2
2
5
2
1
3
9 = -7
1
3
2
2
5
3
9
2
C,2=(-l) l 8 9 =-(-2)= 2
1
1
C43 =(-1) 2 5 2 =-(-8)= 8
1
2
3
1
3
C44 - 2
5
2 =-7
1
1
3
8
-4
3
2
det(A)
0
8
-7
2
C22=l
8
9=-l
1
2
2
1
3
0
-1
2
-1
0
0
adj(A) = -
1 -7
0 -1
6
1
1
6
0
-4
1
8 9 =-(-3) = 3
2
0
0
3
3
0
2
A-'
C2, =(-1) 3
-1
The matrix of cofactors is
C,4=(-l) l 3 8=-(-6)=6
1
2-7
3
1
-7
■-4
3
0
-f
2
-1
0
0
-7
6
53
0
0-1
0
1
8
-7
SSM: Elementary Linear Algebra
Chapter 2: Determinants
25.
4
5
1
2
1
5
2
A=
4
-3
I
-2
-1
0
0
A| -
0
4
5
1
5
4
4
-3
det(A,)=(-3)-2
-1
5
+2
det(/\) = -2
0-3
= -2(20-5)+ 2(4-55)
1
= -132
4
1
At — 2 -2
0
'2 5 o'
4
0-3
/l,= 3 1 2
1
5
2
det(A,)= -2
2
5
1
5
2
5
3
1
2
-2
det(A2)= ,4
0
1
-3
4
A3= 2
1
-2
4
0
0
4 2 o'
3
2
1
2
4
4
2
11
3
+2
A-| =
=-2(4-2)+ 2(12-22)
det(A|)_30_ 30
det(A) ”-11 ” 11
_det(A2)
= -24
-^2 =
4
5
2
A. = 1 1
1
3
1
5
1
•^3 ==
1
3
det(A^)=4 5
1
5
2
5
1
-II
5
2
1
3
38
det(A3) _ 40 _ 40
det(A)~-l l~ 11
30
The solution is
+
11
38
X2=-
11
40
^3 =-
= 12
11
det(A^.) _ -36 _ 3
3
-1
1
A = -1
7
-2
2
6
-1
det(A) ”-132 ~ II
29.
det(A^,) _ -24 _ 2
det(A) ”-132”IT
det(AJ_ 12 _
z=
Note that the third row of A is the sum of the
1
first and second rows. Thus, det(A)= 0 and
Cramer’s rule does not apply.
det(A) ”-132” 11
3
2
1
The solution is X = —, y=—, z =11
11
31.
1
27.
38
det(A) ”-1 1 ’" 11
= 4(l-15)-ll(5-10)+(15-2)
X=
4
-2
det(A3)=4 J _2 = 4(6+4)= 40
2
det(A,,)=-2
1
2
= 38
= -36
= 11
+(-3)
=(0+8)-3(-2-8)
+2
= -2(10-5)+ 2(2-15)
A
= -3(-4-6)= 30
-3
1
A= 2
-1
0
4
0
-3
2
det(A)=
4
-I
0
+(-3)
1
-3
2
-1
4
1
3
7
7
3
1
1
A=
-5
8
1
2
Use row operations and cofactor expansion to
simplify the determinant.
=(0+4)-3(-l + 6)
= -1 1
54
Section 2.3
SSM: Elementary Linear Algebra
f
0
det(A)=
-3
-3
-7
0
4
-4
-5
0
-4
-12
-6
-3
= -(l) 4
-12
-3
det(2A)
1
-6
-3
2^ det(A)
-7
=- 4
-4
-5
0
-16
-11
8(-7)
1
-3
-5
= -f-3-16
-7l
56
“^-16
d
a
= 3(44-80)+ 4(33-1 12)
= -424
A\
1
-1
(d) det((2A)-')=
-7
-4 -5
-4
(e) det b
h
c
i
a
b
= det g
f
d
6
1
1
3
1
-1
1
a
7
-3
-5
8
1
3
1
2
= -det d
g
-6
-3
7
-4
-5
-12
-6
0
det(Ap = 0
-24
1
3
= -(l)
-24
-3
7
-4
-5
-12
-6
b
c
e f
h
i
= -det(A)
=7
37. (a) det(3A)= 3^det(A)= 27(7)= 189
2
-6
c
h i
e f
e
4
0
7
\ ^
2
-3
1
(c) det(2A“‘)= 2^det(A“‘)= 8 —
I
(b) det(A“')=
det(A)
7
-6
-3
7
-8
-4
-5
(c) det(2A''')= 2^det(A“‘)= 8 - =-
0
0
-34
\1) 7
-6
-3
-8
-4
-(-34)
/1 N
Q
1
-I
(d) det((2A)-‘)=
det(2A)
= -(-34)(24-24)
1
=0
2^ det(A)
det(Ap
0
det(A)
-424
1
=0
3’ =
8(7)
1
33. If all the entries in A are integers, then all the
cofactors of A are integers, so all the entries in
adJ(A) are integers. Since det(A)= 1 and
56
True/False 2.3
1
A-'
adj(A)= adj(A), then all the entries
det(A)
(a) False; det(2A)= 1? det(A) if A is a 3 x 3 matrix.
-I
in A ' are integers.
(b) False; consider A =
3
35. (a) det(3A)= 3-’det(A)= 27(-7)= -189
-1
0
1
0
and B =
Then det(A)= det(B)= l,but
det(A + 8)^0^ det(2A)= 4.
1
(b) det(A“')=
det(A)
1
0
-7
7
55
0
-1
SSM: Elementary Linear Algebra
Chapter 2: Determinants
(c) True; since A is invertible, then
(b)
1
1
and
det(A“‘)=
-4
2
3
3
3
3
-4
2
det(A)
1
1
det(A''‘B4)=
1
1
^-4 2
■ det(S)• det(A)= det(B).
det(A)
=-3^ >
0 6
(d) False; a square matrix A is invertible if and only
= -3(l)(6)
if det(A)^ 0.
= -18
(e) True; since adj(A) is the transpose of the matrix
of cofactors of A, and (B^ = B for any matrix
3. (a)
B.
-1
5
0
2
-3
1
2
-1
-.=-F1
1
1
= -(2+l)-3(-5-4)
(f) True; the ij entry of A • adj(A) is
= 24
a,lCyi + aj2Cj2 +•••+ ai^Cj^. If t ^jthen this
sum is 0, and when i -jthen this sum is det(A).
Thus, A • adj(A) is a diagonal matrix with
diagonal entries all equal to det(A) or
(b)
-1
5
2
-1
5
2
0
2
-1
0
2
-1
-3
1
1
0
-14
-5
(det(A))/„.
(g) True; if det(A) were not 0, the system Ax = 0
would have exactly one solution (the trivial
-1
5
2
0
2
-1
0
0
-12
=(-l)(2)(-12)
solution).
= 24
(h) True; if the reduced row echelon form of A were
3 0-1
/„, then there could be no such matrix b.
1
5. (a)
(i) True; since E is invertible.
-1
-1
=3^ ’-0
4
0 4
(j) True; if A is invertible, then so is A
1
2
4
2
2
= 3(2-4)-(0+4)
and so
= -10
both Ax = 0 and A“'x = 0 have only the trivial
3 0-1
solution.
(b)
-1
0
(k) True; if A is invertible, then det(A)^0 and A
-1
is invertible, so det(A)- A
1
1
4
2
1
0
1
= adj(A) is
1
1
3 0-1
4
2
1
0-3-4
invertible.
0
4
1
1
2
(1) False; if the k\h row of A is a row of zeros, then
every cofactor Cjj =0 for / ^ k, hence the
0-3
cofactor matrix has a row of zeros and the
0
corresponding column (not row) of adj(A) is a
column of zeros.
0
1
-4
10
3
10
= -(l)(-3) -—
3;
Chapter 2 Supplementary Exercises
1. (a)
-4
2
3
3
= -10
= ^(3)-2(3)= -18
56
SSM: Elementary Linear Algebra
3
6
0
-2
3
1
Chapter 2Supplementary Exercises
3
7. (a)
1
^1 =30
0
2
-2
-2
-6
2
-9
4
-2
2
1
1
-1
-9
-2
4
-2
3
1
0
-9
2
1 -I
2
-2
2
N
1
=3 3
4
+2
-2
1
2
1
1
1
-6 -2
/
-1
3
1
+4
-2
2
-9
2
-9 -2j ^
2 -2
-(-1)
-2
3
-9
2
= 3[3(-2+ 2)+ 2(1 + 4)]-6[-2(-2+ 2)-(2+9)+4(-2-9)]-[-(-6-2)+(-4+ 27)]
= 30+ 330-31
= 329
3 6
0
1
1
0
1
4
-2
3
1
3
6
0
1
2
-9
2
-2
2
-2
3
1
0
-1
-9
2
-2
(b)
1
1
4
1
0
-1
1
0
3
-1
6
0
6
3
-2
0
2
-11
11
1
0
0
3
0 0
5
-14
31
0 0
7
3
1
0
-1
0
3
-1
6
0 0
5
-14
0 0
0
329
15
329
= -(l)(3)(5) 15
= 329
9. Exercise 3:
-1
5
2
-1
0
2
-1
0
1
-3
-3
2 -1
5
2-1 0
5
2
1
1 -3
=[(-1)(2)(1)+(5)(-l)(-3)+(2)(0)(1)]-[(2)(2)(-3)+(-1)(-1)(1)+(5)(0)(1)]
= 13-(-ll)
= 24
Exercise 4:
-1
-2
-3
-1
-2
-3 -1
-2
-4
-5
-6 = -4
-5
-6-4
-5
-7
-8
-9
-8
-9-7
-8
-7
=[(-l)(-5)(-9)+(-2)(-6)(-7)+(-3)(-4)(-8)]-[(-3)(-5)(-7)+(-1)(-6)(-8)+(-2)(-4)(-9)]
= -225-(-225)
=0
57
SSM: Elementary Linear Algebra
Chapter 2: Determinants
Exercise 5;
3 0-1
0
4
3
0
2
0
3
4
0
1 I
1
20
4
=[(3)(1)(2)+(0)(1)(0)+(-1)(1)(4)]-[(-1)(1)(0)+(3)(I)(4)+(0)(1)(2)]
= 2-12
= -10
Exercise 6:
-5
1
3
0
4
1
-2
-5
1
3
0
2
3
0
1
-2
2
1
-2
2=
2
4-5
1
=[(-5)(0)(2)+(1)(2)(1)+(4)(3)(-2)]-[(4)(0)(1)+(-5)(2)(-2)+(1)(3)(2)]
= -22-26
= -48
11. The determinants for Exercises 1,3, and 4 were evaluated in Exercises 1,3, and 9.
Exercise 2;
7
-2
-6
= 7(-6)-(-l)(-2)= -44
The determinants of the matrices in Exercises 1-3 are nonzero, so those matrices are invertible. The matrix in
Exercise 4 is not invertible, since its determinant is zero.
5
b-3
b-2
-3
13.
= 5{-3)-{b-3)(b-2)
= -\5-b^ +5b-6
= -b-+5b-2l
0
15.
0
0
0
-3
0 0
0
-4
0
0 0-1
0
0
0
2
0
0
0
5
0
0
0
0
=(5)(2)(-l)(-4)(-3)
= -120
-4
17. Let A =
3
2
3 ■
Then det(A)=-18. The cofactors of A are C|
|
=3, C], =-3, C21 =-2, and C
3
-3
-2
-4
= -4. The
matrix of cofactors is
A
3
-1
adj(A)=
det(A)
-1
19. Let A =
0
-3
5
-2
18 -3 -4
i
i
6
9
1
2
6
9
2
2-1 . Then det(A) = 24.
1
The cofactors of A are Cj I = 2+1 = 3, C|2 =-(0-3)= 3, C|3=0+6 = 6, C21 =-(5-2)=-3, C22=-l+6 = 5,
C23 =-(-l + 15)= -14, C3,=-5-4 = -9, C32 =-(1-0) = -1, and C33 =-2-0 =-2.
58
SSM: Elementary Linear Algebra
The matrix of cofactors is
Chapter 2 Supplementary Exercises
3
3
6
-3
5
-14
-9
det(A)= — + — = 1
25
-2
25
4
X
5
3
-3 -9
3
I
A-'
aclj(/l)= —
24
det(/l)
3
5
-1
6
-14
-2
5
3
4
det(A_,0 =-^+-)'
i
3
i
3
i
_5
i
24
24
1
7
I
4
12
12
Ay =
X
5
i
3
3
6
0
1
-2
3
1
4
1
0
1
2-2
4
4
3
5
5
5
5
a
27.
C||=10, C|2=55, C,3=-21, C,4=-31,
C21 = -2, C22 -11, C2^=10, C24 = 72,
A=
1 1 P
a p 1
1
C3|=52, C32 =-43, C33 =-175, C34 =102,
a
det(A)=0
C41 =—27, ^742 = 16, 6*43 =—42, and
0
P-a
0 p-a \-a^
C44=-15.
0
The matrix of cofactors is
55
-21
-2
-11
70
52
-43 -175
-27
, 3
2
Then det(A)= 329. The cofaetors of A are
10
3
The solution is x'=-xH—y, y'= —x+-y.
23. Let A =
-9
4
4
det(Ay)=-y--x =--x +-y
16
72
102 '
-42
p-a
~ P-a l-a^
=0-(P-af
=-(P-af
-31'
-15
det(A)= 0 if and only if a = P, and the system
A-'
has a nontrivial solution if and only if det(A) = 0.
adj(A)
det(A)
10
-2
52
-27
55
-I I
-43
16
329 -21
70
-175
-42
102
-15
-31
72
10
2
52
27
329
329
329
329
55
ii
43
IL
329
329
329
329
1
10
25
47
31
329
47
47
72
329
102
329
29. (a) Draw the perpendicular from the vertex of
angle y to side c as shown.
47
15
3
0
_4
5
3
5
a
construction gives the other two equations in
the system. The matrix form of the system is
X
5
^2
b
c = C| +C2 =acosP+ bcosa. A similar
329
25. In matrix form, the system is
'1 _4lr ,n
5
C]
Then cosa = — and cosy5 = —, so
6
5
59
c
b
cos a
a
c 0 a
cosP = b
b
cos y
a 0
c
Chapter 2: Determinants
SSM: Elementary Linear Algebra
0
c
b
det(A)= c
0
a
b
a
0
c
31. If A is invertible then A ' is also invertible and
a
^b 0
c
0
b
a
det(A) 56 0, so det(A)A~* is invertible. Thus,
adj(A)= det(A)A^' is invertible and
(adj(A)r' -(det(A)A-'r'
+b
1
= -c(0-ab)+ b(ac-0)
= 2abc
Aa
(A-‘r'
det(A)
a
c
b
1
b
0
a
det(A)
c
a
0
A
= adj(A-')
det(A«)=-c^ 0
a
b
1
since adj(A ’)=det(A ')(A ’) *
-a
b
a
= -c{0-ac)-a{a^ -b^)
= a{c^ +b^-a^)
33. Let A be an « X n matrix for which the sum of
the entries in each row is zero and let X| be the
det(A^)
cosflr =
A.
det(A)
n X 1 column matrix whose entries are all one.
det(A)
0
a{c^ +b^ -a^)
0
because each entry in the
Then Axj =
2abc
2
+b^ -a
0
2bc
0
a
product is the sum of the entries in a row of A.
That is, Ax = 0 has a nontrivial solution, so
det(A) = 0.
b
(b) A^ = c b a
b
c
det(A^)=
0
c
a
b
0
c
b
b
c
35. Add 10,000 times the first column, 1000 times
the second column, 100 times the third column,
+b
-a
and 10 times the fourth column to the fifth
column. The entries in the fifth column are now
= -a{Q-ab)+ bic^-b^)
= b{a^+c^-b^)
21,375, 38,798, 34,162, 40,223, and 79,154,
respectively. Evaluating the determinant by
expanding by cofactors of the fifth column, gives
det(A^)
cos^ =
21,375C,5 +38,798C25 +34,162C35
+ 40,223C45+79,154C55.
det(A)
b{a^ +c^-b^)
Since each coefficient of Cj^ is divisible by 19,
2abc
this sum must also be divisible by 19.
Note that the column operations do not change
a^+c^-b'^
2ac
O
c
the value of the determinant—consider them as
row operations on the transpose.
a
Ay= c 0 b
b
a
c
c
b
det(A^)=-c^ ^
c
0
b
a
+a
= -c{c^ -b^)+ aiac-0)
= cia^+b^-c^)
cos/=
det(A^) _ c(a^+^2 _ ^2^
det(A) ^
2abc
+b^ -c^
2ab
60
Chapter 3
Euclidean Vector Spaces
Section 3.1
(f)
V
Exercise Set 3.1
I
X
I
Z
1. (a)
I
---I--,-(-3,4,-5)
“ -X-1
I
r-
--■-^(3,4, 9
r
3. (a)
I
.y
I
A
H—I—I—I—H—
Z
(b)
v«(-3.4,5)
.--1
I
-1
f
y
I
(b)
I
X
I
I
I
X
(c)
✓
I
I
(3,^,5)t--
V
(c)
-- - 1
X
I
t
I
I
.y
I
X
Z
(d)
Z
(d)
I
y
r- -
Tt
’\
I
i
I
- -f
I
«
I
V
I
I
I
+
I ✓"
u- .1.^"
(3,4,-5)
X
(e) (-3,-4,5)
^ - -f - - -1
r--
61
SSM: Elementary Linear Algebra
Chapter 3: Euclidean Vector Spaces
9. (a) Let the terminal point be B{b^, i>2)- Then
(e)
(ii| -1, ^>2 -1)=(L 2) so b^ =2 and 62 = 3.
The terminal point is B(2, 3).
(b) Let the initial point be A(fl|, a2, 133). Then
V
(-1-«!,-1-a2, 2-133)=(1, 1, 3),
=-2, ^2 =“2, and 03 =-1. The initial
X
point is A{-2,-2,-1).
z
(f)
11. (a) One possibility is u = v, then the initial point
(x, y, z) satisfies
(3 - X,0- y, -5 - z)=(4,-2,-1), so that
X = -1, y = 2, and z = -4.
One possibility is (-1, 2, -4).
H—h H—h
(b) One possibility is u = -v, then the initial
point (x, y, z) satisfies
(3- X,0- y, -5 -z)=(-4, 2, 1), so that
X = 7, y = -2, and z = -6.
One possibility is (7,-2,-6).
5. (a) /^P2 =0-4,7-8)=(-!,-!)
.V
13. (a) u + w =(4,-l)+(-3,-3)=(L-4)
(b) V - 3u =(0, 5)-(12,-3)= (-12, 8)
(c) 2(u-5vv)= 2((4,-!)-(-!5,-15))
(b) fjP2=(-4-3,-7-(-5))=(-7,-2)
= 2(19, 14)
=(38, 28)
(d) 3v-2(u + 2w)
=(0, 15)-2((4,-l)+(-6,-6))
=(0, 15)-2(-2,-7)
=(0, 15)+(4, 14)
=(4, 29)
(c) /^P2=(-2-3,5-(-7),-4-2)
(e) -3(w-2u+v)
= -3((-3,-3)-(8,-2)+(0, 5))
=(-5, 12,-6)
= -3(-l I,4)
=(33, -12)
(f) (-2u - v)- 5(v + 3w)
=((-8, 2)-(0,5))-5((0,5)+(-9,-9))
=(-8,-3)- 5(-9,-4)
=(-8,-3)+(45, 20)
=(37, 17)
—>
15. (a) v-w =(4, 7,-3, 2)-(5,-2, 8, 1)
7. (a) /^P2 =(2-3, 8-5)=(-1,3)
=(-l, 9,-11, 1)
(b) PiP2=(2-5, 4-(-2), 2-1)=(-3, 6, 1)
(b) 2u + 7v =(-6, 4, 2, 0)+(28, 49, -21, 14)
=(22, 53, -19, 14)
62
Section 3.1
SSM: Elementary Linear Algebra
(c) -u +(v -4w)=(3,-2,-1,0)+((4, 7,-3, 2)-(20,-8, 32,4))
=(3,-2,-1,0)+(-16, 15, -35,-2)
=(-13, 13,-36, -2)
(d) 6(u-3v)= 6((-3, 2, 1, 0)-(12, 21, -9, 6))
= 6(-15,-19, 10,-6)
=(-90,-114, 60,-36)
(e) -v-w=(^,-7,3,-2)-(5,-2, 8, 1)
=(-9,-5,-5,-3)
(f) (6v - w)-(4u + V)=((24, 42, -18, 12)-(5, -2, 8, 1))-((-12, 8, 4,0)+(4, 7,-3, 2))
=(19,44, -26, 11)-(-8, 15, 1,2)
=(27, 29, -27,9)
17. (a) w-u =(^,2,-3,-5, 2)-(5,-1, 0, 3,-3)
=(-9, 3,-3,-8, 5)
(b) 2v + 3u =(-2, -2, 14, 4, 0)+(15, -3, 0, 9, -9)
=(13,-5, 14,13,-9)
(c) -w + 3(v - u)=(4,-2, 3, 5, -2)+ 3((-l, -1, 7, 2, 0)-(5,-1, 0, 3, -3))
=(4,-2, 3, 5, -2)+ 3(-6, 0, 7, -1, 3)
=(4, -2, 3, 5,-2)+(-18,0, 21,-3,9)
=(-14, -2,24,2,7)
(d) 5(-v + 4u - w)= 5((1, 1, -7,-2,0)+(20,-4,0, 12, -12)-(-4, 2,-3,-5, 2))
= 5(25,-5,-4, 15,-14)
=(125,-25,-20, 75,-70)
(e) -2(3w + v)+(2u + w)= -2((-12, 6, -9,-15, 6)+(-l, -1, 7, 2, 0))+((10, -2,0, 6, -6)+(^, 2, -3,-5, 2))
= -2(-13, 5, -2,-13, 6)+(6, 0, -3, 1, -4)
=(26,-10, 4, 26, -12)+(6, 0, -3, 1, -4)
=(32,-10, 1, 27, -16)
1
(f) -(w-5v + 2u)+V =-((-4, 2, -3, -5, 2)-(-5, -5, 35,10, 0)+(10, -2, 0, 6, -6))+(-l, -1, 7, 2, 0)
2
2
38, -9,-4)+(-l, -1, 7, 2, 0)
9
3
-,-2 +(-l,-1, 7, 2, 0)
-2
^,^,-12,-^.
2 2
2’
19. (a) v-w =(4, 0,-8, 1, 2)-(6,-1,-4, 3,-5)
=(-2, 1, -4,-2, 7)
(b) 6u + 2v =(-18, 6, 12, 24, 24)+(8, 0, -16, 2, 4)
=(-10, 6,-4, 26, 28)
63
SSM: Elementary Linear Algebra
Chapter 3: Euclidean Vector Spaces
(c) (2u-7w)-(8v + u)=((-6, 2, 4, 8, 8)-(42, -7,-28, 21, -35))-((32, 0, -64, 8, 16)+(-3, 1, 2, 4, 4))
=(-48, 9, 32, -13, 43)-(29, 1, -62, 12, 20)
=(-77, 8,94,-25, 23)
21. Let X =(j:|,
+3>+4>+5)-
2u-v + x =(-6, 2, 4, 8, 8)-(4, 0, -8, 1, 2)+(xy, X2, X3, X4, X5)
=(-10+ Xj, 2+ X2,12+X3, 7+ X4, 6+ X5)
7x + w =(7x|, 7x2,’^^3>
7x5)+(6, -1, -4, 3, -5)
=(7X|+6, 7x2 “1’ "^^3 “4, 7x4+3, 7x5 -5)
Equate components,
g
-10+Xj = 7x|+6^ X] = —
3
1
2+ X2 =7X2-1^X2 = 2
8
I2+ X3 =7x3 -4 => X3 =-
7+ X4 = 7x4 +3=> X4 =^
11
6+ X5 =7x5 -5
X=
Xj = —
r 8 J_ 8 2 n'
3’ 2’ 3’ 3’ 6 .
23. (a) There is no scalar k such that kxi is the given vector, so the given vector is not parallel to u.
(b) -2u =(4,-2,0,-6,-10,-2)
The given vector is parallel to u.
(c) The given vector is 0, which is parallel to all vectors.
25. aa + b\ ={a, -a, 3a, 5a)+(2b, b, 0,-3b)
= ia + 2b, -a + b, 3a, 5a-3b)
=(1,-4, 9, 18)
Equating the third components give 3a = 9 or a = 3. Equating any other components gives b = -l. The scalars are
a = 3,b = -\.
27. Equating components gives the following system of equations.
c,+3c2
-Ci+2c2 +C3=1
C2 +4c3 = 19
1
3
The reduced row echelon form of -1
2
0-1
1
1
0
1
4
19
is
The scalars are C] =2, C2 = -1, C3 = 5.
64
1
0
0
2
0
1
0
-1
0 0
1
5
Section 3.2
SSM: Elementary Linear Algebra
True/False 3.1
29. Equating components gives the following system
of equations.
(a) False; vector equivalence is determined solely by
-C|+ 2c2 + 7c3 + 6C4 =0
3cI
+ C3 +3c4 =5
2C| +4c2 + C3 + C4 =6
—C2 +4c3 + 2C4 = -3
length and direction, not location.
(b) False;(a, b) is a vector in 2-space, while (a, b, 0)
is a vector in 3-space. Alternatively, equivalent
vectors must have the same number of
The reduced row echelon form of
-1
2
7
6
0
1
3
0
1
3
5
0
2
4
1
6
2
-3
0 0 0
1
0 0
components.
1
1
(c) False; v and
IS
0-1
4
0 0
1
0-1
0 0 0
1
1
are parallel for all values of k.
(d) True; apply Theorem 3.1.1 parts (a),(b), and (a)
again; v +(u +w)= v-i-(w + u)
The scalars are C| = 1, C2 = 1, C3 = -1, C4 = 1.
=(v +w)+ u
=(w-I-v)-I-u.
31. Equating components gives the following system
of equations.
(e) True; the vector -u can be added to both sides of
—2c| —3c2 "H C3 =0
9ci +2c2 -i-7c3 =5
6C| -t-C2-(-5c3 =4
the equation.
(f) False; a and b must be nonzero scalars for the
The reduced row echelon form of
"-2 -3
1 O'
9
2
7
5
6
1
5
4
is
statement to be true.
1
0
1
0
0
1
-1
0
0 0
0
1
(g) False; the vectors v and -v have the same length
and can be placed to be collinear, but are not
equal.
From the bottom row, the system has no
solution, so no such scalars exist.
(h) True; vector addition is defined component-wise.
(i) False;(k + m)(u + v)=(k + m^u +(k + m)v.
1
33. (a) -F(2 =-(7-2,-4-3, 1-(-2))
1
0) True; x = —(5v-i-4w)
=^(5,-7, 3)
8
5 _7 3
,2’ 2’2.
(k) False; for example, if V2 = -V| then
+^2''2
I
Locating ~PQ with its initial point at P
C!| — CI2 —
gives the midpoint.
— ^2'
Section 3.2
24,34-Z ,-2+^2;
2
2
15
2’
2’ 2
Exercise Set 3.2
1. (a)
21 9
(b) ^Pe
=^(5,-7,3)=
4
4
L4
4
||v|| = 74^+(-3)2 =V^=5
4
V
ui =
3
Locating — PQ with its initial point at P
4
f4
3
has the
same direction as v.
21
15
2+ —,3+
V
4
direction opposite v.
4’ 4
65
4 3
has the
V
,-2-t--
l 4
1
U2 =-
9
^ _9 J_
4’
1
—
^=-(4,-3)=-,
IIV 5
1^5 5
gives the desired point.
4
+^2'''2 as long as
e
Chapter 3: Euclidean Vector Spaces
SSM: Elementary Linear Algebra
(b) ||v|= V2^+2-+2^ =^/i2=2^/3
I
V
u, =
(2, 2, 2)=
v|| 273
has the same direction as v.
173’ 73’ 73
V
ll2 = V
(2, 2, 2)
2^3
7^’ 75’ 75
has the direction opposite v.
(c)
V =71^+0^+22 + 12+3^ =715
V
U| =
1
(1, 0, 2, 1, 3)
711
1
2
1
3 1
TIs’ ’TIf’Tls’TlI
has the same direction as v.
V
U2 =V
1
(1, 0, 2, 1, 3)
715
3 5
I
7i5’ °’ 715’ Tis ’ 7i5
has the direction opposite v.
3. (a) ||u + v|=||(2,-2,3)+(l,-3,4)||
H|(^-5.7)1
= >/32+(-5)2+72
= 755
(b)
u + V
= [(2,-2, 3)11+11(1,-3, 4)
= ^2^+(-2)2+32 +Vi2+(-3)2+42
= 717+726
(c) ||-2u + 2v||= ll(-4, 4,-6)+(2, -6, 8)|1
=IH-2. 2)11
= V(-2)2+(-2)2+22
= 715
= 275
66
Section 3.2
SSM: Elementary Linear Algebra
(cl) 3u-5v + w||
=||(6, -6, 9)-(5, -15, 20)+(3, 6, -4)
= 1(4, 15,-15)[
= V4‘'+15^+(-15)-= V4^
5. (a) ||3u-5v + w = |(-6, -3, 12, 15)-(15, 5, -25, 35)+(-6, 2, 1, 1)||
= |(-27, -6, 38, -19)||
= V(-27)^ +(-6)^ +38^ +(-19)^
= V2570
(b) ||3u||-5||vi+|| = |(-6,-3, 12, 15)||-5|(3, 1,-5, 7)||+||(-6, 2, 1, 1)||
w
= V(-6)^ +(-3)“ +12^ +15^ -5^3^ +1^ +(-5)2 +7^ + V(-6)^ +2^+1^+1^
= 7414-5784 +742
= 3746-107^+742
(c)
U
V
= -a/(-2)^+(-I)^+4^+5^(3, 1, -5, 7)
= -746(3, 1,-5, 7)
= 746^3^+1^+(-5)2+7^
= 746784
= 27^
7. |/(:v|| =|jt|||(-2, 3, 0, 6)
= A'|V(-2)2+32+0+62
= 749
= 7 /t
7 L' = 5 if k =—, so k = — or k =-—
7
7
7
9. (a) u-v = (3)(2) +(1)(2) +(4)(-4)= -8
u
[2=32+12+42=26
vf =22 +22 +(-4)2 =24
u= u
V V=
(b) u ■ V =(1)(2)+(l)(-2)+(4)(3)+(6)(-2)
=0
|2
u u = u|I
V V=
.I
=l- + l-+4-+6 =54
v|2 = 22 +(-2)2 +32 +(-2)2 = 21
67
SSM: Elementary Linear Algebra
Chapter 3: Euclidean Vector Spaces
11. (a) <i(u, v)=||u-v||
=
+{3-0f +(3-
=^2^+3^+(-1)2
= Vi4
(b) t/(u, v)=||u-v
= V(0-(-3))^ +(-2-2)2 +(-1-4)2 +(1-4)2
= yl3^+{-4f+i-5f+i-3f
= V59
(c) diu, v)= u-v
-\/(3-(^))2 +(-3-1)2 +(-2-(-1))2 +(0-5)2 +(-3-0)2 +(13-(-11))2 +(5-4)2
=
+(-4)2 +(-1)2 +(-5)2 +(-3)2 +242 + 12
= V^
U
V
u
V
13. (a) cos^ =
(3)(l)+(3)(0)+(3)(4)
V32+32+3271^+02+42
15
^/27^/^7
Since cos ^is positive, ^is acute.
U
V
U
V
(b) cos^ =
(0)(-3)+(-2)(2)+(-l)(4)+(l)(4)
Vo2 +(-2)2 +(-1)2 +12^(_3)2 +22+42+42
-4
76^/45
4
76n/45
Since cos 0is negative, ^is obtuse.
u
V
(c) cos d =
(3)(-4)+(-3)(1)+(-2)(-l)+(0)(5)+(-3)(0)+(13)(-l 1)+(5)(4)
732 +(-3)2 +(-2)2 +o2 +(-3)2 +132+527(-4)2 + l2 +(-1)2 +52 +o2 +(-1 1)2 +42
-136
7^71^
136
T^Tl^
Since cos 6* is negative, ^is obtuse.
68
Section 3.2
SSM: Elementary Linear Algebra
15. The angle ^between the vectors is 30°.
a b
u
V
23. (a) cos^ =
v||
u
=||a||||b||cos(9 =9 ■ 5cos 30° = 45^
(2)(5)+(3)(-7)
+?,^
17. (a) u • (v • w)does not make sense because
+i-7f
-11
V ■ w is a scalar.
(b) u • (v + w) makes sense,
II
^/%2
(c) ||u • v|| does not make sense because u • v is
a scalar.
u
V
U
V
(b) cos^ =
(d) (u • v)-||u|| makes sense since both u • v
and
u
(-6)(4)+(-2)(0)
are scalars.
V{-6)^+(-2)^V4^+0^
-24
19. (a)
(-4, -3)
(^,-3)
24
sVio
3
4 _3'
VTo
5’ 5,
U
(b)
(1,7) _ (1,7) _(1,7)
(L7)|| ^,2+72
V
(c) cos6> =
7
bV2’5V2
(I)(3)+(-5)(3)+(4)(3)
sj\^+{-5f+4^yj?,^+3^+3^
(c)
(-3,2,7^)
(-3, 2, 7^)
-3, 2, 73
^(_3)2+2^+(J5f
0
=0
(-3, 2, S]
U
V
(d) cos^ =
Tib
'_3 _1 S
(-2)(l)+(2)(7)+(3)(-4)
4’ 2’ 4
V(-2)^ +2^ +3^
+7^ +(-4)^
0
(d)
(1, 2, 3, 4, 5)
TnT^
id, 2,3, 4,5)1
=0
(1, 2, 3, 4, 5)
25. (a) u-v = (3)(4)+(2)(-l) =10
7i^+2^+3^+4^+5^
(1, 2, 3, 4, 5)
u
7^
1
2
3
4
|v|= 732+2^^42+(-1)2
= 713717
5
= 14.866
7^’7^’7^’7^’7^
21. Divide v by v| to get a unit vector that has the
V
same direction as v, then multiply by m: m-r.—.
V
69
Chapter 3: Euclidean Vector Spaces
SSM: Elementary Linear Algebra
Section 3.3
(b) u ■ V = (-3)(2)+(])(-!)+(0)(3) = -7=7
u
v|| = yj{-3f + \~ +0^ yj2-+(-lf +3^
Exercise Set 3.3
= VioVi4
1. (a) u-v = (6)(2) + (l)(0) +(4)(-3)= 0, sou
and V are orthogonal.
-1 1.832
(c) u-v = (0)(1)+(2)(1)+(2)(1)+(1)(1) =5
(b) u-v = (0)(l) + (0)(l) +(-l)(l)= -l ^0,so
u and V are not orthogonal.
ll«lll|v||
= V02+22+22+i2Vi2 +|2+i2 + i2
(c) u • V =(-6)(3)+ (0)(1)+ (4)(6)= 6 ^ 0, so u
and V are not orthogonal.
= V9V4
=6
(d) u ■ V =(2)(5)+ (4)(3) +(-8)(7)= -34 ^ 0,
so 11 and v are not orthogonal.
27. It is a sphere of radius I centered at
(^b- yo’ ^o)-
3. (a) v,-V2=(2)(3)+(3)(2)= 12^0
The vectors do not form an orthogonal set.
True/False 3.2
(b) V|-V2=(-l)(l)+(l)(l)= 0
(a) True; ||2v||= 2 v .
The vectors form an orthogonal set.
(b) True
(c) V| V2=(-2)(1)+(1)(0)+(1)(2)=0
V| V3=(-2)(-2)+(1)(-5)+(1)(1)= 0
V2-V3=(1)(-2)+(0)(-5)+(2)(1)=0
(c) False; 0=0.
V
The vectors form an orthogonal set.
V
(d) True; the vectors are — and —
V
V
(d) V, ■ V2 =(-3)(1)+(4)(2)+(-1)(5)= 0
u
V| • V3 =(-3)(4)+(4)(-3)+(-l)(0)= -24
V
(e) True; cos—= —
3
2
u
Since V| • V3
V
0, the vectors do not form an
orthogonal set.
(f) False;(u • v)+ w does not make sense because
5. Letx =(A'|,jr2> -^3)-
u ■ V is a scalar, u ■ (v + w)is a meaningful
expression.
u ■ X = (1)(a-,)+(0)(x2)+(1)(a3)= a-| + A3
V • X =(0)(A|)+(1)(A2)+(1)(A3)= A2 + A3
(g) False; for example, if u =(1, 1), v =(-1, I), and
Thus Aj = A2 = -A3. One vector orthogonal to
w = (I, -1), then u • V = u • w = 0.
both u and V is X =(1, 1,-1).
(h) False; let u =(l, l)andv =(-l, I), then
(1. L-1)
X
u ■ V =(1)(-1)+ (1)(1) = 0. 11 • V = 0 indicates
Hi v/l^ + l^+(-l)2
that the angle between u and v is 90°.
(L L-1)
(i) True; if u =(t(|, M2) and v =(V|,V2) then
I
M|, M2 > 0 and V|, V2 < 0 so H|V|, U2V2 < 0,
1
1
hence U\V\ +M2V2 < 0.
1
The possible vectors are ±
(j) True; use the triangle inequality twice.
u
U+V + W < U + V + w
70
v/3’V3’ Sj
Section 3.3
SSM: Elementary Linear Algebra
u a
1. Afi =(-2-l, 0-1, 3-l)=(-3,-1, 2)
19. (a) ||proj,,u
a
a
|2
5C =(-3-(-2), -1-0, 1-3)=(-1, -1, -2)
u a
rii“
C/l =(l-(-3), l-(-l), 1-1)=(4, 2, 0)
ua
ABBC =(-3)(-l)+(-1)(-1)+(2)(-2)=0
all
(l)M)+(-2)(-3)
Since AB ■ BC = 0, AB and BC are orthogonal
and the points form the vertices of a right
triangle.
V(-4)2+(-3)2
2
9. -2(x-(-l))+ l(y-3)+ (-l)(z-(-2))= 0
-2(x+ l) + 0'-3)-(z + 2)= 0
2
5
11. 0(x - 2)+ 0(j’- 0)+ 2(z - 0)= 0
2z = 0
u
a
(b) ||projau||=
13. A normal to 4x - y + 2z = 5 is n| =(4, -1, 2).
(3)(2)+(0)(3)+(4)(3)
A normal to 7x - 3y + 4z = 8 is n2 =(7, -3, 4).
722+3^+3^
Since n| and n2 are not parallel, the planes are
not parallel.
V22
15. 2y = 8x - 4z + 5 => 8x - 2y - 4z = -5
21. u ■ a =(6)(3) +(2)(-9)= 0
A normal to the plane is n, =(8,-2,-4).
The vector component of u along a is
1
1
-z+-y => X — y — z = 0
2
4
4
0
u a
2
prpiaU =
a
a
A normal to the plane is n2 = 1,
\
4’ 1)'
y(3,-9)=(0, 0).
Ta =
The vector component of u orthogonal to a is
u -proJaU =(6, 2)-(0, 0)=(6, 2).
Since n| =8112, the planes are parallel.
23. u ■ a =(3)(1)+ (1)(0) +(-7)(5)= -32
17. A normal to 3x - y + z-4 = 0 is n| =(3, -1, 1)
a ^ =1^+0^+5^ =26
and a normal to x + 2z = -1 is n2 =(1, 0, 2).
n, ■n2=(3)(l) + (-l)(0) + (l)(2) = 5
The vector component of u along a is
u
Since iij and 112 are not orthogonal, the planes
are not perpendicular.
a
a
projau =
2
a
-32
26
(1. 0, 5)
16
—, 0,
13
80^
13
The vector component of u orthogonal to a is
u-projaU = (3, 1,-7)55
11
13’ ’’ 13
71
Chapter 3: Euclidean Vector Spaces
SSM: Elementary Linear Algebra
25. u a =(l)(0)+(l)(2)+(l)(-l)= 1
31. The line is 4x + y -2 = 0.
l|af-o2+22+(-1)2=5
D=
axp+byp+c
+b^
The vector component of u along a is
u a
projau =
4(2)+(l)(-5)-2
|2 ®
1
=-(0,2.-l)
I 5
5)
33. The plane is x + 2^ - 2z-4 = 0.
The vector component of u orthogonal to a is
u-projaU =(l, 1, 1)-
D=
oxq + byQ +czQ+d
yja^+b^+c^
5
5
(l)(3)+ 2(l)-2(-2)-4
3 6
1,-,- •
Vi^+2^+(-2)2
I 5 5)
5
27. u ■ a =(2)(4)+(l)(-4)+(1)(2)+(2)(-2)= 2
||af = 4^ +
5
+2^+i-lf = 40
3
The vector component of u along a is
u a
35. The plane is 2x + 3y - 4z - 1 = 0.
a
projaU =
a
|2
D=
=^(4,-4,
2,-2)
40
^a^+b^+c^
2(-l)+ 3(2)-4(l)-l
1_'
V2^+3^+M)2
~ ,5’ 5’ lO’ 10/
-1
The vector component of u orthogonal to a is
u
-projaU =(2, l, l,2)-
oxq + byQ +czQ+d
1 _]_ J
5’ 10’ 10,
1
^/29
^'9 6
,5’ 5’ 10’ lOy
29. D =
37. A point on 2x -y -z = 5 is Po(0, -5, 0). The
distance between Pq and the plane
axQ+byQ+c
-4x + 2y + 2z= 12 is
yla^ +b^
^(0)+ 2(-5)+ 2(0)-12
4(-3)+ 3(l)+4
D=
^{-4f+2^+2^
4^
-22
-5
V24
22
=1
2V6
11
39. Note that the equation of the second plane is -2
times the equation of the first plane, so the
planes coincide. The distance between the planes
is 0.
72
Section 3.4
SSM: Elementary Linear Algebra
Section 3.4
41. (a) a is the angle between V and i, so
VI
cos or =
Exercise Set 3.4
V
(tj)(l)+(fe)(0)+(c)(0)
V
1. The vector equation is x = Xq +rv.
1
(X, y)=(-4, 1)+ f(0, -8)
Equating components gives the parametric
equations.
X = -4, y = 1 - 8r
a
V
(b) Similar to part (a), cos/? = —
3. Since the given point is the origin, the vector
equation is x = fv.
(x,y, z)= r(-3,0, 1)
and
V
c
cos;'=
Equating components gives the parametric
equations.
X = -3f, y = 0,Z = r
V
a
b
Q
—jj =(cos a, cos p. cos y)
V
5. For t = 0, the point is (3, -6). The coefficients of
t give the parallel vector (-5,-1).
V
7. For t = 0, the point is (4, 6).
(d) Since —r is a unit vector, its magnitude is
X =(4 - 4r,6- 6r) + (-2r, 0)=(4 - 6r,6-6f)
v||
The coefficients of t give the parallel vector
1, so yjcos^ a+co&^ P+cos^ y = 1 or
(-6,-6).
cos^ cr+cos^ y5+cos^ y= 1.
9. The vector equation is x = Xg + Vj +12 V2.
(X, y, z)=(-3, 1, 0)+ r,(0, -3, 6)+f2(-5, 1, 2)
43. v-(A:|Wi +^2^2)= v-(fciWi)+ v-(^2'''2)
= ^l(v-Wi)+/:2(v-W2)
= ki(0)+ k2(0)
Equating components gives the parametric
equations.
x =-3-5/2, y = i
=0
+^2’ 2=6/|+2/2
11. The vector equation is x = Xg+/1V1+/2V2.
(X, y, z)=(-l, 1, 4)+/,(6, -1, 0)+ r2(-l, 3, 1)
True/False 3.3
(a) True; the zero vector is orthogonal to all vectors.
Equating components gives the parametric
equations.
(b) True;(hi) •(mv)= kin(u ■ v)= kin(0)= 0.
x = -1 +6/i-/2, y-l-/i+3/2, z = 4+/2
(c) True; the orthogonal projection of u on a has the
same direction as a while the vector component
of u orthogonal to a is orthogonal
(perpendicular) to a.
13. One vector orthogonal to v is w =(3, 2). Using
w, the vector equation is (x, y)= /(3, 2)and the
parametric equations are x = 3r, y = 2/.
15. Two possible vectors that are orthogonal to v are
(d) True; since projj,u has the same direction as b.
V| =(0, 1, 0) and V2 =(5, 0, 4). Using V] and
V2, the vector equation is
(e) True; if v has the same direction as a, then
projaV = V.
(x, y, z)= /](0, 1, 0)+ (5, 0, 4) and the
parametric equations are x = 5/2, y = /|,
(f) False; for a =(1, 0), u =(1, 10) and v =(1, 7),
z = 4/2.
proJaU = proJaV.
(g) False; u + v|<||u + v .
73
SSM: Elementary Linear Algebra
Chapter 3: Euclidean Vector Spaces
1
1
1
1
0
1
17. The augmented matrix of the system is 2 2 2 0 which has reduced row echelon form 0 0
[3 3 3 oj
I
0
0 0
0 0 0 0
The solution of the system is x^ = -r-s, X2 = r, x^ = s, or x =(-r- 5, r, s) in vector form.
1
Since the row vectors of 2 2 2 are all multiples of(1, 1 , I), only r| needs to be considered.
3
3
3
r| - x =(1, 1, l)'(-r-^, r, s)
= l(-/-5)+!(/•)+IW
=0
1
19. The augmented matrix of the system is
I
0
0
3
ii
7
7
5
1
2
-1
0
-2-1
3
2
0
which has reduced row echelon form
1
f 0
I -f -7 » ■
1
3
19
The solution ot the system isxi= —•/•
3
x= —r
7
19
7
7
2
3
s — t, a'9 = — r + — s + — t, x-i = r, X4 = s, X5 = t, or
7
8
2
1
3
5—t,—r+ — s + — t,r,s,t
7
7
7
7
.
7
7
7
.t
in vector form.
For this system, r, =(1, 5, 1, 2, -1) and r2 =(1, -2,-1, 3, 2).
( 2
1
3 ^
, 3
19 „
ri -x=
'
—r
5 — t +5
(7
7
7
r-\—s-\—t +!(/•)+2(5)-1(/)
\ 1
7
1)
-3 _19 _8 _10 /•+ —^+
5 15 t + r + 2s — t
~ 1’ 1 ^ 1
7
7
7
0
f3
19
8^
1
1
f 2
1
3
7
7
7 y
r2'X = l —r——s-— t -2 -—r+ — s + — t -l(;')+ 3(5)+ 2(r)
1
3
2
^ — t + — r — i’--t-r + 3s + 2t
—/•
7
=0
4
19
7
7
7
7
7
21. (a) A particular solution of jc + y + z = 1 is (1, 0, 0).
The general solution of x + y + z = 0 is x = -5 - f, y = 5, z = f, which is 5(-l, 1, 0)+ r(-l, 0, 1) in vector form.
The equation can be represented by (1, 0,0)+ if-1, 1,0)+ f(-l, 0, 1).
(b) Geometrically, the form of the equation indicates that it is a plane in
, passing through the point Ffl, 0,0)
and parallel to the vectors (-1, 1,0) and (-1,0, 1).
23. (a) If X = (x, y, z) is orthogonal to a and b, then a-x=x + y + z = 0 and b • x = -2x + 3y = 0.
The system is
x+y +z = 0.
=0
-2x + 3y
(b) The reduced row echelon form of
1
1
-2
3
1
0
.
IS
0 0
1 0 I 0
0
1
so the solution will require one
I
3
parameter. Since only one parameter is required, the solution space is a line through the origin in R'.
74
Section 3.5
SSM: Elementary Linear Algebra
(c) From part (b), the solution of the system is
3
-r,
5
X2 = s, xj =t, X4 =0 and a particular solution
2
y = — t, z = t, or
5
of the nonhomogeneous system is x^ =-^,
2
t
3
— t, —r,t in vector form. Here,
5
5
X=
^2 = 0, X3 = 0, x^=\.
True/False 3.4
r, =(1, 1, 1) and r, =(-2, 3, 0).
3
—t +1
5
n1 - x = 1
5
where Xq is the given point and v is the given
2^
3
r, • X = -2
(a) True; the vector equation is then x = Xq +rv
-r +1(0 =0
vector.
t +3 —t +0(0 = 0
5 y
(b) False; two non collinear vectors that are parallel
25. (a) The reduced row echelon form of
3
2
6
4-2 0
-3
-1
-2
"i f 4 0
0
1
to the plane are needed.
is
0
0
0
0 0
0
0
0
(c) True; the equation of the line is x = fv where v is
any nonzero vector on the line.
0
(tl) True; if b = 0, then the statement is true by
Theorem 3.4.3. If b 0, so that there is a
The solution of the system is
2
nonzero entry, say bj of b, and if Xg is a
X| =-—y+ —t, X2=s, X'3=t.
solution vector of the system, then by matrix
multiplication, r,- Xo=6; where r,- isthe/th
(b) Since the second and third rows of
■ 3
6
row vector of A.
2
2
4 -2
4 are scalar multiples of
(e) False; a particular solution of Ax = b must be
-3 -2 1 -2J
used to obtain the general solution of Ax = b
from the general solution of Ax = 0.
the first, it suffices to show that x^ = 1,
a'2 =0, X3 = 1 is a solution of
3x| + 2x2 ~ -^3 = 2.
(f) True; A(X|-X2)= AX|- Ax2 = b -b = 0.
Section 3.5
3(1)+ 2(0)- I = 3- 1 = 2.
Exercise Set 3.5
(c) In vector form, add (1,0, 1) to
2
I
— A + —t, y, t
3
3
Xi
^
1. (a) vxw =(0, 2,-3)x(2, 6, 7)
to get
2 -3 _0 -3 0 2
1
=1-2.+-t,
X2 = y, X3 = 1 +r.
6
3
4
1
2
3
6
8
2
5
7
9
12
3
10
13
1 I i 0
IS
1
4
7’ 2 6
1
(b) ux(vxw)
=(3, 2,-l)x(32,-6,-4)
3
0
0
0
1
1
0
0
0 0
0
2
-1
A general solution of the system is
^1
2
=(14+ 18, -(0+6), 0-4)
=(32, -6,-4)
27. The reduced row echelon form of
3
7’
-6
-4’
3
-1
3
2
32 -4 ’ 32 -6
=(-8-6,-(-12+32),-18-64)'
1
-^2=^> •^3=^ ^4 = '-
=(-14, -20,-82)
From the general solution of the
nonhomogeneous system, a general solution of
4
the homogeneous system is x, = —.
3
t.
3
75
SSM:Elementary Linear Algebra
Chapter 3: Euclidean Vector Spaces
(c) uxv =(3, 2,-l)x(0, 2,-3)
2
-1
3
-1
3
1)
2
0 -3’ 0 2
2 -3’
The parallelogram can be considered in
being determined by u =(3, 2,0)and
=(-6+ 2, -(-9-0), 6-0)
=
9, 6)
v =(3, 1,0).
(uxv)xw
uxv =
9
-4
6
6
-4
3 0
2 0 _
1
=(^,9, 6)x(2, 6, 7)
^ 6 7’
.
as
0’
3
2
3 0’ 3
1
=(0,0,-3)
9
2 7’ 2 6
|| = Vo2+o2+(-3)2 = 3
ux V
=(63-36, -(-28-12), -24-18)
=(27, 40,-42)
The area is 3.
->
13. /lfi =(l, 4) and AC =(-3, 2)
3. u X V is orthogonal to both u and v.
uxv =(-6, 4, 2)x(3,1, 5)
4
1
-6
2
5’
2
-6
4
3
1
3 5’
The triangle can be considered in
as being
half of the parallelogram formed by u =(1,4,0)
and V =(-3, 2, 0).
=(20-2,-(-30-6),-6-12)
=(18, 36,-18)
4
0
1
1
0
4
ux V =
3 .
0’ -3
.
_ 2
v2 0’
=(0,0, 14)
5. u X V is orthogonal to both u and v.
uxv =(-2,1, 5)x(3, 0,-3)
1
5 _-2
0-3’
uxv
5 -2 1
3 -3’
= Vo^+0^+14“ =14
The area of the triangle is -ilux v||= 7.
3 0
=(-3-0,-(6-15),0-3)
=(-3, 9,-3)
15. Let u = P^P2=(-1, -5, 2) and
7. uxv =(l,-1, 2)x(0, 3, 1)
-1
3
2
1’
1
0
2
1
-1
r0
3
y = P^P2={2, 0, 3).
uxv =
0
=(-1-6, -(1-0), 3-0)
=(-7,-1, 3)
0
2
-1
0
-1
-5
3’ 2
0
/
||uxv||= 7(-15)^+7^+10^ =^/w
The area of the triangle is ^||ux v
9. uxv =(2, 3, 0)x(-1, 2,-2)
3
2
2
3’
=(-15, 7, 10)
|uxv V(-7)^+(-1)^+32 =s/^
The area is 4^-
2 -2’
-5 2 _ -1
2
3
-2’ -1
2
2
-6
17. u (vxw)=0
=(-6-0, -(^-0),4+ 3)
=(-6, 4, 7)
2
|= V(-6)^+4^+7^ =Vim
2
2
4 -2
2
-4
= 2^
-2+2-6
2 -4
4
uxv
4yn
2
-2
= 2(-12)+ 2(4)
The area is VlOl.
= -16
The volume of the parallelepiped is -16=16.
11. P^P2 =(3, 2) and P3P4 =(-3, -2) so the side
determined by P\ and P2 is parallel to the side
determined by P3 and
but the direction is
opposite, thus P^P2 and P^P^ are adjacent sides.
76
Section 3.5
SSM: Elementary Linear Algebra
-1
-2
19. u-(vxw)= 3
1
(b)
0-2
5
-4
||;^||= V(-1)2+22+22
=3
Let X be the length of the altitude from
0
vertex C to side AB, then
0 + 2 -1
^3
5
-4
-2
5
-4
2^IUs
= -12+ 2(14)
= 16
2
X=
->
The vectors do not lie in the same plane.
-2
0
6
1
-3
1
-5
-1
1
21. u (vxw)=
3
AB
29. (u + v)x(u-v)
=(u +v)xu-(u+ v)x V
1
=(uxu)+(vxu)-((uxv)+(vxv))
= 0+(vxu)-(ux v)-0
=(vxu)+(vxu)
= 2(vxu)
-3
=-2-' !"^-5
-1
= -2(-2)+6(-16)
= -92
23. u-(vxw)=
a
0 0
0
b
0 = abc
0
0
c
37. (a) a = Pe =(3,-l,-3)
b = PB =(2,-1, 1)
25. Since u • (v x w)= 3, then
det
c= PS =(4,-4, 3)
U,
1(2
M3
3
V|
V2
V3 = 3.
2
1
W|
VV2
VV3
4-4
3
a(bxc)=
M|
M2
_V|
^ 3
W3 =-3
V2
-3
= 3-‘ '+ 1 2 1_ 2 -1
M3
(a) u •(w X v)= det W| vv2
1
4
3
4
= 3(l)+ l(2)-3(-4)
V3_
= 17
(Rows 2 and 3 were interchanged.)
1
The volume is
(b) (v X w) ■ u = u ■ (v X w)= 3
W|
W2
W3
(c) w ■(u X v)= det M|
1(2
M3 = 3
V]
V2
V3
17
-|a-(bxc)|=-6
6
(b) a = PQ =(l,2,-l)
b = PP =(3, 4, 0)
(Rows 2 and 3 were interchanged, then rows
1 and 2 were interchanged.)
—>
c = PS =(-l, -3, 4)
27. (a) Let u = AB =(-1, 2, 2) and
1
2
-1
a-(bxc)= 3
4
0
-1
-3
4
v=:4C =(l, 1,-1).
2
2
1
-1 ’
2
-1
= -l ^ S4' 2
-1 -3
3 4
2
uxv =
1
1
= -(-5)+4(-2)
=(-f, L-3)
ux V
=-3
1
= V(-4)2+1^+(-3)^ =V^
The area of the triangle is
The volume is
ux V
2
77
3
-|a-(bxc)|
=6
6
2
-4
SSM: Elementary Linear Algebra
Chapter 3: Euclidean Vector Spaces
(d) u ■ w =(-2)(2)+(0)(-5)+(4)(-5)= -24
Triie/False 3.5
||w||-=22+(-5)2+(-5)2=54
(a) True; for nonzero vectors ii and v,
|u|
u
vllsin^ will only be 0 if 0= 0, i.e..
w
-w
proj^u =
if u and v are parallel.
Iw|
-24
(2, -5,-5)
(b) True; the cross product of two nonzero and non
collinear vectors will be perpendicular to both
vectors, hence normal to the plane containing the
54
12
-—(2,-5,-5)
27
vectors.
(c) False; the scalar triple product is a scalar, not a
-2
0
3
-1
6
2
-5
-5
(e) u (vxw)=
vector.
(d) True
(e)
(f)
4
6
3
-1
= -2-5 -5 ■^'^2 -5
False; for example
(ixi)xj = 0xj = 0
ix(ixj) = ixk = -j
= -2(35) + 4(-13)
= -122
(f)
False; for example, if v and w are distinct
vectors that are both parallel to u, then
-5v + w = (-15, 5,-30) + (2,-5,-5)
= (-13, 0,-35)
uxv = uxw = 0, but V ^ vv.
(u • v)w
= ((-2)(3) + (0)(-l) + (4)(6))(2, -5,-5)
Chapter 3 Supplementary Exercises
1. (a)
= 18(2,-5,-5)
= (36,-90,-90)
3V - 2u = (9, - 3, 18) - (-4, 0, 8)
= (13,-3, 10)
(-5v + w)x(u-v)w
J 0 --^5
-90
(b) ||u + v + wi
-90’
-13
36
-35
-90’
-13
0
36
-90
= (-3150,-2430, 1170)
= ||(-2 + 3 + 2, 0-1-5, 4 + 6-5)||
= 1(3. -6, 5)1
3. (a)
= ^j3-+{-6f+5-
3V - 2u = (-9, 0, 24, 0) - (^, 12, 4, 2)
= (-5,-12, 20, -2)
= yllQ
(b) ||u+v + w||
(c) The distance between -3u and v + 5w is
||-3u - (V + 5 w)|| = ||-3u - V - 5 w||.
= |l(-2-3 + 9, 6 + 0 + 1, 2 + 8-6, l + 0-6)||
= 1(4, 7, 4, -5)||
ll-3u-v-5w||
= 742+72+42+(-5)2
= ||(6, 0,-12)-(3,-1, 6)-(IO, 25,-25)1
=vr^
= ||(-7, 26, 7)|
= ^j(-7)-+26^+^= y[7T4
78
SSM: Elementary Linear Algebra
Chapter 3 Supplementary Exercises
(c) The distance between -3u and v + 5w is ||-3u -(v +5w)||=||-3u- v -5w||•
||-3u-v-5w||=||(6, -18,-6, -3)-(-3, 0, 8, 0)-(45, 5, -30, -30)||
= |(-36, -23,16, 27)1
= V(-36)^ +(-23)^ +16“+ 27^
= V2810
(d) u ■ w =(-2)(9)+(6)(1)+(2)(-6)+(l)(-6)
= -30
||wl|“ =9^+1^+(-6)^+(-6)“ =154
u
proj^u
w
w
|2
|W|
-30
(9, 1, -6,-6)
154
15
(9, 1, -6,-6)
77
5. u =(-32,-1, 19), v =(3,-l,5), w =(l,6, 2)
u-v =(-32)(3)+(-l)(-l)+(19)(5)= 0
u • w =(-32)(1)+(-1)(6)+ (19)(2) = 0
v-w =(3)(l) +(-l)(6)+(5)(2) = 7
Since v • w 0, the vectors do not form an orthogonal set.
7. (a) The set of all such vectors is the line through the origin which is perpendicular to the given vector,
(b) The set of all such vectors is the plane through the origin which is perpendicular to the given vector,
(c) The only vector in
that can be orthogonal to two non collinear vectors is 0; the set is (0), the origin.
(d) The set of all such vectors is the line through the origin which is perpendicular to the plane containing the
two given vectors.
9. True; ||u + vf =(u + v)•(u + v)
= U U +U
|u
2
Thus, if ||u + v||
v+v U+V
V
Ip + U'V+V'U +||v|P
Ip +||v|p. u ■ V = V • u = 0, so u and v are orthogonal.
u
->
11. Let S be S(-l, 52, 53). Then u = PQ =(3, 1, -2) and \= RS =(-6, 52 -1, 53 -1). u and v are parallel if
u X v = 0.
1
-2 _ 3 -2 3
1
■^2 “ 1 L"! “' '
-^3 “ 1 ’ “6 52-1
= (53 -1 + 2(52 -1). - (3(53 -1) -12), 3(52 -1) + 6)
iix V =
= (252 + L3 “ 3, -353 + 15, 352 + 3)
If u X v = 0, then from the second and third components, 52 = -1 and 53 = 5, which also causes the first
component to be 0. The point is 5'(-l, -1, 5).
79
SSM:Elementary Linear Algebra
Chapter 3: Euclidean Vector Spaces
21. One point on the line is (0,-5). Since the slope
13. u = Pe =(3, 1,-2), v = P/? =(2, 2,-3)
of the line is m =
u
V
cosd =
j, the vector(1, 3)is parallel
to the line. Using this point and vector gives the
vector equation (x, y)-(0,-5)+ r( 1, 3).
Equating components gives the parametric
equations x-t,y = -5 + 3t.
|u| v|
(3)(2)+(l)(2)+(-2)(-3)
73^+1^ +{-2f ^|2^+2^ +i-3f
14
23. From the vector equation,(-1, 5,6)is a point on
the plane and (0,-1, 3) x (2,-1,0) will be a
normal to the plane.
VhViv
14
17
(0,-1, 3)x(2, -1, 0)
-1 3 _o 3 0 -r
15. The plane is 5x - 3)' + z -i- 4 = 0.
-1 0’ 2 0’ 2 -1^
|5(-3)-3(l)+3+4|_ 11
D=
=(3, 6, 2)
^5
A point-normal equation for the plane is
3(x-h l)+ 6(y-5)+ 2(z-6)=:0.
17. The plane will contain u = PQ =(1, -2,-2)
25. Two vectors in the plane are
and v = P/? =(5,-1,-5).
u = PQ = {-\{), 4,-1) and
Using P(-2, 1, 3), the vector equation is
{x, y, z)
v = P^ =(-9, 6,-6).
=(-2, 1, 3)+ r,(l, -2,-2)+ r2(5, -1, -5).
A normal to the plane is u x v.
Equating components gives the parametric
UX V =
(4 -\ _-\0 -I -10 4'
6 -6’
equations x =-2-I- -I-5^2, >' = l-2ri-r2,
-9 -6’ -9 6 /
=(-18, -51, -24)
z = 3 —2fj —5t2-
Using point P{9,0, 4), a point-normal equation
for the plane is -18(x- 9)- Sly - 24(z-4)= 0.
19. The vector equation is (x, y)=(0,-3)+ t(8, -1).
Equating components gives the parametric
equations x = 8f, y = -3- r.
29. The equation represents a plane perpendicular to
the xy-plane which intersects the xy-plane along
the line Ax + By = 0.
80
Chapter 4
General Vector Spaces
Section 4.1
Exercise Set 4.1
1. (a) u+v =(-l, 2)+(3, 4)
=(-1 + 3, 2+4)
=(2, 6)
3u = 3(-l,2)=(0,6)
(b) The sum of any two real numbers is a real number. The product of two real numbers is a real number and 0 is
a real number.
3.
5.
(c)
Axioms 1-5 hold in V because they also hold in R?'.
(e)
Let u =(«[, «2) with Uj
0. Then lu = 1(M|, «2)=(0- M2)
The set is a vector space with the given operations.
The set is not a vector space. Axiom 5 fails to hold because of the restriction that x>0. Axiom 6 fails to hold for
k<0.
2
2
2
7.
The set is not a vector space. Axiom 8 fails to hold because {k + m) ^ k +m .
9.
The set is a vector space with the given operations.
11.
The set is a vector space with the given operations.
23.
u +w = v + w
(u + w)+(-w)=(V + w)+(-w)
U +[W +(-w)]= V +[w +(-w)]
25.
Hypothesis
Add -w to both sides.
Axiom 3
u +0= v +0
Axiom 5
U= V
Axiom 4
(1) Axiom 7
(2) Axiom 4
(3) Axiom 5
(4) Axiom 1
(5) Axiom 3
(6) Axiom 5
(7) Axiom 4
True/False 4.1
(a) False; vectors are not restricted to being directed line segments.
(b) False; vectors are not restricted to being «-tuples of real numbers.
(c) True
81
Chapter 4: General Vector Spaces
SSM: Elementary Linear Algebra
(d) False; if a vector space V had exactly two
elements, one of them would necessarily be the
(d) This is a subspace of P^.
zero vector 0. Call the other vector u. Then
5. (a) This is a subspace of 7?”.
0 + 0 = 0, 0 + u = u, and u + 0 = u. Since -u
must exist and -u ^ 0, then -u = u and
(b) This is not a subspace of 7?°°, since for
1
u + u = 0. Consider the scalar —. Since — u
2
1, L'v is not in the set.
2
1
(c) This is a subspace of Z?”.
would be an element of V, — u = 0 or — u = u. If
2
2
(d) This is a subspace of Z?”.
— u = 0, then
2
1
n
1
1
7. Consider Z:|U +Z:2V =(a, b, c). Thin
u = lu= —H— ll=—UH—u =0+0 = 0.
2 2j
2
2
(OL, + k2,-2ki+3k2, 2k,-k2)=(a, b,c).
Equating components gives the system
If — u = u, then
0
ki = a
-2k, + 31:2 =b or Ax = b where A= -2
2
2k,-L'2 =c
2
1
n
1
I
u = lu = —+ — u =—u+ —u = u + u = 0.
2
2;
2
2
Either way we get u = 0 which contradicts our
assumption that we had exactly two elements.
3
a
'
k 1
X = ' , and h= b
c
(e) False; the zero vector would be 0 = 0 + Ox which
The system can be solved simultaneously by
does not have degree exactly 1.
2
0
reducing the matrix -2
Section 4.2
2
2
Exercise Set 4.2
which reduces to
1. (a) This is a subspace of Z?^.
3 0 0
2
3
4
0
5
0
5
1
0
2
4
0
1
2
3 0 0 . The
0 0
0
0
0
0
0
results can be read from the reduced matrix.
3
(b) This is not a subspace of R', since
(a|, 1, l)+(a2. L 1) =(c, +^2. 2, 2) which
(a) (2, 2, 2)= 2u + 2v is a linear combination of
u and V.
is not in the set.
(c) This is a subspace of Z?"^.
(b) (3, 1, 5)= 4u + 3v is a linear combination of
(d) This is not a subspace of R^, since for k ^ 1
(c) (0, 4, 5) is not a linear combination of u and
u and V.
k(a + c + \)it ka + kc + \, so k{a, b, c) is not
V.
in the set.
(d) (0, 0,0)= On + Ov is a linear combination of
u and V.
(e) This is a subspace of R^.
9. Similar to the process in Exercise 7, determining
3. (a) This is a subspace of Pj.
whether the given matrices are linear
combinations of A, B, and C can be
(b) This is a subspace of Z^.
accomplished by reducing the matrix
■ 4
1 0 6 0 6 -f
(c) This is not a subspace of Z3 since
2
0
0
5
k{aQ +a|A'+ a2X“ +a2X^) is not in the set
-2
0-1
2
1
0
3
7
for all noninteger values of k.
-2
3
4
0
8
82
Section 4.2
SSM: Elementary Linear Algebra
1
0
0
0
1
0
2
0
0
0
0
1
-3
0
1
0
0
0 0
0
0
0
1
I
0
2 0
The matrix reduces to
6
= \A + 2B-3C is a linear combination of A, B, and C.
(a)
0
(b)
(c)
6
0
3
8
-1
(cl)
0
= OA + Ofl + OC is a linear combination of A, B, and C.
0 0
= 1A + 2S + IC is a linear combination of A, B, and C.
5
is not a linear combination of A, B, and C.
7
3
11. Let b =(bf, b2,
be an arbitrary vector in R'.
2k1
(a) If L'|V| +L'2V2 +^3V3 =b, then (2k^, 2k\ +L'3, 2k^ +3^2 +%)=((’!> ('2’
2
0
= b1
+ /C3 — ^>2 •
o*" ^^1
21:| + 3/c2 +^3 = ^3
0
Since det 2 0 1 =-6 0, the system is consistent for all values of/?|, ii2 > and ^3 so the given vectors
2
span
3
I
.
(b) If l'|V|+/:2V2+^A''3 ='t)’ then (2L'i+4^:2+8L'3,-^1+^2 “^3’^^1 + ^^'2+8^3)=(I?l, ^>2- (^3) or
2k\ +4/^2 +8^3 = /?|
-k\ +L'2 -k2=b2.
3/t| + 2k2 +8^3 = b^
'2 4
Since det -1
1 = 0, the system is not consistent for all values of , /?2> and b^ so the given vectors
_ 3 2
do not span R^.
(c) If /:|V| +L'2V2 +^'3V3 +^4V4 =b, then
i3k^ + 2k2 + 5/C3 + k^, ky — 3^2 ~2^:3 +4k^, 4kj +5^2 "*■ ^^3 ~k^) = (i>|, i>2> (^3) or
3^1 + 2^2 + 5/^3 +k^ = ii|
/tj — 3/^2 — 2^3 + 4^4 = b^ .
4k^ + 5/^2 “T ^k^ ~ k^ = /?3
3
Reducing the matrix
2
1 -3
4
5
5
-2
9
The system has a solution only if
1 b,
4 172
-1 ^73
Isads to
1
-3
0
1
0 0
-2
4
62
-1
0
0
-IIi,,+^Z,2+Z73
+^(>2 +^3 = 0 or 17i>[ =7^72+11^73. This restriction means that the
span of V|, V2, V3, and V4 is not all of R^, i.e., the given vectors do not span R^.
83
Chapter 4: General Vector Spaces
SSM: Elementary Linear Algebra
(d) If LjV,+^2^2+^3V3+^4V4 =b, then
(^1 + 3^2 4^3 3^4 >
+4^2 +3^3 + 3^:4, 6L[ +^2 +^3 +^4)~(^1 > ^2’ ^3)
+3^2 + 4^3 +3L4 = b^
2L|+4^2 + 3^3 +3L4 = ^>2 •
6Lj +^2 +^3 +^4 = ^3
1
1 3 4 3 /?,I
The matrix 2 4 3 3 62 reduces to
6 1 1 1 ^^3
I
0 0
39
0
1
0
0 0
1
16
39
17
39
Thus, for any values of b^,b2, and b^,, values of k\, k2, k^, and k^ can be found. The given vectors span
R^.
13. Let p = Gq+C'i'r+6i2'r^ be an arbitrary polynomial in P2. If A:iP|+L2P2+%P3+^4P4 =p, then
(^1 +31:2 +5^:3 -21:4)+(-1:] +^2 -^3 -21:4)x+(21:| + 41:3 + 21:4
= Gq +GiJC+ G2X^ or
k^ + 3^2 +51:3 -21:4 = Gq
-ky +k2 -1:3-21:4 = G| .
2k1
+41:3 + 21:4 = 02
1 3 5 -2 Go
Reducing the matrix -1 1 -1 -2 a, leads to
2 0
4
The system has a solution only if -
Pi’ P2’ P3’
2 G2_
1
3
5
-2
0
1
1
-1
0 0 0
0
Go
1
1
-iGo+|G,+G2_
Gq +-^ Gj + G2 =0 or gq = 3g|+2g2. This restriction means that the span of
P4 is not all of P^, i.e., they do not span P3.
-1
1
3
-1
15. (a) Since det(A)= 0, reduce the matrix
1
0 0 . The matrix reduces to
2 -4-5 0
1
set is X =-— t,
1
0
0
1
0
i 0
. The solution
f 0
0 0
0 0
3
y = -—t, z = t which is a line through the origin.
1
-2
(b) Since det(A)= 0, reduce the matrix -3
6
-2
4
3 0
9 0 . The matrix reduces to
-6 0
1
-2
0
0
0 0
0
0 0 0
1
0 . The solution
set is X = 2r, y = r, z = 0, which is a line through the origin.
(c) Since det(A)= -1
0, the system has only the trivial solution, i.e., the origin,
(d) Since det(y4) = 8 0, the system has only the trivial solution, i.e., the origin.
1
-1
1
(e) Since det(/4) = 0, reduce the matrix 2 -1
4
3
1
11
0
1
0
3
0
0 . The matrix reduces to
0
1
2
0 . The solution set
0
0 0
is X = -3r, y = -2t,z~t, which is a line through the origin.
84
0 0
Section 4.3
SSM: Elementary Linear Algebra
1
-3
1
0
1
(f) Since det(/l) = 0, reduce the matrix 2 -6 2 0 . The matrix reduces to 0
3
-9
3 0
0
-3
1
0
0 0 0 . The solution set
0 0 0
is X - 3y + z = 0, which is a plane through the origin.
17. Let f =/(x) and g = g(x) be elements of the set. Then
f +g= f(f(x)+ g(x))dx
= ff{x)dx+ f g{x)dx
=0
and f + g is in the set. Let k be any scalar. Then kt = kf{x)dx = k ^f{x)dx = A: ■ 0=0 so AT is in the set. Thus
the set is a subspace of C[a, b],
True/False 4.2
(a) True; this is part of the definition of a subspace,
(b) True; since a vector space is a subset of itself,
(c) False; for instance, the set W in Example 4 contains the origin (zero vector), but is not a subspace of
(d) False;
(e) False; if b
.
is not a subset of R^.
0, i.e., the system is not homogeneous, then the solution set does not contain the zero vector,
(f) True; this is by the definition of the span of a set of vectors,
(g) True; by Theorem 4.2.2.
(h) False; consider the subspaces W\ ={(x, y)|y = x} and VV'2 ={(x, >')|y = -x} in R^. v, =(1, 1) is in IT, and
V2 =(1, -1) is in W2, but V| + V2 =(2, 0) is not in W\ UVF2 ={(x, y)|y = ±x}.
(i) False; let v, =(1, 0), V2 -(0, 1), u, =(-l, 0), and U2 =(0, -1). Then (v,, V2} and {u,, U2) both span R^ but
{U], U2}^t{v,, V2).
(j) True; the sum of two upper triangular matrices is upper triangular and the scalar multiple of an upper triangular
matrix is upper triangular,
(k) False; the span of X-1,(x-1)^, and (x-1)'^ will only contain polynomials for whichp(l)= 0, i.e., not all of
Section 4.3
Exercise Set 4.3
1. (a) U2=-5u|, i.e., U2 is a scalar multiple of U].
(b) Since there are 3 vectors and 3 > 2, the set is linearly dependent by Theorem 4.3.3.
(c) P2=2p|, i.e., P2 is a scalar multiple of P[.
85
Chapter 4: General Vector Spaces
(d) B = -A, i.e.,
SSM: Elementary Linear Algebra
is a scalar multiple of A.
3. (a) The equation L'i(3, 8, 7, -3)+ A:2(L 5, 3, -l)+ ^3(2, -1, 2, 6)+ L'4(l, 4, 0, 3)=(0, 0, 0, 0) generates the
homogeneous system.
3^1 + + 2/^3 + =0
8/^1 + 5/^0 — ^3 +4^4 =0
=0
7^1+ 3/^2 "^2^3
—3A^I — /c2
3^4 “0
3
1
2
1
8
5
-1
4
7
3
2
0
-3
-1
6
3
Since det
= 128
0, the system has only the trivial solution and the vectors are linearly
independent.
(b) The equation A:|(0, 0, 2, 2)+ k2(?>, 3, 0, 0)+ ^3(l, 1, 0, -1)=(0, 0, 0, 0) generates the homogeneous system
3k^ + ^3 =0
3k2+kj =0
2k I
=0
2k I
-L'3 =0
The third equation gives k^ =0, which gives k^^^O in the fourth equation. ^3=0 gives k2=0 in the first
two equations. Since the system has only the trivial solution, the vectors are linearly independent.
(c) The equation A:|(0, 3, -3, -6)+ ^2(“2, 0, 0, -6)+ L'3(0, -4,-2,-2)+ A:4(0, -8, 4, -4)=(0, 0, 0, 0)
generates the homogeneous system
=0
-2k-,
3k I
-4/t3-8it4 =0
-2/t3+4/t4 =0'
-6/t| -6/t2-2A:3-4/t4 =0
-3k I
“0-2
■
0
Since det -3^
0
-4
0
-2
-6
-6
-2
0
^ = 480 ^ 0, the system has only the trivial solution and the vectors are linearly
-4
independent.
(d) The equation L'i(3, 0, -3, 6)+/:2(0. 2, 3, l)+ ^3(0, -2,-2, 0)+ ^4(-2, 1, 2, 1) =(0, 0, 0, 0) generates the
homogeneous system
3k 1
-2L'4=0
2/^2 — 2^3 + k^ =0
-3^'|+3/t2-2yt3+2L'4=0'
6L'| + k2
+ L'4=0
3
0
0
0
2
-2
-3
3
-2
6
1
0
Since det
-2
2 = 36 ^ 0, the system has only the trivial solution and the vectors are linearly
independent.
86
Section 4.3
SSM: Elementary Linear Algebra
5. If the vectors lie in a plane, then they are linearly
dependent.
3
2
2^1 + 6A:2 +
=0
= 0.
-2k^ +k2
4yt2-4lt3 =0
2 ‘ 2 ^
7
9. The equation
2
+^2''2 +^3''3
generates
the homogeneous system
U,--L'2 --^'3 -0
2
Since det -2
3
1: ''3 =-3''1 + 3''2
generates the homogeneous system
6
^
,
(a) Theequation ^,Vi+^'2^2+^3V3 =0
2
7
I 22 2
0 = -121^0, the
— k, +Xk2 —kn = 0.
0 4 -4_
2'
vectors are linearly independent, so they do
not lie in a plane.
2 2 ^
--k,--k2 + U2=0
2
(b) The equation k^v^ +L'2''2 +^3^3 = 0
■
2
X
1
1
X
i2 =1(4^3
_3^_])
4
I
>
X
2
2
2
det
generates the homogeneous system
2
—6^1 + 3^:3 +4^:3 =0
7/i| + 2/^2 — ^3 ~ 0.
2/t|+4/^2+21^3=0
_l
2
,
|(^-1)(2^+ 1)2
4
'-6 3
4
1
Since det
7
2
-1 = 0, the vectors are
For X = 1 and X =-—, the determinant is zero
2
.2 4
2
and the vectors form a linearly dependent set.
linearly dependent, so they do lie in a plane.
11. Let {v„, v^, ..., v,j) be a (nonempty) subset of
7. (a) The equation I'lV]+1:2V2+1:3V3 = 0
5. If this set were linearly dependent, then there
generates the homogeneous system
would be a nonzero solution (k^, k/^,
61:2+41:3 =0
-71'3 =0
3k I
expanded to a nonzero solution of
k\ + 5k2 +1:3 = 0 ’
—k^ +1:2 + ^^3 ~ ®
L'|V| +^2''2
6
4
0
3
0-7
0
1
5
1
0
-1
1
3
0
1
0
0
0
1
s »
0
0
0
0
0
0
0
0
^k^\^ =0 by taking all other
coefficients as 0. This contradicts the linear
0
The matrix
l'„) to
=0. This can be
k^\^ +kjj\lj H
independence of S, so the subset must be linearly
independent.
reduces to
13. If S is linearly dependent, then there is a nonzero
solution {k\. k2, ■■■, k^) to
+A:2V2+--- + ^rV^ = 0. Thus
(l'l, 1:2, ...,
. Since the system has a
0, 0, ..., 0)
^ nonzero solution
to
l'lVi+l:2V2+- - - + l:,v,+l:,+iV,+i+--- + l:„v
nontrivial solution, the vectors are linearly
dependent.
n
= 0
so the set {vj, V2, ..., v^,
..., v,,) is
linearly dependent.
(b) The solution of the system is k^ =
3
15. If {V|, V2, V3} were linearly dependent, there
would be a nonzero solution to
^2 =“^0
h
3
2
3
t = -:
7
Vi =-v,
I
7 -
^l''l +^'2''2+^3''3
^3
(since Vj
and V2 are linearly independent). Solving for
3
7
V3
87
Chapter 4: General Vector Spaces
SSM: Elementary Linear Algebra
determinant was simplified by adding the first
V3 gives V3 =- — V]
V2 which contradicts
^3
row to the third row.
%
that V3 is not in span {vj, V2}. Thus,
True/False 4.3
{V|, V2, V3) is linearly independent.
(a) False; the set {0} is linearly dependent.
19. (a) If Vj, V2, and V3 are placed with their
(b) True
initial points at the origin, they will not lie in
the same plane, so they are linearly
independent.
(c) False; the set {vj, V2} in
where v, =(0, 1)
and V2 =(0, 2) is linearly dependent.
(b) If V|, V2, and V3 are placed with their
(d) True; suppose {Arvj, k\2, k\^} were linearly
initial points at the origin, they will lie in the
same plane, so they are linearly dependent.
dependent, then there would be a nontrivial
Alternatively, note that V|+ V2 = V3 by
solution to ^i(A:v,)+^2(^2)‘''^3(^3)= ®
placing V| along the z-axis with its terminal
which would give a nontrivial solution to
point at the origin.
^l''l +^2''2 +^3''3 ~ b-
21. The Wronskian is
(e)
= -xsinjc-cosx.
1
-smx:
IT(0)= -cos 0 = -1, so the functions are linearly
independent.
1
X
23. (a) IT(x)=o
1
True; consider the sets {vj, V2), {V|, V2, V3),
..., {v,, V2, ..., v„}. If {v,, V2) is linearly
dependent, then yj = L'Vj for some k and the
condition is met. If V2 ^ kv^ then there is some
2 < A: < n for which the set {V|, V2, ..., v^_|} is
linearly independent but {V|, V2,..., Vj(.} is
linearly dependent. This means that there is a
nontrivial solution of
0 0 e-'
Since W(x)=
a,V|+a2V2+---+ at_iV^_i+a^v^ =0 with
^0 and the equation can be solved to give
is nonzero for all x, the
vectors are linearly independent.
as a linear combination of V|,
V2,
1 X x^
(b) IT(x)=o
1
2x
0
0
2
=2
(f) False; the set is
Since W{x)= 2 is nonzero for all x, the
1
1
O
O’ l
1
1
0
O ’ O
0
0
l ’ l
1
0
O ’ O
1
l ’
vectors are linearly independent.
0 0
and
sinx
25. iy(x)= cosx
cosx
xcosx
-sinx
cosx-xsmx
1
1
sinx
cosx
xcosx
cosx
—sinx
cosx-xsmx
0
0
-2sinx
1
1
|-sinx -cosx -xcosx-2sinx
0
+1
0 0_
1
0
+(-l)
1
0
1
0
+(-l)
0
1
0
1
'O 0
0 0
so the set is linearly dependent.
sinx
cosx
cosx
-sinx
= -2sinx
(g) True; the first polynomial has a nonzero constant
term, so it cannot be expressed as a linear
= -2sin x(-sin^ x-cos^ x)
combination of the others, and since the two
= 2sinx
others are not scalar multiples of one another, the
three polynomials are linearly independent.
Since 2sin x is not identically zero, the vectors
are linearly independent, hence span a threedimensional subspace of F{-o°, 00). Note that the
88
Section 4.4
SSM: Elementary Linear Algebra
(h) False; the functions /| and /2 are linearly dependent if there are scalars
and ^2 such that
k\f\ix)+ k2f2(x)=0 for all real numbers x.
Section 4.4
Exercise Set 4.4
1. (a) The set 5 = {uj, U2, U3} is not linearly independent; U3 = 2U| +U2.
(b) The set 5' = {U|,U2} spans a plane in R^, not all of R^.
(c) The set 5 ={pi,p2) does not span P2; x^ is not a linear combination of pj and P2.
(d) The set S = {A, B, C, D,E] is not linearly independent; E can be written as linear combination of A, B, C, and
D.
3. (a) As in Example 3, it is sufficient to consider det
1
2
3
0
2
3 = 6^^0.
0
0
3
Since this determinant is nonzero, the set of vectors is a basis for R^.
3
2
1
5
4 = 26^0. Since this determinant is nonzero, the set of vectors is a
-4
6
8
(b) It is sufficient to consider det
1
basis for R^.
2
4
(c) It is sufficient to consider det -3
1
1
1
0
-7 = 0. Since this determinant is zero, the set of vectors is not
linearly independent.
2
(d) It is sufficient to consider det 6
4
4
-1
-1
2 = 0. Since this determinant is zero, the set of vectors is not
5
linearly independent.
5. The equations q ^ 6
3
Cl
6
0
3 -6
+ C2 -1
3c1
^
0
0
+ C3 -12
1
0
0
0
+ C4 -1
2
0
0
-8
1
0
a
b
+ C3 -12 -4 + C4 -1
2
c
d
0
0
-1
3cI
+ C4 =0
=0
6c| -C2 -8C3
.3ci-C2-12c3 -C4=0
-6c1
-4c3 + 2C4 =0
and
-8
+ C4 =a
=b
6c[ -C2 -8C3
3Ci-C2-12c3 -C4=C
-6cI
-4C3 + 2C4 = d
89
and
generate the systems
Chapter 4: General Vector Spaces
SSM: Elementary Linear Algebra
which have the same coefficient matrix. Since
“ 3
det
0
o
-8
0
3
-1
-12
-1
-6
0
^
2
6
q -Acj +7c3 = 5
2cj +5c2 -8C3 = -12. The matrix
3c| + 6c2 +9c3 =3
f
= 48
0 the matrices
“1 -4
are linearly independent and also span M22, so
they are a basis.
7
5“
2
5
■8
-12
3
6
9
3
reduces to
“^ 0 0-2“
0
7. (a) Since S' is the standard basis for
(w)5 =(3,-7).
0
0 0
1
0 .
1
The solution of the system is q = -2,
C2 = 0, C3 = 1 and (v)^ =(-2, 0, 1).
(b) Solve C|U| +C2U2 = w, or in terms of
components, (2q +3c2, -4C| +802) = (1, 1).
Equating components gives
The matrix
1
2
3
1
-4
8
1
11. Solve CjAi+C2A2+C3A3+C4/\4 = A. By
2C| + 3c2 =1
inspection, C3 =-l and C4 =3. Thus it remains
-4c| + 80, = 1 '
to solve the system
reduces to
o which can be
q+C2=0
solved by adding the equations to get C2 = 1,
I
0
0 ^
1
282 . Thus, (w)5 = ( A
28
14
3
from which q =-l. Thus, (A)^ =(-l, 1, -1, 3).
14j
13. Solving C|A|+C2A2+C3A3+C4A4 =
(c) Solve C|U| +C2U2 = w, or in terms of
components, (q, C| + 2c2) = (a, b). It is
C| A| + C2 A2 + C3A3 + C4A4 —
clear that q = a and solving a + lc-) =b
(w) 5 = a.
b
c
d
0
0
0 ’
, and
qA| +C2A2 +C3A3 +C4A4 = A gives the systems
b-a
tor C2 gives C2 =
a
0
. Thus,
q +C2
b-a']
=0 q
= 0 C| + C2
= a
"Land
= c
= 0 ’ C| + C2 + C3
C| + C2 + C3 + C4 = 0 C| + C2 + C3 + C4 = c/
q+C2+C3
2
9. (a) Solve C|V| +C2V2 +C3V3 = v, or in terms of
C|
components,
q+C2
=0
(C| + 2c2 + 3c3 , 2c2 + 3c3 , 3c3 ) = (2, -1, 3).
q +C2+C3
= 1 ■
Equating components gives
C| + 2c2 + 3c3 = 2
Cl + Cj + C3 + C4 = 0
2c2 +3C3 = -1 which can easily be
3c3=3
It is easy to see that det
solved by back-substitution to get q = 3,
1
0
0
0
1
1
0
0
1
1
1
0
1
1
1
=1
0, so
{A| , A2, A3, A4} spans M22- The third system
is easy to solve by inspection to get Cj = 1,
C2 =-2, q3 = 1. Thus, (v)^ =(3, -2, 1).
(b) Solve C|V| +C2V2 +C3V3 = V, or in terms of
C2 =-l, C3 =1, C4 =-l, thus,
components,
A = A| - A2 -1- A3 - A4.
(q -4c2 -l-7c3,2q -l-5c2 -8c3,3C| +6C2 +9C3)
= (5,-12, 3).
15. Solving the systems C|P| -1- C2P2 + C3P3 = 0,
Equating components gives
2
CiPi+C2P2+C3P3 = cn-i>jc4-c.r , and
CiPi -1-C2P2 +C3P3 =p gives the systems
90
Section 4.5
SSM: Elementary Linear Algebra
True/False 4.4
=a
=0 C|
= b , and
C| +C2
= 0, C| + C2
C] + C2 + C3 =0 C| + C2 + C3 = c
C|
(a) False; {vi,...,v„} must also be linearly
independent to be a basis.
=7
q +C2
(b) False; the span of the set must also be V for the
set to be a basis.
C| + C2 + C3 =2
1
0 0
It is easy to see that det 1
1
1
1
0 =1
(c) True; a basis must span the vector space.
0, so
(d) True; the standard basis is used for the
{Pl>P2’P3! spans /2- The third system is easy
coordinate vectors in /?”.
to solve by inspection to get C| =7, C2 = -8,
(e) False; the set {1 + x+x^
C3=3, thus, p = 7p|-8p2+3p3.
2
3
4
2
+ x‘^,
3
4
3
4
4, .
X + X +X +X , X +X'+X, X +X , X j lS
17. From the diagram, j= U2 and
a basis for P^.
V^.
U| =(cos30°)i +(sin30°)j=
2 '’^2'’’
Section 4.5
1 ,
1 + —u,
tor 1 m terms or u 1
Solving u
Exercise Set 4.5
2
1
2
and U2 gives
1. The matrix
-1
-2
0
(a) (>/3, 1) is V3i +j.
1
0
^/3i+j= V3f^
U|-
^3
0-1
1
_0 0
U2 +U2
0
2 0
1
reduces to
0
0
0 0 . The solution of the system is
0
0
Xi =t, X2= 0, X3 = t, which can be written as
= 2U|-U2 +U2
(X[, X2, X3)=(r, 0,0 or (x,, X2, X3)= r(l, 0, 1).
= 2u 1
Thus, the solution space has dimension 1 and a
The xy'-coordinates are (2, 0).
basis is (1,0, 1).
(b) (1,0) is i which has xy'-coordinates
2
3/3’
1
-4
3
-1
0
3. The matrix ^;
-8
6
-2
0
reduces to
'1 -4 3 -1 0
0
(c) Since (0, l)= j= ii2, the x'y'-coordinates
. The solution of the syste m
which can be written as
(X|, x,, X3, X4)=(4/--35--i-r, r, 5, t) or
(d) (a, b) is ai + 6j.
(2
1
U|-
(X|, X2, X3, X4)
U2 +b\X2
2
f,
= -pau|+ b-~r
= r(4, 1, 0, 0)+ s(-3, 0, 1, 0)+ f(l, 0, 0, 1).
Thus, the solution space has dimension 3 and a
a
v3
0 0
is X] =4r-3j'+ r, X2 = r, X3 = 5, X4=r,
are(0, 1).
a\ + bi = a
0 0
I,
“2
(2
The x'y'-coordinates are —p
S’
basis is (4, 1,0, 0),(-3, 0, 1, 0),(1,0, 0, 1).
a
S)
91
SSM:Elementary Linear Algebra
Chapter 4: General Vector Spaces
5. The matrix
1
0
2
1
3
0
1
0
5
0
0
1
1
0
(c) As in part (b), there are
reduces to
n(n + l)
di mension
2
1 0 0 , so the system has only the trivial
1
entries on
2
or above the main diagonal, so the space has
0 0 0
0 0
«(« +!)
0
11. (a) Let Pi = p^ix) and pj - P2M be elements
solution, which has dimension 0 and no basis.
of IT. Then (p,+^2)0)= Pi(l)+ P2(l)=0
7. (a) Solving the equation for x gives
2
JC =
5
.
so pi +P2 is in W.
.
{kp^){\)= k(pi(\))= k-0 =0 so /cpi is in VT.
— y-— z, so parametric equations are
2
5
x = —r-
3
Thus, VT is a subspace of P2 ■
3 ^> 3^ =
r,z = s, which can be
(b),(c) A basis for W is {-\ + x, -x+ x^} so the
written in vector form as
dimension is 2.
,
, f2 5
{x, y, z)= —r — s, r, s
3
V3
=r
3
A basis is
13. Let u, =(1, 0, 0, 0), U2 =(0, 1, 0, 0),
2
f 5
')
-, 1, 0 +5 —,0,1 .
(2
U3 =(0, 0, 1, 0), U4 =(0, 0, 0, 1). Then the set
3
{V|, V2, Ui, U2, U3, U4} clearly spans
W
1, 0 , -
3
0,1 .
. The
equation
3
qv,+C2V2 +^iUi + A:2U2 +/:3U3 +/:4U4 =0
leads to the system
(b) Solving the equation for x gives x = y, so
parametric equations are x= r,y = r, z = s,
=0
Cl -3c2 + k^
=0
-4ci +8c2
+^2
= 0'
2ci -4c2
+^3
-3ci +6c2
+ A:4=0
which can be written in vector form as
(x, y, z)=(/-, r, s)= r(l, 1, 0)+ ^(0, 0, 1).
A basis is (1, 1,0),(0,0, 1).
1
-3
1
-4
8
0
(c) In vector form, the line is
(x,y, z)=(2r,-r, 40 = r(2,-1,4).
A basis is (2,-1,4).
0 0 0 0
1
0 0 0
reduces
The matrix
2 -4
-3
(d) The vectors can be parametrized as a = r,
b = r + s, c = s or
(a, b, c)=(r, r + i', s)= r(l, 1, 0)+ ^(0, 1, 1).
6
1
0
-2
0 0
0
1
-1
0 0
0 0
0
1
0
0 0
0 0
1
to
A basis is (1, 1,0),(0, 1, 1).
9. (a) The dimension is n, since a basis is the set
0 0
1
0 0 0
-1
0 0
1
0
0
40
40I
{/4|, A2,..., A,j} where the only nonzero
From the reduced matrix it is clear that Ui is a
entry in A,- is q,- = 1.
linear combination of Vj and V2 so it will not
be in the basis, and that {vi, V2, U2, U3} is
(b) In a symmetric matrix, the elements above
the main diagonal determine the elements
below the main diagonal, so the entries on
and above the main diagonal determine all
linearly independent so it is one possible basis.
Similarly, U4 and either U2 or U3 will produce
a linearly independent set. Thus, any two of the
the entries. There are
n +(n-l)+ ---+ l =
vectors (0, 1, 0,0),(0, 0, 1, 0), and (0,0, 0, 1)
n(n + l)
can be used.
entries on or
2
above the main diagonal, so the space has
dimension
n{n +\)
2
92
Section 4.6
SSM: Elementary Linear Algebra
(i) True; the set of nxn matrices has dimension
15. Let V3 =(a, b, c). The equation
C|V| +C2V2 +C3V3 =0 leads to the system
C|
+ ac3 =0
-2c| +5c2 +bc^ = 0.
3ci -3c2 + CC3 =0
The matrix
1
0
0
1
1
0
a
0
-2
5
b
0
3
-3
c
0
n , while the set {/, A, A~,
+1 matrices.
(j) False; since P2 has dimension three, then by
Theorem 4.5.6(c), the only three-dimensional
reduces to
subspace of P2 is P2 itself.
Section 4.6
0
a
la^lb 0
0 0 -^a+^b+c 0
A" ) contains
Exercise Set 4.6
and the
1. (a) Since {U|, U2) is the standard basis 5 for
homogeneous system has only the trivial
9
3
solution if --a +-b +c ^0.
3
R^, [w], =
Thus, adding any
vector V3 =(a, b, c) with 9a -3b-5c^O will
create a basis for R^.
(b)
-7
2
3
1
0
-4
8
0
1
reduces to
1
1 0 f -
True/False 4.5
0
28
I
1 1
. Thus PiB-^S -
14
(a) True; by definition.
2
1
7
28
1
i
7
14
and
17
(b) True; any basis of R
independent vectors.
2
will contain 17 linearly
3
[y/h = Pa^sl^h = I
28
1
i
1
1
28
3
14
7
14
17
(c) False; to span R , at least 17 vectors are
required.
1
(c)
0
1
1
0
1
2 0
1
1
1
1
1
2
2
0
(d) True; since the dimension for R^ is 5, then
1
linear independence is sufficient for 5 vectors to
0
reduces to
Thus Pi
0
-
1 1
be a basis.
2
and
2.
[w]5 - PB_>5[w]g
(e) True; since the dimension of R^ is 5, then
spanning is sufficient for 5 vectors to be a basis.
14 ihi
(f) True; if the set contains n vectors, it is a basis,
while if it contains more than n vectors, it can be
a
reduced to a basis.
(g) True; if the set contains n vectors, it is a basis,
3. (a) Since 5 is the standard basis fl for P2,
while if it contains fewer than n vectors, it can be
4
enlarged to a basis.
(P)5 =(4,-3, 1) or [p]^ = -3
1
(h) True; the set
0 1
Ti oi r 1 0 0 1
■ 0 1 ’ 0 -1 ’ 1 0 ’ -1 0
form a
basis for M 22-
93
0
SSM: Elementary Linear Algebra
Chapter 4: General Vector Spaces
1
(b)
1
0
1
0
1
0
0
1
0
1
1
1
0
1
1
1
2
0
2
1
1
0 0
1
1
0 0
[B]e = Ps-,e[B]s =
I
0
0 0
1
-1
reduces to
0 0
0 0
1
0 0
I
1
I
2
2
2
I
I
I
7
1
0
6
0 0 0
1
3
11
_1
10
2
--
0
0 0
15
. Thus
-1
6
9
2
3
1
1
_i
2
2
2
I
i
i
9
2
2
i
1
i
9
9
9
Pb-^S -
15
and B =
and
7. (a)
[pis '1
3 ■
6
2
4
1
2
-1
3
reduces to
-1
i _i
9
9
9
I
1
1
2
->
9
2
1
0
13
I
10
“
2
0
1
1
9
2
2
0
2
1
-1
2
4
(b) 3
-1
2
-1
0
0
1
-
0
5
i
so f*
5
reduces to
5
or (p)5 =(0, 2, -1).
0
-5
-2
J1
2
SO Pb^b' =
0
1
-2
2
2
1
2
3
5. (a) From Theorem 4.6.2, Ps-^B - 0 2 3
0 0 3_
(c)
2
4
1
0
2
-1
0
1
reduces to
where B is the standard basis for R^. Thus,
'1
2 3
6
0
3
4
2
I
10
0
10
[w]g =Fs_^g[w]5 =0 2 3
0
0
16
12
i
1
5
5
so P,
E^B -
[w]g =
1
i
(b) The basis in Exercise 3(i3) is the standard
3
10
5
3
1
i
-5
5
basis B for P2, so [q]g =[q]5 = 0 and
5j^
17
4
10
q =3+ 4x2.
5
[wig' = /z?^B'[w]g
(c) From Theorem 4.6.2,
■-1 1 0 O'
0
-2
1
1
0
0
0
0
1
0
-4
0
0
0
1
-7
standard basis for
9
5
1
1
5
5
where E is the standard basis for R^.
and w =(16, 10, 12).
Ps-^E -
i
10
where E is the
■
Thus,
94
5
17
2
10
13
2
5
Section 4.6
SSM: Elementary Linear Algebra
1
-I
1
0
(0 3
-1
0
1
I
reduces lo
0
1
I
2
2
0
]
, 2 I
1 1
2
2
= -2
-3
so Pe^B'-
2
5
1 1
2
-9
2
1 1
2
9
I
-5
6
7
2
2
[w]b' = /’£^B'[w]£
23
2
i
i
2
2
3
1 1
■5
2
6
2
1
9. (a)
-5
-5
-1
2
2
1
0
1
-1
2
2
1
1
-3
1
0
0
3
2
_5
0
1
0
-2
-3
i
0
0
1
5
reduces to
1
0
0
0
0
1
0
-3
2
0
0
1
1
0
0
0
1
0
0
0
1
1
2
Pe^b'- -1 ^
_ I
1
1
0
0
0
1
0
1
1
0
0
1
0
0
0
1
2
2
2
1
I
2
2
3
2
-1
0
2
3
1
5
2
0
2
I
3
1
-1
0
2
6
SO
11. (a) The span of f| and f2 is the set of all linear
combinations af| +M2 =asinx + &cosx
and this vector can be represented by {a, b).
Since gj = 2f| +f2 and g2 = 3f2, it is
where £ is the
sufficient to compute det
.
2
1'
0
3
= 6. Since
this determinant is nonzero, g| and g2
form a basis for V.
_5'
1
1
1
= _ 1
_ 1
3
0
-5
2
[wJb =P£^g[w]£
9
1
2
_5
1
2
2
23
1
1
-5
7
reduces to
3
standard basis for
2
i -2
1
1
1
i
1
Pe^B -
2
i
2
6
1
I
_1
2
1
2
2
0
1
1
[wy =
I
1
0
2
so
2
2
2
0
-2
reduces to
5
-2
1
1
SO
2
1
(b)
I
2
i
2
2
5
I
2
1
6
3
Pb-^b' -
-1
(c)
-7
3
1
1
3
-4
-5
(b) Since B can be represented as {(1, 0), (0, 1)}
2 o'
Pb'-^b - 1 3
^2 -5
9
-9
(c)
-5
2
0
1
0
1
3
0
1
SO Pb^b’ -
95
1
0
0
1
reduces to
i 0
1
i
6
3_
i 0
i
i
6
3
SSM: Elementary Linear Algebra
Chapter 4: General Vector Spaces
1
2
(d) [h]B =
since B is the standard basis
15. (a)
-5
2
1
1
2
3
3
4
for V.
so
[h]g'- P
_1
1
6
-5
3j'-
■
2
1
(b)
2
3
2
5
3 0
1
1
0
1
1
0
1
(d)
reduces to
-40
16
9
13
-5
-3
-40
16
9‘
13
-5
-3 .
5
-2
-1
4
=
2
5
-2
3
_
2
1
0
2
5
0
1
-1
-3
2 5
_] _3 ’
1
0
so
1
0-3
2
reduces to
2
3 0
0
1
-3
2
2
-1
1
2
-1
where 5 is the
standard basis {(1,0),(0, 1)) for R^. Since
2
w is the standard basis vector (0, 1) for R ,
-1
^ ■ 2 5ir 2
(d) [w]g =P5^b[w]5
16
9
5
-1
13
-5
-3
-3
1
5
-2
-1
1
1
'-40
3
-1
2
so Ps^B^ =
5 -2 -1_
Ps^B -
3
1
0 0
0 8 0 0
1
1
B|—>02
0
0 0
1
so A
1
0 0
-2
reduces to
0 8
1
5
-1
3
13. (a) Pb_,s= 2 5 3
1
3
1
-2
(b)
1
0
=
i n2
-
1
0
reduces to
-3
-1
-239
1
77
(e)
30
“l 25 33]
1
0
8
1
0
3 4 0
1
so Ps^B^ =
-239
0
0
1
4-1
-3
1
-3
77
4-1
=Ps^B2^^h =
30
5
[wJb, =Pb^-^b,Mb2
-3
-[3 5ir 3'
1
-1
4
3
(e) [w]^ = -5
0
[w]b =Ps^bMs
=
1
reduces to
4
[w]s ='P0-45[w]b
= 2
1
■^0
16
9
3
13
-5
-3
-5
5
-2
-1
0
■200
64
25
96
-2
-1
-3
1
2
5
3
SSM: Elementary Linear Algebra
3
I
-1
2
2
1
1
1
0
1
-1
2
-5
-3
2
1
1
1
17. (a)
1
0
0
3
2
0
I
0
-2
-3
0
0
1
5
3
reduces to
3
1
1
1
-5
-3
(c)
1
-1
0
0
9
0
1
0
]
6
0
0
1
2
^
Pb,^B^ - -2 -2
i
2
I
2
1
1
1
0
0
1
-1
2
0
1
0
1
1
1
0
0
1
0
0
0
1
0
0
1
5
2
2
_1
_1
2
2
2
-I
0
2
so
2
i
2
2
_1
2
1
1
1
1
2
2
1
_1
2
2
2
1
-5
8
-5
7
2
23
2
6
2
2
9
1
_i
1
2
2
2
-1
0
2
where 5 is the
V
19. (a)
|V
I >
Standard basis for R^.
e
X
■>
ei
= Ps-^B, Ms
1
1
1 , SO
1
1
reduces to
I
■ t 2
i
-■ i -2
reduces to
1
Ps-^B,
I
[w]g^
2
1
0
1
3
0
1
6
2
0 0
2 0 0
1 i
Ps^B, = -1 ^
2
2
1
so
1
0 0
1
5
(b)
Section 4.6
2
2
tr-5
i
i
1
2
2
2
-1
0
2
From the diagram, it is clear that
=(cos2^, sin 20.
-5
■^A
9
\
^2
-9
X
-5
\
From the diagram, the angle between the
- 3 2 f
9
i
-9
= -2 -3
5
1
3
6
positive }3-axis and \2 is
Jt
2 —6 =71-20 so the angle between
-5
7
V2 and the positive x-axis is
2
23
71
71
- 71-2(9-- =20— and
2
29
2
6
71
[V2]5 = COS 20— ,sin 202J
= (sin20 -cos20
Thus by Theorem 4.6.2,
Pb-^S -
97
cos 2^
sin 2(9
sin 2(9
-cos 2(9
7t
2JJ
SSM: Elementary Linear Algebra
Chapter 4: General Vector Spaces
Section 4.7
21. If w is a vector in the space and [w]g' is the
coordinate vector of w relative to B\ then
Exercise Set 4.7
[wig =f’[w]^' and [w]c =G[w]g, so
[w]c = G(/’[w]b')= QP['iy]g. The transition
1. Row vectors: r| =[2 -1 0 1],
r2=[3 5 7 -1], r3=[l 4 2 7]
matrix from B' to C is QP. The transition matrix
2
from C to B' is {QP)~' =
5
Column vectors: C| = 3 , C2 =
4
23. (a) By Theorem 4.6.2, P is the transition matrix
fromB= 1(1, 1,0),(1,0, 2),(0, 2, 1)) to S.
(b)
1
0
1
1
0
2
0
0
2
1
0
C3= 7 , C4 =
0 0
1
0
0
1
0
2
reduces to
7
1
3
1
0
0
0 0
1
4
.S
5
I
1
0
0
I
9
5
5
5
5
9
2
1
.S
5
1
3. (a) ^
-2
B=
1
0
0
I
1
, so
3
10
-6 ■
4
.S,
Thus, P is the transition matrix from 5 to
4
.10^ reduces to
-6
_ 1 2'
(b)
5’5’“5j’U’~5’ sj’
^ 1 1 L]
1
1
1
0
1
0
1
0 reduces to 0
1
1
0
2
1
3
2
0
0 0
1
2-1
0
so the system Ax = b is inconsistent and b is
not in the column space of A.
5’ 5’ 5.
5
27. B must be the standard basis for R", since, in
(c)
particular [e,]g =e,- for the standard basis
9
3
1
vectors e|, 63, ..., e„.
True/False 4.6
1
1
0
0
0
1
0
0 0
1
(a) True
reduces to
1
-1
1
-3 , so
1
5
9 -3
3
(b) True; by Theorem 4.6.1.
+
1
(c) True
(d) True
1
-I
1
2
1
-1
0
1
-1
1
0
(d)
(e) False; consider the bases 5 = {e,, 63, 63} and
1
= {263, 3e|, 5e2l for R^, then
‘0 3 o'
=0 0 5 which is not a diagonal
2
0
0
0 0
0
0
reduces to
x.
1 , so the system A X2 = b
0
•^'3
has the solution X| =1, X2 = t -1, x^=f-
0 0
-I
2
matrix.
Thus
0
0
(f) False; A would have to be invertible, not Just
square.
98
+(f-l)
1 +t
-1
1
Section 4.7
SSM: Elementary Linear Algebra
(e)
^^3 =s, JC4 = t, or in vector form
1
2
0
1
4
0
1
2
1
3
1
2
1
3
5
0
1
2
2
7
reduces to
1
0 0 0
0
1
2
-1
0
1
0
13
0
1
0
0
0
-7
2
1
0 0 0
1
4
1
0
0
2
0
2
1
5
1
2
7
0
1
1
5. (a)
-3
1
2
-6
2
1
-3
1
0
0 0
or in vector form, x =
2
+r
0
2
3
-1
4
reduces to
0
2
7
_i
6
5
5
5
11
5
5
5
0 0
0
0
0
0 0
0
0
0
6 7
1
' 5 5
5
X, = — + — s+ — t,
4
3
^
5 5
5
3_
1 -2
l
vector form, x =
7 , and the solution of
0 0 0
and the solution ot
Xj= — + — s—f, xt,=s, XA=t, orm
1 -2 reduces to
i
3
0-5 -5
4
1
7
5
1 3
o
-1
.
.■
Ax = b IS
1 ■
■1 0
2
0
3
1 0
1
1
The general form of the solution of Ax = 0 is
(b)
1
4-7
3
1
0
-2
, and
the solution of Ax = b is X| = 1 + 3f, X2 = t,
1
1
0
1 .2 -3
2
reduces to
x =t
0
0
3
2
(d)
1
0
+4
-7
+ 13
-2
+t
+X
X=r
= -26
0
The general form of the solution of Ax = 0 is
, so
0
3
0
+t
0
1
4
-2
+s
+r
X=
-26
0 0
0 0
-1
0
Ax = b is X| = -2-r, X2 = 7-r, xj =t, or
-2
in vector form, x =
7
0
^
1
I
5
5
5
7
5 +s
4
3
5 +t
5
0
1
0
0
0
The general form of the solution of Ax = 0 is
-1
. The
+t
^
6
1
x =s
7
I
5
5
4
3
5 +r
5
0
general form of the solution of Ax = 0 is
0
-1
1
x = t -1
7. (a) A basis for the row space is r| =[1 0 2],
1
r2=[0 0 1]. A basis for the column
(c)
2
1
-2
2
-4
2
4
-2
-1
2
-1
-2
1
3 -6
3
6
-3
2
space is Cj = 0 , C2 = 1
reduces to
1
2
-1
0
1
-2
0 0
0
0
0
0 0 0
0
0
0 0
0
0
0
, and the solution of
Ax = b is X| =-l + 2/--^-2r, X2 = r.
99
0
Chapter 4: General Vector Spaces
SSM:Elementary Linear Algebra
(b) A basis for the row space is
(b) A basis for the row space is
r, =[1 -3 0 0], r2=[0 1 0 0],
r, =[1 -3 0 0], r2=[0 1 0 0],
0
0
1
A basis for the column space is Cj =
A basis for the column space is C] =
0
0
0
-3
-3
1
‘=2 =
0
1
C2 =
0
0
0
0
(c) A basis for the row space is
r,=[l 2 4 5], r2=[0 1 -3 0],
(c) A basis for the row space is
ri=[l 2 4 5], r2=[0 1 -3 0],
r3=[0 0 1 -3], r4=[0 0 0 1],
A basis for the column space is Cj =
r3=[0 0 1 -3], r4=[0 0 0 1],
1
1
0
0
0
A basis for the column space is C] =
0
0
0
2
4
5
2
4
5
1
-3
0
1
-3
0
C2 = 0 . C3 =
1 , C4 = -3
1
0
0
0
0
0
0
C2 = 0 . ^3 =
0
1 , C4 = -3
0
0
1
0
0
0
(d) A basis for the row space is
ri=[l 2 -1 5], r2=[0 1 4 3],
(d) A basis for the row space is
r, =[1 2 -1 5], r2=[0 1 4 3],
r3=[0 0 1 -7], r4=[0 0 0 1].
r3=[0 0 1 -7], r4-[0 0 0 1],
A basis for the column space is C| =
1
1
0
0
A basis for the column space is cj =
0
0
C2 =
0
2
-1
5
2
-1
1
4
3
1
4
0
> ‘=3 =
0
1
. C4 =
0
C2-
-7
1
0
- C3 =
0
1
r2 =[0 0 1]. A basis for the column
-7
1
1
1
0
2
-2
2-1
3
2
2
0
3
, C4 =
11. (a) A row echelon form of 2
space is Cj = 0 , C2 = 1
0
5
0
9. (a) A basis for the row space is rj =[1 0 2],
1
0
IS
-4-3
1
1
-4
-3
0
1
-5
-2 . Thus a basis for the
0 0
1
1
2
subspace is (1, 1, -4,-3),(0, 1, -5,-2),
1
0, 0, 1, — .
2
100
Section 4.7
SSM: Elementary Linear Algebra
(b) A row echelon form of
-1
1
-2 0
3
3
6 0
9
0
0
(b) det(A)= det(B) = 5
A and B are invertible so the systems Ax = 0
and fix = 0 have only the trivial solution.
That is, the null spaces of A and fi are both
the origin.
The second row of C is a multiple of the
first and it is clear that if 3x + y = 0, then
(x, y) is in the null space of C, i.e., the null
space of C is the line 3jc + y = 0.
is
3
1
-1
2
0
0
1
0
0 . Thus a basis for the
0
0
1
1
6
subspace is (1,-1,2, 0),(0, 1, 0,0),
n
0,0,1,— .
6J
X
The equation £)x = 0 has all x = ^
(c) A row echelon form of
1
1
0
1
1
1
0
0
0
1
-2
0
2
2
0 -3
0
3
as
solutions, so the null space of D is the entire
jcy-plane.
0
IS
True/False 4.7
(a) True; by the definition of column space.
0 0
1
(b) False; the system Ax = b may not be consistent.
1
. Thus a basis for the
0 0
0
1
1
0 0
1
(c) False; the column vectors of A that correspond to
the column vectors of R containing the leading
1 ’s form a basis for the column space of A.
subspace is (1, 1,0, 0),(0, 1, 1, 1),
(0, 0, 1, 1),(0, 0,0, 1).
(d) False; the row vectors of A may be linearly
1
dependent if A is not in row echelon form.
0 0
15. (a) The row echelon form of A is 0
1
0 ,
1
_0 0 0
(e) False; the matrices A =
3
1
3
and fi =
2 6
0 0
so the general solution of the system Ax = 0
is X = 0, y = 0,z = t or t
0
have the same row space but different column
0 . Thus the null
spaces.
1
(f) True; elementary row operations do not change
space of A is the z-axis, and the column
the null space of a matrix.
0
(g) True; elementary row operations do not change
the row space of a matrix.
space is the span of C| = 1 , C2 = 0
0
0
which is all linear combinations of y and x,
i.e., thexy-plane.
(h) False; see (e) for an example.
(i) True; this is the contrapositive of Theorem 4.7.1.
(b) Reversing the reasoning in part (a), it is
■q 0 O'
clear that one such matrix is
0
1
0
0
0
1
0)
row equivalent to /„. However, since fi is
singular, it is not row equivalent to /„, so in
17. (a) If the null space of a matrix A is the given
line, then the equation Ax = 0 is equivalent
to
3 -sirxi^r
0
0
0
for A is
oJLyJ
3a
-5a
_3b
-5b
False; assuming that A and fi are both nxn
matrices, the row space of A will have n vectors
in its basis since being invertible means that A is
reduced row echelon form it has at least one row
of zeros. Thus the row space of fi will have
and a general form
fewer than n vectors in its basis.
where a and b are real
numbers, not both of which are zero.
101
Chapter 4: General Vector Spaces
SSM: Elementary Linear Algebra
Section 4.8
7. (a) Since rank(A)= rank[/\|b]. the system is
consistent, and since nulIity(A) = 3-3 = 0,
there are 0 parameters in the general
Exercise Set 4.8
solution.
1. The reduced row echelon form of A is
0 -I -f
0
17
1
0 0
1
A^= 2
(b) Since rank(A)< rank A b , the system is
, so rank (A)= 2.
7
7
0
0
inconsistent.
(c) Since rank(A)= rank A b , the system is
-3 -2
4
^ ^ and the reduced row
consistent, and since nullity(A) = 3-1=2,
there are 2 parameters in the general
0
2
solution.
2
echelon form of A^ is
1
0
1
0
1
1
0 0 0
(d) Since rank(A)= rank A b , the system is
, so
consistent, and since nullity(A) = 9-2 = 7,
there are 7 parameters in the general
0 0 0
solution.
rank(A^)= rank(A)= 2.
(e) Since rank(A) < rank A b , the system is
3. (a) Since rank (A)= 2, there are 2 leading
variables. Since nullity(A) = 3 - 2 = 1, there
is 1 parameter,
inconsistent.
(f) Since rank(A = rank A b , the system is
(b) Since rank(A)= 1, there is 1 leading
variable. Since nullity(A) = 3-1 =2, there
consistent, and since nullity(A) = 4-0 = 4,
there are 4 parameters in the general
solution.
are 2 parameters.
(c) Since rank(A)= 2, there are 2 leading
variables. Since nullity(A)= 4-2 = 2, there
are 2 parameters.
(g) Since rank(A)= rank A b , the system is
consistent, and since nullity(A) = 2-2 = 0,
there are 0 parameters in the general
solution.
(d) Since rank(A)= 2, there are 2 leading
variables. Since nullity(A) = 5-2 = 3, there
1
are 3 parameters.
(e) Since rank(A)= 3, there are 3 leading
variables. Since nullity(A) = 5-3 = 2, there
9.
The augmented matrix is
are 2 parameters.
1
0
1
b1
1 -2 b2
1
1 Z?3 , which is
I -4 b^
1
0
-3
5
3b2-2b^
h ~t>\
row equivalent to 0 0 ^3-4^2+^^! - Thus,
0 0 b^+bo — 2bi
0 0 b^-?,b2+7bi
5. (a) Since A is 4 X 4, rank(A)+ nullity(A) = 4,
and the largest possible value of rank(A) is
4, so the smallest possible value of
nullity(A) is 0.
(b) Since A is 3 x 5, rank(A)+ nullily(A) = 5,
and the largest possible value of rank(A) is
3, so the smallest possible value of
nullity(A) is 2.
the system is consistent if and only if
3b^ -4b2 + bT,
-2b^ +b2
+ b^
=0
=0
Ib^ -8/72
+^5 =0
which has the general solution b^ = r, b2= s,
(c) Since A is 5 x 3, rank(A)+ nullity(A) = 3,
and the largest possible value of rank(A) is
3, so the smallest possible value of
nullity(A) is 0.
/>3=-3r+ 45, b^=2r-s, b^=-7r + ?,s.
102
Section 4.9
SSM: Elementary Linear Algebra
11. No, since rank(/4) + nullity(/\) = 3 and nullity(A)
= 1, the row space and column space must both
be 2-dimensional, i.e., planes in 3-space.
(e) True; if the rows are linearly dependent, then the
rank is at least 1 less than the number of rows, so
13. The rank is 2 if;- = 2 and 5=1. Since the third
(f) False; if the nullity of A is zero, then Ax = b is
since the matrix is square, its nullity is at least 1.
column will always have a nonzero entry, the
consistent for every vector b.
rank will never be 1.
(g) False; the dimension of the row space is always
equal to the dimension of the column space.
17. (a) The number of leading 1 ’s is at most 3,
since a matrix cannot have more leading I’s
than rows.
(h)
(b) The number of parameters is at most 5,
False; rank(A)= rank(A^) for all matrices.
(i) True; the row space and null space cannot both
since the solution cannot have more
have dimension 1 since
parameters than A has columns.
dim(row space) -i- dim(null space)= 3.
(c) The number of leading 1 ’s is at most 3,
1
(j) False; a vector w in IT
since the row space of A has the same
dimension as the column space, which is at
need not be orthogonal
to every vector in F; the relationship is that V'^
is a subspace of W'^.
most 3.
(d) The number of parameters is at most 3 since
Section 4.9
the solution cannot have more parameters
than A has columns.
Exercise Set 4.9
0
1
1
2
and B =
19. Let A =
0 0
. Then
1. (a) Since A has size 3x2, the domain of
2 4,
rank(A)= rank(S)= 1, but since A^ =
0
0
0
0
and the codomain of
is
.
(b) Since A has size 2x3,the domain of
and 5^ =
5
is
is
10
R^ and the codomain of 7^ is R^.
10 20j’
rank(A^)= 0 rank(B^)= 1.
(c) Since A has size 3x3,the domain of 7^ is
R^ and the codomain of 7^ is R^.
True/False 4.8
(a) False; the row vectors are not necessarily
(d) Since A has size 1 x 6, the domain of 7^ is
linearly independent, and if they are not, then
R^ and the codomain of 7^ is 7'.
neither are the column vectors, since
dim(row space)= dim(column space).
3. The domain of 7is R^ and the codomain of 7is
(b) True; if the row vectors are linearly independent
then nullity(A) = 0 and rank(A)= n = the number
of rows. But, since rank(A)+ nullity(A) = the
number of columns, A must be square.
7(1 -2)=(1 -2,-(-2), 3(1)) =(-1,2, 3)
5. (a) The transformation is linear, with domain
R^ and codomain R^.
(c) False; the nullity of a nonzero m x n matrix is at
most n - 1.
(b) The transformation is nonlinear, with
(d) False; if the added column is linearly dependent
domain R" and codomain R .
on the previous columns, adding it does not
affect the rank.
(c) The transformation is linear, with domain
R^ and codomain R^.
103
Chapter 4: General Vector Spaces
SSM: Elementary Linear Algebra
(d) The transformation is nonlinear, with
domain
and codomain
0 0
1
.
0
0
(d) The matrix is 0 0
7. (a) T is a matrix transformation;
0
0 0 0‘
Ta =
1
1
1
1
0 0
0
0 0
0-1 0
0 0 0 ’
13. (a) AfX =
(b) T is not a matrix transformation since
no,0,0) 0.
-1
1
-1
5
0
1
4
4
Tf-l, 4)=(-(-!)+4,4)=(5,4)
(c) 7is a matrix transformation;
2
=r^ -4 o'
2
1
2
0
1
1
-2
0 OJL-3
0
(b) Arx= 0
0 -5 ■
0
7(2, 1, -3)=(2(2)-(1)+(-3), 1 +(-3), 0)
=(0,-2,0)
(d) 7is not a matrix transformation, since for
1,
T{k{x,y, z))=(k^y^,kz)
Tt kT{x, y, z)
15. (a)
={ky^,kz).
3
5
0
1
0
2
2
0 -5 = -5
The result is (2,-5,-3).
1
0 0
2
2
(b) 0-1 0 -5 = 5
-1
_0 0 ij|_ 3J [3
1
9. By inspection, 7=4
1
0
_0 0 -lj[ 3J L-3_
(e) 7is not a matrix transformation, since
7(0,0,0)=(-l,0)9i0.
3
-1
The result is (2, 5, 3).
2-1
For (X, y,z)=(-1, 2, 4),
=3(-l)+5(2)-4 = 3
-1
0
0
0
1
0
(c)
2
-2
-5 = -5
0 0 lj[ 3
W2 =4(-l)-(2)+4 = -2
W3=3(-1)+ 2(2)-(4)= -3
3
The result is (-2, -5, 3).
so7(-l,2,4)=(3,-2,-3).
■3
Also,
5
4-1
3
2
r 3
1
17. (a)
2 = -2
-1
4
1
0
0
0
1
0
-2
-2
1 =
1
0 0 oJ[ 3J [ 0_
-3
The result is (-2, 1, 0).
11. (a) The matrix is
0
1
-1
0
1
(b)
3
1
-1
1
0
0
-2
-2
0
0
0
1
0
0
0
1
3
3
The result is (-2, 0, 3).
(b) The matrix is
7
2-1
0
1
-1
(c) The matrix is
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
(c)
0
0
0
0
-2
0
1
0
1
0
0
1
3
0
3
The result is (0, 1, 3).
104
Section 4.9
SSM: Elementary Linear Algebra
1
19. (a)
0
0
0
cos30°
-sin 30°
0
sin 30°
0
0
-2
-2
1
1
04
1
2
2
cos 30°
» I
2
-2
S-2
2
\+2S
2
The result is -2,
S-2 l + 2^/3
2
2
1
cos45°
0
sin 45°
0
1
0
1
-sin 45°
0
cos45°
2
(b)
-2
1
0
-2
VI
0
I
0
‘VI
0
1
1
2
VI
0
1
2V2
The result is (0, 1, 2V2).
cos90°
(c)
-sin 90° 0
-2
0-1 0
-2
0 0
1
sin 90°
cos90°
0
1
0
0
1
2
0
0
2
1
-1
-2
2
The result is (-1, -2, 2).
1
1
21. (a)
0
0
-2
0
0 cos(-30°) -sin(-30°)
0 sin(-30°) cos(-30°)
0
2
0
1
2
-7
1
1
VI
2
2
2
0
-2
V3+2
2
-I+2VI
2
The result is -2,
V3+2 -I +2V3
2
’
-2
S
2
105
Chapter 4: General Vector Spaces
SSM: Elementary Linear Algebra
I
cos(-45°)
(b)
0 sin(-45°)
0
1
0
-2
Vi
1
0
2
I
0
-sin(-45°) 0 cos(-45°)
1
“Vi
0
1
2
0
^/i
-2
^/i
-2V2
0
The result is (-2V2,1, o)(c)
cos(-90°) -sin(-90°) 0
sin(-90°) cos(-90°) 0
0
-2
0
1
-1
2
0
0
1
0
-2
0 0
0
2
2
2
The result is (1, 2, 2).
25. —
(2, 2, 1)
2 2 j_
llvll 722+2^+12
3’ 3’ 3
=(a, b, c)
cos 180° = -1, sin 180° = 0
The matrix is
(if(l-(-l))+(-l) (f)(f)(l-(-l))-5(0) (f)(^)(l-(-l))+(|)(0) f(2)- l
+
(ff(l-(-l))+(-l) (f)(5)(l-(-l))-(|)(0) = |(2)
(f)(2)(l-(-l))-f(0) (l)(y(l-(-|))+l(0) (i)-(!_(-|))+(-i)
f(2) I(2)
|(2)-l f(2)
f(2) |(2) i(2)-l
I
9
4
9
I
9
1
9
9
4
9
1-1
9
9
29. (a) The result is twice the orthogonal projection of x on the x-axis.
(b) The result is twice the reflection of x about the x-axis.
9
9
31. Since cos2^ = cos ^-sin 9 and sin 29= 2 sin 9co% 9, the matrix can be written as
cos 2^
-sin 2^
sin 2^
cos 26*
which
has the effect of rotating x through the angle 29.
33. The result of the transformation is rotation through the angle /9followed by translation by xg. It is not a matrix
transformation since 7(0)= Xq
0.
35. The image of the line under the operator 7is the line in /?" which has vector representation x = 7(xo)+ r7(v),
unless 7is the zero operator in which case the image is 0.
106
Section 4.10
SSM: Elementary Linear Algebra
True/False 4.9
3. (a) The standard matrix for
(a) False; if A is 2 x 3, then the domain of
is
1
1
-1
and
is
the standard matrix for T2 is
r\
(b) False; if A is m x n, then the codomain of
1
is
3
0
2
4
(b) The standard matrix for 7’2o7’| is
'3 olTi
2
(c) False; 7’(0) = 0 is not a sufficient condition. For
4
1
1
3
3
6 -2 ■
-1
instance if T: R" —>/?' is given by 7’(v)=||v||
The standard matrix for 7j o T2 is
then 7(0)= 0, but 7is not a matrix
"1
1
transformation.
(d) True
i]r3 o'
-1
2
5
4
1
-4
4
(C) 7,(72(X|, X2))=(5X| +4x2• ^1 -4^2)
72(7|(X|, X2))=(3xi +3x2 - 6x1 -2x2)
(e) False; every matrix transformation will have this
property by the homogeneity property.
5. (a) The standard matrix for a rotation of 90° is
(f) True; the only matrix transformation with this
property is the zero transformation (consider
X = y ^ 0).
cos90°
-sin 90°
sin 90°
cos90°
0
-1
and the
0
standard matrix for reflection about the line
. To
(g) False; since b 0, then 7(0)= 0 + b = b?i0, so
7cannot be a matrix operator.
y = X IS
(h) False; there is no angle 6>for which
0
1
1
0
r
0 ■
1
0
0
0
0
cos^ = sin0 = —.
2
(b) The standard matrix for orthogonal
0 0
(i) True; when a = 1 it is the matrix for reflection
projection on the y-axis is
about the x-axis and when a = -\ it is the matrix
and the
0
1
standard matrix for a contraction with factor
for reflection about the y-axis.
1 .
k=- is
Section 4.10
2 LO i
Exercise Set 4.10
2 °lro 0
1. The standard matrix for TgoT^ is
'2 -3 3ir 1 -2
BA= 5
0
1
4
1
0 i
o'
-3
1
0
.0 i
(c) The standard matrix for reflection about the
6 1 7j[5 2 4
5 -1
0
0
1
0
0
-I
and the standard matrix
x-axis IS
21'
4
10
for a dilation with factor A' = 3 is
.45 3 25_
The standard matrix for 7^ o7g is
"1 -21 -30]r25 -30 3'
AB= 4
1
5 2 4j[6
-3
-5
-15
44
-1 1
1 7_
11
45
107
3
0
1
0
3
0
0
3
0
-1
0
-3
3
0
0
3 ■
Chapter 4: General Vector Spaces
SSM:Elementary Linear Algebra
-1
7. (a) The standard matrix for reflection about the yz-plane is
1
0
0 0
1
0
0 0
1
and the standard matrix for orthogonal
0 0
projection on the xz-plane is 0 0 0 .
[o 0 1_
1
0 0
-1
0 0 0
0
1
0
0 0
0 0
1
1
-1 0 o'
0 0
0 0 0
0 0
1
1
cos45°
(b) The standard matrix for a rotation of 45° about the y-axis is
0
sin 45°
0
cos45°
0
0
-sin45°
0
0
I
"VI
VI 0
the standard matrix for a dilation with factor ^ = VI is
1
VI 0 0
0 VI 0
0 0 VI
VI
1
VI
1
1
VI
0
-1
0
1
VI
0
reflection about the yz-plane is
0
0 0
1
1
9. (a) [7,]=
1
0
0 0
1
-1
0
0
0 0 0
0
0 0
0 0
1
0
0 0
1
0 0
1
0
0 0 0
0 0
and [T2]=
0
1 ■
0 0
[7j][7’2]=[7I][7i]=
VI
1
0 0
1
0
0 0 0
1
and
= 0 VI 0
0
-1
0 0
0
1
o’
(c) The standard matrix for orthogonal projection on the xy-plane is 0
0
0
W 0 •
0 VI
1
-1
1
1
0
0
0
0
1
VI
0 0
so 7| O ^2 = ^2 O 7j.
108
and the standard matrix for
Section 4.10
SSM: Elementary Linear Algebra
cos(^l+^2) -sin(0,+^2)
sin(^l +02) cos(^i +$2)
(b) From Example 1, [7’2o7'[] =
[7ior2]=[r,][72]
cos^l -sin6>| cos^2 -sin ^2
sin^l
cos^i sin ^2 cos ^2
cos^l cos^2 “S'f* ^1 ^2 -cos^i sin ^2
0^ cos^2
sin 6'| cos02 + cos 0^ sin 02 -sin 0^ sin 6*2 +cos0^ cos 02
cos(0^ + 6*2) -sin(6’, +02)
sin(6*| + 6*2) cos(^| +02)
=[72 o7,]
Thus 7] o 72 = 72 o 7|.
1
(c) [7,]=
0
and [72]=
0 0
1
[7',][7'2]=
[7'2][7,]=
0086*
-sin^
sin^
COS0
0
COS0
-sin (9
0 0
sin^
cos6>
cos^
-sin6>
0
0
cos9
-sin^
sin^
COS0
COS0
0
sin6>
0
1
0
0 0
7j 0T2 *T2 °7,
11. (a) Not one-to-one because the second row (column) is all O’s, so the determinant is 0.
(b) One-to-one because the determinant is -1.
(c) One-to-one because the determinant is -1.
(d) One-to-one because the determinant is
^ 0.
(e) One-to-one because the determinant is 1.
(f) One-to-one because the determinant is -1.
(g) One-to-one because the determinant is
13. (a) [7]=
1
2
-1
1
det[7] = 1 -I- 2 = 3
1
1
-2
3 1
1
7 ’(vv|, W2)=\^
3
v3
^ 0.
0, so 7is one-to-one.
— Wi
i
-1
3
3
i
I
3
3
2
1
VV^9, —W]+ — W9
3 “ 3
3
109
Chapter 4: General Vector Spaces
(b) m= -2t
SSM: Elementary Linear Algebra
-6
3
det[7] = 12- 12 = 0, so r is not one-to-one.
(c) m =
0
-1
-1
0
det[7] = 0- I = -1 ^ 0, so r is one-to-one.
I
1
[T~']= —
-1
0
1
0
1
0
-1
0
-I
T (w^, W2)= i-W2,-w^)
id) [T]=
3
0
-5
0
det[7] = 3(0)-(-5)(0)= 0,soT is not one-to-one.
15. (a) The inverse is (another) reOection about the x-axis in
^
(b) The inverse is rotation through the angle —
.
in r\
4
1
0
(c) The inverse is contraction by a factor of — in R .
3
(d) The inverse is (another) reflection about the yz-plane in R^.
(e) The inverse is dilation by a factor of 5 in R^.
17. Let u =(X|, Ji), V =(^2, ^2)> ^nd k be a scalar.
(a) 7'(ii + v)= TfX| -r j:2, >', -r >’2)
=(2(x, +X2)+(y, +>’2).(X| +a;2)-(>’! +>”2))
=((2X| + >’|)-r(2x2 + >’2)- (-^1 - Ji)+i^2 -3'2))
= r(u)+ T(v)
T(l'u)= r(Lx,,^y,)
=(2Lx| -t-^y,, Ax| -Ly,)
= 1:(2X| + y,, X, - y,)
= kT{u)
7is a matrix operator,
(b) T(ku)= T(kx^, ky^)
=(Ax| -H 1, Lyi)
k(x^ +1, y,)
7is not a matrix operator, which can also be shown by considering 7(u + v).
110
Section 4.10
SSM: Elementary Linear Algebra
(c) 7(u + v)= 7’U|+X2, >’|+y2)
=(}'! +}'2- 3'i +J2)
= r(u)+r(v)
T(ku)=T(kx^, ky^)
=(ky\’ ^}'|)
= k(y], y\)
= kT(u)
r is a matrix operator.
(d) 7’(u+v)=7(X|+^2, )'!+}’2)
T is not a matrix operator, which can also be shown by considering T{ka).
19. Let u =(X|, y|, 2]), v =(^2, y2> ^2)’
^
^
(a) r(u + v)= r(X|+X2, y|+>>2. Z|+^2)
=(0,0)
= Tin)+ T{y)
T{ku)= T{kx\,ky^,kz\)
=(0, 0)
= k{0, 0)
= kT{u)
7is a matrix transformation.
(b) 7(u + v)= 7(X|+X2, yi + y2-Zi+Z2)
=(3(X| +X2)-4(y, +^2). 2(X| +X2)-5(Z[ +Z2))
=((3X| -4y,)+(3x2 -4y2),(2X| -5z,)+(2x2 -5z2))
= 7(u)+ 7(v)
T(ku)= T{kx\, ky^, kz\)
=(3fcr| -4^1, 2b:|-Sfe,)
= ^'(3X| -4y|, 2x, -5Z|)
= kT{u)
7is a matrix operator.
21. (a)
-1
0
0
1
0
(b)
0
0
0
0
1
1
-1
0
0 0
0
0 0
(c)
1
0 0
-1
0
0
1
1
0
3 0
0
3
0
0
0
3
0
1
0
0
3
0
0
3
0
23. (a) 7^(e,)=(-l, 2,4),7^(e2)=(3, 1,5), and 7^(e3)=(0, 2,-3)
111
Chapter 4: General Vector Spaces
SSM: Elementary Linear Algebra
(b) 7^(61+62+63)
(b) True; use C| = C2 = 1 to get the additivity
=(-1 +3+0, 2+ 1 + 2, 4+5-3)
=(2, 5, 6)
property and C| = /:, C2 =0 to get the
homogeneity property.
(c) r^(7e3)= 7(0, 2,-3)=(0, 14,-21)
(c) True; if 7is a one-to-one matrix transformation
with matrix A, then Ax = 0 has only the trivial
solution. That is 7(x - y)= 0 => A(x - y)= 0
25. (a) Yes; if u and v are distinct vectors in the
domain of T^ which is one-to-one, then
7j(u) and 7[(v) are distinct. If 7|(u) and
7i(v) are in the domain of T2 which is also
one-to-one, then 72(7i(u)) and 72(7j(v))
are distinct. Thus 72o7| is one-to-one.
=> X = y, so there are no such distinct vectors.
(d) False; the zero transformation is one example.
(e) False; the zero transformation is one example.
(f) False; the zero transformation is one example.
(b) Yes; if 7,(x, y)=(x, y, 0) and
Section 4.11
72(x, y, z)=(x, y), then 7) is one-to-one
Exercise Set 4.11
but 72 is not one-to-one and the
1. (a) Reflection about the line y = -x maps(1,0)
composition T2 °T] is the identity
transformation on
to (0,-1)and (0, 1) to (-1,0), so the
0 -f
which is one-to-one.
However, when
is not one-to-one and T2
is one-to-one, the composition T2 °T\ is not
one-to-one, since 7|(u)= 7,(v) for some
distinct vectors u and v in the domain of T^
and consequently 72(7|(u))= 72(7](v)) for
standard matrix is
0 ■
(b) Reflection through the origin maps(1, 0)to
(-1,0) and (0, 1) to (0, -1), so the standard
. . f-l
01
matrix is
0
those same vectors.
(c) Orthogonal projection on the x-axis maps
27. (a) The product of any mxn matrix and the
n X 1 matrix of all zeros (the zero vector in
(1, 0)to (1,0) and (0, 1) to (0, 0), so the
R”) is the m x 1 matrix of all zero, which is
standard matrix is
1
0
0 0 ■
the zero vector in R"‘.
0
(d) Orthogonal projection on the y-axis maps
"7
(b) One example is 7(xi, X2)=(xf +X2, X]X2)
(1,0) to (0, 0) and (0, 1) to (0, 1), so the
0 O'
standard matrix is
29. (a) The range of 7 must be a proper subset of
0
1 ■
/?", i.e., notall of /?”. Since det(A)= 0,
3. (a) A reflection through the xy-plane maps
there is some b] in R" such that Ax = b|
(1,0,0)to (1,0,0),(0, 1,0) to (0, 1,0), and
(0,0, 1) to (0, 0,-1), so the standard matrix
is inconsistent, hence b| is not in the range
of 7.
"1 0
is
(b) 7must map infinitely many vectors to 0.
0
1
o'
0 .
0 0-1
True/False 4.10
(b) A reflection through the xz-plane maps
■y
y
(1, 0,0)to (1, 0, 0),(0, 1,0) to (0,-1, 0),
and (0, 0, 1) to (0,0, 1), so the standard
(a) False; for example 7: R —> R given by
7(xi, X2) = (X|^ +X2, XjX2) is not a matrix
transformation, although
matrix is
7(0) = 7(0, 0) = (0, 0) = 0.
0
0
0-1
0
0
112
0
1
Section 4.11
SSM: Elementary Linear Algebra
(b) The geometric effect is expansion by a
factor of 5 in the y-direction and reflection
(c) A reflection through the yz-plane maps
(1,0, 0) to (-1,0, 0),(0, 1,0) to(0, 1,0),
and (0, 0, 1) to (0, 0, 1), so the standard
■-1
0
0'
0
1
0 .
0
0
matrix is
about the x-axis.
(c) The geometric effect is shearing by a factor
of 4 in the x-direction.
1
13. (a)
5. (a) Rotation of 90° about the z-axis maps
(1,0, 0) to(0, 1,0), (0, 1,0) to (-1,0, 0),
and (0, 0, 1) to (0, 0, 1), so the standard
'0
-1
1
0
0
0
matrix is
0‘
(b)
0 .
1
0
0
1
(c)
7.
5
0
1
0
1
0
1
0
5
2
5
cos 180°
-sin 180°
0
1
sin 180° cosl80°j[l 0
■-1
OlfO
0
1
ll
0
0
0
0
matrix is
0
0
1
-1
(c) Rotation of 90° about the y-axis maps
(l,0,0)to(0,0,-1), (0, l,0)to(0, 1,0),
and (0, 0, I) to (1, 0, 0), so the standard
'o o f
-3
1
2
(1,0, 0) to (1,0, 0), (0, 1,0) to (0, 0, 1), and
(0, 0, 1) to (0, -1, 0), so the standard matrix
'1 0
o'
0
i 0 = i 0
0 5J 0
1
(b) Rotation of 90° about the x-axis maps
IS
0
0
1
0 .
-1
0
0
0
0
_ 0 uk
0 ’
■-3
oiro
0
_ 0 iJLi
1
0
1
0
1
1
0
1
0
0
I
0
1
0
-3
0
1
0
1
1
, the inverse
transformation is reflection about y = x.
(b) Since for 0 < ^ < 1,
-3
’
15. (a) Since
-3
0 ’
1
0
0
k
n-1
1
k
0
0
1
n-l
= i 0
0
and
1
0
, the inverse
“ 1
transformation is an expansion along the
-3
same axis.
The image is a rectangle with vertices at (0, 0),
(c) Since
(-3,0), (0, 1), and (-3, 1).
0
r- s
0
-r—f
-1
0
0
1
n-l
-I-1
-1
1
0
0
-1
-1
0
0
1
and
the inverse
transformation is a reflection about the same
coordinate axis.
9. (a) Shear by a factor of /c = 4 in the y-direction
h as matrix
1
(d) Since k^O,
O'
4
1
k
(b) Shear by a factor of 1: = -2 in the x-direction
has matrix
1
0
0
1
n-l
-|-l
1
k
0
1
0
1
0
-k
1
.
1
and
, the inverse
transformation is a shear (in the opposite
direction) along the same axis.
-2
1 ■
11. (a) The geometric effect is expansion by a
factor of 3 in the x-direction.
113
SSM: Elementary Linear Algebra
Chapter 4: General Vector Spaces
23. (a) To map (x, y, z) to (x + kz, y + kz, z), the
0 k'
17. The line y = 2x can be represented by It
J.
standard matrix is
(a)
1
3
t
t + 6t
It
0
1
It
0+2r
2t
I
1
2
(b)
7
0
t
t
0 ^J[2r
t
matrix is
0
1
0
k
I
Shear in the yz-direction with factor k maps
(x, y, z) to {x,y + kx,z + kx).
The standard matrix is
y = X.
(c)
I
1
0 .
The image is x = r, y = r whieh is the line
0
k .
(b) Shear in the xz-direction with factor k maps
(x, y, z) to (x + ky, y, z + ky). The standard
■ 1 k o’
this is the line y= —x.
7
]
0 0
The image is x = It, y = It, and using
t = — X,
0
t
It
_1 0jL2r
t
1
0
0
k
1
0
k
0
I
Triie/False 4.11
The image is x = 2r, y = t, which is the line
(a) False; the image will be a parallelogram, but it
will ngt necessarily be a square.
2
(d)
(b) True; an invertible matrix operator can be
-I
0
t
-t
0
1
2t
2t
expressed as a product of elementary matrices,
each of which has geometric effect among those
listed.
The image is x = ■t, y = 2t, which is the line
y = -2x.
(c)
True; by Theorem 4.1 1.3.
(d)
True;
I
(e)
COS 60°
-sin 60°
sin 60°
COS 60°
t
2
t
2
2t
1
2
2t
0
0
n-l
1
-1
0
0
2
l-2^/3
-1
0
0
1
-I-1
-1
0
0
1
9
n/3+2
, and
0
1
0
1
0
1
1
0 ■
2
XU
■
■
The image is
,
•
and using t =
)' =
V3+2
I-2V3
’-2^3
x = —-— 6 }' =
n/3 + 2
(e)
f.
2
1
1
1
-1
maps the unit square to the
parallelogram with vertices (0, 0), (1, 1), (1, -1),
2 ^ X, t his is the line
and (2, 0) which is not a reflection about a line.
I-2V3
8 + 5V3
X or y = -
False;
1
(f)
X.
1 1
False; ^
^ maps the unit square to the
parallelogram with vertices (0, 0), (1,2), (-2, 1),
19. (a)
3
1
X _ 3x+y _ x'
6
2
y
6x+2y
and (-1, 3) which is not a shear in either thexory-direction.
3’
Thus, the image (x', y') satisfies y'=2x'
(g)
for every (x, 3’) in the plane,
True; this is an expansion by a factor of 3 in the
v-direction.
(b) No, since the matrix is not invertible.
Section 4.12
Exercise Set 4.12
1. (a) A is a stochastic matrix.
114
Section 4.12
SSM: Elementary Linear Algebra
(b) A is not a stochastic matrix since the sums of
9
so ^ = —. The steady-state vector is
the entries in each column are not I.
(c) A is a stochastic matrix.
Ife)
q=
(d) A is not a stochastic matrix because the third
I7
17
column is not a probability vector due to the
negative entry.
3. X,1 = Px0 -
17
9.
9. Each column of E is a probability vector and
3
1
0.5
0.6
0.5
0.55
8
2
i
6
0.5
0.4
0.5
0.45
i
3
J.
3
8
18
i
4
0.5
0.6
0.55
0.545
0.5
0.4
0.45
0.455
0.5
0.6
0.545
0.5455
1
1
0.5
0.4
0.455
0.4545
2
2
1
I
1
4
3
<72
0
2
i
0
I
L9'3j
0
24
X, = PX| =
, so P is a regular stochastic
9
matrix. The system (/- P)q = 0 is
X3 = PX2 =
X4 - PX3 =
0.5
0.6
0.5455
0.54545
0.5
0.4
0.4545
0.45455
0.5455
0.5454
X4 = P\q =
0
4
0.5
0
3
The reduced row echelon form of /- P is
0.4545 0.4546J[0.5
1
0
0
1
0
0
3
0.54545]
0.45455
-i. , so the general solution of the
0
5. (a) P is a regular stochastic matrix.
4
= — s, 9'2 =^ ^3 = s. Since q
system is
(b) P is not a regular stochastic matrix, since the
4
must be a probability vector,
entry in row 1, column 2 of P^ will be 0 for
3
11
\ = q\+q2+q-i = ^' s, so ^ = —
all A' > 1 .
(c) P is a regular stochastic matrix since
"li i'
P2 =
25
5
J_
4 ■
25
5
m
The steady-state vector is q =
7. Since P is a stochastic matrix with all positive
entries, P is a regular stochastic matrix.
3 _2lr
_2
4
3
4
3
II
probability that something in state 1 stays in
state 1.
-|
3
_ 0
0 ■
1 <72
3J'-
f 0
3
1 1
11. (a) The entry 0.2 =/j|1 represents the
The system (/- P)q = 0 is
4
4
II
(b) The entry 0.1 = p\2 represents the
probability that something in state 2 moves
-I 0
reduces to
1
0
L 4 f 0
0
to state 1 .
so the
0
0.2
(c)
8
general solution is q\ =-5, ^2
9
Since q
0.1
1
0.2
0.8 0.9j[o
0.8
The probability is 0.8.
17
must be a probability vector 1 =
-1-^2 = —
0.2
9
(d)
0.1
0.5
0.15
0.8 O.9JL0.5
0.85
The probability is 0.85.
115
SSM: Elementary Linear Algebra
Chapter 4: General Vector Spaces
13. (a) Let state 1 be good air. The transition matrix is P =
0.93
1
(b)
0
0.77
1
0.93
0.07 0.23j[o
0.07
0.95
0.55
0.05
0.45 ■
The probability is 0.93 that the air will be good.
(c)
3 0
0.922
0.858
0
0.858
1
0.078
0.142
1
0.142_
P
The probability is 0.142 that the air will be bad.
(d) P
0.2
0.95
0.2
0.63
0.8
0.05 0.45j[o.8
0.37
0.55
The probability is 0.63 that the air will be good.
15. Let state 1 be living in the city. Then the transition matrix is P =
0.95
0.03
0.05
0.97
. The initial state vector is
100
Xo =
(in thousands).
25
0.95
0.03
100
95.75
0.05
0.97
25
29.25
(a) Pxq =
P^x0 -
P^xo =
P\=
P^Xq =
0.95
0.03? flOO'
91.84
0.05
0.97
33.16
25
0.95
0.03?[100
88.243
0.05
0.97
36.757
25
0.95
0.03?[100'
84.933
0.05
0.97
25
40.067
0.95
0.03?[100
81.889
0.05
0.97
43.111
25
Year
1
2
3
4
5
City
95,750
91,840
88,243
84,933
81,889
Suburbs
29,250
33,160
36,757
40,067
43,111
(b) The system (/- P)q = 0 is
0.05 -0.03]
-0.05
_|"0
0 ■
0.03][^2
1
-0.6
0
0
The reduced row echelon form of /- P is
so the general solution is qy = 0.6s, ^2 “
1
1=
+^2 = Lbi', s =
1.6
= 0.625 and the steady-state vector is q =
0.6(0.625)
0.625
Since
0.375
0.625 ■
Over the long term, the population of 125,000 will be distributed with 0.375(125,000)= 46,875 in the city
and 0.625(125,0(X)) = 78,125 in the suburbs.
116
SSM: Elementary Linear Algebra
17. The transition matrix is P =
Chapter 4 Supplementary Exercises
_L
i
1
10
5
5
1
Column 3: 1 —
5 10 “2
4.
J. 1
5
10
5
2. ±
-L
1
i
10
10
2
5
P= 1
23
100
19
50
11
50
(a) p2 0
il
JL
29
50
20
50
0
43
27
1
100
100
5.
23
100
1
0
17
43
1
5
10
2
_L
3
^
10
5
lOj
3
i
i
10
10
5
1
7
1
5
10
2
_3
5
J_
10
100
23
The probability is
5
J.
The system (/- P)q = 0 is
50
0
1
10
1
100
L
10
9l
0
<72
0
373
0
The reduced row echelon form of /-P is
(b) The system (7- P)q = 0 is
9
'l 0 -f
3
1
10
5
5
0
0
-4
1_
_i
5
10
5
<72
0
0 0 oj
4 [<73
0
<72 = s, <73 =
__L _1
10
2
+ <72 +^3 = 35,
1
46
0
Since 1 =
5_
The reduced row echelon form of 7- P is
1
1 -1 , so the general solution is (7, = s,
5=
j
3
— and the steady-state vector is q = -^
47
1
0
, SO the general solution is
1
3
47
0 0
0
46
21. Since q is the steady-state vector for P,Pq = q,
66
hence
<7l = — 5, <72=—5, ^3=5.
47
47
Since \ = ci\+q2+q^ -
and
5, 5 =
47
P^ q = P*^"‘(Pq)= P^“'q = P^'^(Pq)= ■ ■ ■ = q
and P*^q = q for every positive integer k.
47
159
159
the steady-state vector is
q=
'46/' 47 y
46
47U59/
-53
47
47
159
159
True/False 4.12
47ll59j
159
M/'JI'l = 22
(a) True; the sum of the entries is 1 and all entries
are nonnegative.
(b) True; the column vectors are probability vectors
0.2893
35
and
(c) 120q = 120 0.4151 = 50
[o.2956
0.2 if
0.84
0.2
0.8
0.16
0.8 ■
0
35
(c) True; this is part of the definition of a transition
Over the long term, the probability that a car
ends up at location 1 is about 0.2893,
matrix.
location 2 is about 0.4151, and location 3 is
(d) False; the steady-state vector is a solution which
about 0.2956. Thus the vector 120q gives
the long term distribution of 120 cars. The
locations should have 35, 50, and 35 parking
spaces, respectively.
is also a probability vector.
(e) True
Chapter 4 Supplementary Exercises
19. Column 1: 1—^ !10
Column 2: 1
10
1
1. (a) u + v =(3+1,-2 + 5,4-h(-2))=(4,3,2)
(-l)u =((-1)3,0,0)=(-3,0,0)
5
J__3 _
To 5~10
117
SSM: Elementary Linear Algebra
Chapter 4: General Vector Spaces
(b) V is closed under addition because addition
I
in V is component-wise addition of real
numbers and the real numbers are closed
1
under addition. Similarly, V is closed under
scalar multiplication because the real
numbers are closed under multiplication.
0
1
0
1
0 , so
0 0 0
1
rank(A)= 2 and nullity(A) = 1.
(c) Axioms 1-5 hold in V because they hold in
1
0
1
0
1
0
1
1
0
1
0
(d) Axiom 7:
k(u + v)= k(i/i + V), «2 +^2’ “.3 +''3)
= a'(M|+V|),0,0)
=(ku^, 0, 0)+(kv^, 0, 0)
0
0
1
0
1
reduces to
(b) /l =
/?l
=
I
1 0 reduces to 0
0
9. (a) A = 0
1
0
1
0
0
1
0
1
0 0 0 0
, so rank(A)= 2 and
0 0 0 0_
-h ky
nullity(A) = 2.
Axiom 8:
(k + m)u =((k + m)u\, 0, 0)
=(L'm,, 0, 0)-t-(OTU|, 0, 0)
(c) The row vectors r,- will satisfy
= ku-\- nm
/
Axiom 9: k{inu)= k{mu\, 0, 0)
=
r, if i is odd
. Thus, the reduced row
r2 if i is even
echelon form will have 2 nonzero rows. Its
, 0, 0)
rank will be 2 and its nullity will be n - 2.
={km)u
11. (a) If Pi and pj are in the set, then
{p\ + P2)(-.«)= P\ i-x)-t- P2{-x)
= P\{x)+P2{x)
= {P\ + P2)ix)
(e) For any u =(M|, 112, ut,) with M2, M3 ^ 0,
lu =(IM|, 0, 0) (m|, M2> M3)= u.
1
3. The coefficient matrix is A = 1
5
so the set is clo,sed under addition.
and
s
Also
s
(/:pi )(-x)= k{p\(-x))= k{p\(x))={kp^ ){x)
det(A)= -(s -1)"(5 -t- 2), so for 5
1, -2, A is
so kp\ is in the set for any scalar k, and the
invertible and the solution space is the origin.
I
HA = I, then A = I
1
I
1
1
1
1
set is a subspace of
I
the subspace
consisting of all even polynomials. A basis
, which reduced to
is {1, x~, x^,
x^'"} where 2m = n if n is
even and 2m = n- \ if n is odd.
1
0 0 0 so the solution space will require 2
(b) If p| and P2 are in the set, then
(At + A2)(0)= f't(0)+ P2(0)= 0- so
Pi -I- P2 is in the set. Also,
(^Pl)(0)= k(p|(0))=/:(0)= 0, so kp\ is in
_0 0 oj
parameters, i.e., it is a plane through the origin.
■ 1
If 5 = -2, then A =
1
1 -2]
-2
1 , which
-2
1
reduces to 0
0
the set. Thus, the set is a subspace of
-1
the subspace consisting of polynomials with
constant term 0. A basis is
I -1 , so the solution space
0 0 oJ
{x, x^, x^,..., x").
will require 1 parameter, i.e., it will be a line
through the origin.
7. A must be an invertible matrix for Av I
Av2,..., Av,j to be linearly independent.
118
SSM: Elementary Linear Algebra
Chapter 4 Supplementary Exercises
, i > j, below the main diagonal is determined by the ajj entry above the main diagonal. A
13. (a) The entry
0
0
basis is ■ 0 0 0
1
0
1
0 0
0
0
0
0 0
1'
0
0 0
0 0
0
0
0
0
1
0
(b) In a skew-symmetric matrix,
0
1
=
0
0if/ = y
0 0
0
0 0
0 0 0
0
0
0
1
0 0 0
0 0
0
1
0
1
0 0
1
. Thus, the entries below the main diagonal are determined
-aji if i ^ j
0
by the entries above the main diagonal. A basis is
0
0
0
1
0
0 0
0 0
0
0 0
0
0
1
0 0 0
-1
0 0
0-1
0
-1
1
15. Every 3 x 3,4 x 4, and 5x5 submatrix will have determinant 0, so the rank can be at most 2. For i < 5 and j< 6,
0
det
a/6
= -a^^a^j so the rank is 2 if any of these products is nonzero, i.e., if the matrix has 2 nonzero
entries other than ajg. If exactly one entry in the last column is nonzero, the matrix will have rank 1, while if all
entries are zero, the rank will be 0. Thus, the possible ranks are 0, 1, or 2.
119
Chapter 5
Eigenvalues and Eigenvectors
Section 5.1
X-3
5. (a)
0
Exercise Set 5.1
1.
4
0
1
1
Ax= 2
3
2
2
X] _ 0
0 0 xi _[0
^ = 3 gives
5
0
^+ lj[x2
-8
0 ■
-8 4 X2
10 = 5x
J 0 4j[l
0
5
reduces to
-8
2 , the
4
0
X corresponds to the eigenvalue X = 5.
A.-3
general solution is Xj
0
3. (a) det
I
1
0
Since
0
^2
o*'
=(l-3)(^+ l)
X+ l
•^1
= X'^-2X-3
i
_
i
2
1
= S 2 , so
^2
1
s
is a basis for
The characteristic equation is
the eigenspace corresponding to ^ = 3.
^^-2>.-3= 0.
■-4 oirx, i_[o'0 .
X = -1 gives
(b) det
A.-10
9
-A
X+2
= (3.-10)(X + 2) + 36
= A,^-8^ + 16
The characteristic equation is
(c)
det
-3
-4
X
0
reduces to
0
X]
= X^-12
X+2
1
-1
X-2
1
0
0
X
(b)
1-10
9
-4
0
the general
is a basis for the
1
X|
_ 0
0
= (^+2)(?i-2) + 7
^ = 4 gives
+ 3 = 0.
Since
-6 9 X, _[0
_-4 6j[x2j^
0 ■
-6
9
1
-4
6
reduces to
2 , the
0
0
general solution is Xj = —5, X2 = s, or
2
The characteristic equation is A. = 0.
^1
_
^2
det
0 ’
X+2\ X2
'y
(f)
0
eigenspace corresponding to ^ = -1.
The characteristic equation is
det
1
0
, so
■^[1
= ^2+3
(e)
Sinc e
X2
0
.^2
The characteristic equation is ^^-12 = 0.
(d) det
0
-8
0
solution is X] =0, X2 = s, or
A,^-8^+16 = 0.
'X
-4
—8
A-1
0
0
A-1
= (A-1)^ = A^-2A + 1
3
3
= S 2 , so
2
is a basis for
1
S
the eigenspace corresponding to A = 4.
The characteristic equation is
(c)
A^-2A+1 = 0.
A -3] xi ^[0
-4
A
X2
X = yl\2 gives
120
0
4v2 -3 lr-*^i]^ro'
-4 V12 U2J Lo.'
SSM:Elementary Linear Algebra
Section 5.1
'yjil -3’ reduces to
Since
-4 ^/^2_
7. (a) The eigenvalues satisfy
1
?l^-6^^ +ia-6=0 or
0
0
(A,- 1)(A,- 2)(X.- 3)= 0, so the eigenvalues
the general solution is
=-p
are A,= 1, 2, 3.
5, X2=S,
(b) The eigenvalues satisfy A,^-2A,=0 or
^1
or
_
=S M , SO yli2 IiS a
.^2
1
X(X+^^2)(X-y^2)=0, so the eigenvalues
1
are A,=0,-^/2, yfl.
basis for the eigenspace corresponding to
A.= Vi2.
(c) The eigenvalues satisfy +8A,^ + A,+8=0
or (A,+8)(A,^ +1)=0, so the only (real)
X =-y/l2 gives
’-Vi2 -3 Ir^ii^ro'
-4
[o’
*-^/^2 -3
-4 -^I^2
Since
eigenvalue is A,= -8.
(d) The eigenvalues satisfy A,^-A,^-A,-2 =0
or (A-2)(A^ + A+1)=0, so the only (real
reduces to
eigenvalue is A = 2.
^/^2 , the general solution is
0
0
(e) The eigenvalues satisfy
A^-6A^+12A-8=0 or (A-2)3=0, so
3
s, X2 = s, or
Xi =-
the only eigenvalue is A = 2.
^1
=S
.^2
3
3
3_e
^/^2 , so
^/i2
1
1
s
(f) The eigenvalues satisfy
i
IS
A^-2A^-15A+36 =0 or
a basis for the eigenspace corresponding to
(A+4)(A-3)^ =0, so the eigenvalues are
x=-^Jr2.
A = -4,3.
(d) There are no eigenspaces since there are no
real eigenvalues.
9. (a) det
(e)
'A oi JT] _ro'
0 A
X2
A
0
-2
0
-1
A
-1
0
0
-1
A+2
0
0
0
0
A-1
0
'A
A = 0 gives
'O 0l Xi
0
0’
1
=(A-l)det -1
0 oJ[x2j“Lo_‘
Clearly, every vector in
1
-2 ■
0
0
0
A
-1
-1
A+2
=(A-l){A[A(A+2)-l]-2(l-0)}
is a solution, so
=(A-1)(A^+2A2-A-2)
is a basis for the eigenspace
= A^+a3-3A2-A+2
corresponding to A = 0.
The characteristic equation is
A'* + A^-3A^-A+2=0.
^1
0
A-lJ X2
0
A-1
(f)
0
0
A = 1 gives
0 Ol xi
0
0 ■
0 0 X2
fy
Clearly, every vector in R is a solution, so
1
0’
0
1
is a basis for the eigenspace
corresponding to A = 1.
121
SSM:Elementary Linear Algebra
Chapter 5: Eigenvalues and Eigenvectors
?i-10
9
0
0
-4
X+2
0
0
0
0
^+2
7
0
0
-1
1-2
(b) det
X +2
0
0
0
X+2
1
= a-10)det
[0
-1 X-2
^
-9det 0
0
0
0
X+2
1
-1
X-2
J ={X -10)(X+2)[{X+2)iX-2)+7]-9(-4)[(X+2){X-2)+7]
=(X^ -8^-20+36)(^^ -4+7)
=(A,^-8X+16)(^^+3)
=:L'^-8^^ +]9X,^-24^+48
-8X,^ +19A^ -24A+48 = 0.
The characteristic equation is
11. (a)
A
0
-2
0
X,
0
-1
A
-1
0
^2
0
0
-1
A+2
0
■^3
0
0
0
0
A-1
X4
0
1
A = 1 gives
1
Since
0-2
0
X,
1
-1
0
X2
0
0-1
3
0
^3
0
0
0
0
X4
0
-1
0
0-2
0
1
1
-1
0
0
1
-3
0
0-1
3
0
0
0
0
0
0
0
0
0
0
0
0
-1
X4 = r or
0
reduces to
2
0
2
0
X2
3s
3
0
3
0
J^3
s
1
0
1
0
X4
t
0
1
0
1
+t
= 5
0-2
0
^1
0
-1
0
X2
0
0-1
0
0
■^3
0
0
0
-3
X4
0
0
-2
0
-2
0
-1
-2
-1
0
0-1
0
0
0
0
-3
0
1
reduces to
X,
-5
-1
^2
0
0
^3
s
1
0
0
-1
-1
0-2
= 3i, X3 = s.
is a basis for the eigenspace corresponding to A = 1.
0
1
0
0
1
0
0
0
0
0
1
0
0
0
0
0
, so
, the general solution is X] = -s,
=0, X3 = s.
is a basis for tbe eigenspace corresponding to A = -2.
1
0
0
0
-1
-1
0
^2
0
0-1
1
0
^3
0
0
0
-2
X4
0
0
, the general solution is X) = 2s,
-1
= s
X4
A = -1 gives
, so
-2
-1
A = -2 gives
X4 = 0 or
0-20
2s
-2
Since
0
122
SSM: Elementary Linear Algebra
0-2
Section 5.1
15. If a line is invariant under A, then it can be
0
-1
0
0
0
expressed in terms of an eigenvector of A.
Since
reduces to
0
0
1
0
2
0
0
1
-1
0
0 0
0
1
0 0
0 0
0
-2
(a) det(l/-A)=
■^3
5
=(l-3)(l-2)
1
X
0
-2
1-1
1
0
-1
1 = 3 gives
1
is a basis
, so
0
0
1-4
-2
-2
= S'
X4
0
for the eigenspace corresponding to 1 = -1
-1
1
-2
2
9
0
0
•^1
0
-4
1+2
0
0
X2
0
0
0
1+2
7
^3
0
-2
1
0
-2
1
1-2
0
0
-6
9
0
0
X
I
0
-4
6
0
0
X2
0
0
0
6
-7
^3
0
0
0
2
X4
0
1 = 4 gives
Since
X4
-6
9
0
0
-4
6
0
0
0
0
6
-7
0
0
X
-2 2j[y,
reduces to
under A. 1 = 2 gives
0
0 ■
1
0
0 ’
so a general
-2
llfx
0
-2 lj[y
can be reduced to
0 ■
2
-1
0
0
, so a
general solution is 2x = y. The line y = 2x is
invariant under A.
(b) det(l/-A) =
" -*=1^
+1
1
Since the characteristic equation 1^ +1 = 0
has no real solutions, there are no lines that
reduces to
are invariant under A.
2
(c)
1 -| 0 0
0
0
1
0
0
0
0
1
0
0
0
0
, the general solution is
X2 = s, X3 = 0, X4 = 0 or
•^1
3
1
2
2
X2
5
X3
0
0
0
X4
0
0
0
= s
1
,
-3
0
1-2
-3 1 X
0
0
l-2j[y
0
0
0
X ^
= (^-2)2
0
_0 oJ[y_
0 ■
^ reduces to
0
1
0
0
0
invariant under A.
23. We have that Ax = lx.
A-' (Ax) = /x = X, but also
13. Since A is upper triangular, the eigenvalues of A
A“'(Ax) = A"'(lx) = l(A''x) so A“'x = ^x.
1
are the diagonal entries; 1, —, 0, 2. Thus the
’ 2
1
Thus —% is an eigenvalue of A *, with
V2
, the
general solution is y = 0, and the line y = 0 is
the eigenspace corresponding to 1 = 4.
eigenvalues of A^ are 1^ = 1,
1-2
1-2
Since
is a basis for
so
det(l/-A) =
1 = 2 gives 0 -3
3
X| = —2
1
solution is X = y. The line y = x is invariant
1-10
(b)
1-1
= 12-51+6
, the general solution is
-2^
s
1
-2
=(l-4)(l-l)+ 2
X| = -2s, X2 = s, X3 =5, X4 =0 or
Xi
A.-4
512
corresponding eigenvector x.
0^ =0, and 2^ =512.
123
SSM:Elementary Linear Algebra
Chapter 5: Eigenvalues and Eigenvectors
5. Since the geometric multiplicity of an eigenvalue
is less than or equal to its algebraic multiplicity,
the eigenspace for ^ = 0 can have dimension 1 or
2, the eigenspace for A, = 1 has dimension 1, and
the eigenspace for X-2 can have dimension 1,
25. We have that Ax = Ax.
Since ^ is a scalar,(s'A)x = s(Ax)= x(Ax)=(sA)x
so sX is an eigenvalue of sA with corresponding
eigenvector x.
True/False 5.1
2, or 3.
(a) False; the vector x must be nonzero to be an
7. Since the matrix is lower triangular, with 2s on
eigenvector—if x = 0, then Ax := Ax for all A.
the diagonal, the only eigenvalue is A = 2.
0 0
(b) False; if A is an eigenvalue of A, then
ForA = 2, 2/-A =
det(A/- A)= 0, so {XI- A)x = 0 has nontrivial
solutions.
(c)
-1
0
, which has rank 1,
hence nullity 1. Thus, the matrix has only 1
distinct eigenvector and is not diagonalizable.
True; since p{0)=0^ +1 0, A = 0 is not an
9. Since the matrix is lower triangular with
diagonal entries of 3, 2, and 2, the eigenvalues
are 3 and 2, where 2 has algebraic multiplicity 2.
0 0 0l
eigenvalue of A and A is invertible by Theorem
5.1.5.
(d) False; the eigenspace corresponding to A
contains the vector x = 0, which is not an
ForA=3, 3/-A= 0
eigenvector of A.
0
(e) True; if0 is an eigenvalue of A, then 0=0 is
9
an eigenvalue of A , so A is singular.
[-1 0 o'
For A = 2, 2/-A=
(f) False; the reduced row echelon form of an
3
0
-1
0 0 which has rank
0
2, hence nullity 1, so the eigenspace
corresponding to A = 2 has dimension 1.
The matrix is not diagonalizable, since it has
eigenvalues that are not 1. For instance, the
8
0
0-1
invertible nxn matrix A is /„ which has only
one eigenvalue, A = 1, whereas A can have
matrix A =
1
2, hence nullity 1, so the eigenspace
corresponding to A = 3 has dimension 1.
'y
9
1 0 which has rank
-1
only 2 distinct eigenvectors.
has the eigenvalues A = 3
11. Since the matrix is upper triangular with
diagonal entries 2, 2, 3, and 3, the eigenvalues
and A = -l.
(g) False; if 0 is an eigenvalue of A, then A is
singular so the set of column vectors of A cannot
are 2 and 3 each with algebraic multiplicity 2.
0 1 0 -l]
be linearly independent.
For A = 2, 2/-A =
Section 5.2
0 0-1
1
00-1
-2
0 0
,. . ,
0 -1
Exercise Set 5.2
1.
rank 3, hence nullity 1, so the eigenspace
corresponding to A = 2 has dimension 1.
Since the determinant is a similarity invariant
1
and det(A)= 2- 3 = -1 while
det(B)= -2-0 = -2, A and B are not similar
For A = 3, 3/- A = 0
matrices.
0
1
1 -1
0
0 0
3. A reduces to
1
1
0
0
0
1
0
0 0
1
1
u- u u
0 -2
0
0
rank 3, hence nullity 1, so the eigenspace
corresponding to A = 3 has dimension 1.
The matrix is not diagonalizable since it has only
2 distinct eigenvectors. Note that showing that
the geometric multiplicity of either eigenvalue is
less than its algebraic multiplicity is sufficient to
show that the matrix is not diagonalizable.
while B reduces to
2 0
0 0
0-1'
1 , so they have different ranks. Since
0 0 0
rank is a similarity invariant, A and B are not
similar matrices.
124
SSM: Elementary Linear Algebra
13. A has eigenvalues
Section 5.2
= 1 and
(the
0
Since
diagonal entries, since A is triangular).
^X-\
(kl -A)n = 0 is
A,] =1 gives
0
0
-6 >.+ lJLx2j [o '
0 0 X, _[O
-6
2
-1
0
0
1
^1
3 , the general
solution is
^2 ~
X,
or
eigenspace corresponding to A, = 2.
[0 0 0
1
1
-2
•^1
corresponding to A| =1.
-2
X2
=5
^3
-2 0 X, ^[0
0 +r 1 , and P2 =
1
0
0 0’
X2
0
=^ 1
0
2
P~'AP = 0 0
1
1
is a basis for the eigenspace
3 0
1
i 0
diagonalizes A.
1
0 1 0j[0 0 3
3
O'
3 0
0 0
17. det(A/-A)=
0
A2 0
0 A2
0
1 -10]
-2 0
0
2 0 0
0
1
1
0
i 0
2 0 -2
0
1
A2
3
A +1
-4
2
3
A-4
0
3
-1
A-3
= x^-6x^h-11x-6
=(x-l)(x-2)(x-3)
15. A has eigenvalues A| = 2 and A2 = 3 (the
The eigenvalues are A] =1, A2 = 2, and A3 = 3,
diagonal entries, since A is triangular).
each with algebraic multiplicity 1. Since each
(A/- A)x = 0 is
■A-2
0
2
X|
0
0
A-3
0
•^2
0
0
0
A-3
•^3
0
0
A] = 2 gives 0-1
0
0
0
A| 01
0
1
0
0
0
-3 lj[6 -lj[i 1
0
0 1
0
'A, 0
corresponding to A2 =-l.
-1
0
diagonalizes A.
, and
0
P ‘AP =
1
0
1
Thus, P =[P| P2]=
0 , P3 = 1
0
^2 = 3. Thus P =[p] P2 P3]= 0
the general
X,
solution is X] = 0, X2 = ^ or
1
0
'1 -2 0
1
reduces to
P2 =
-2
0
is a basis for the eigenspace corresponding to
0 ■
0
-6 0
0
,^3
general solution x^ = -2s, X2 =t, X3 = s, or
and Pi = 3 is a basis for the eigenspace
Since
0
3^1
A2 = 3 gives 0 0 0 X2 = 0 which has
=S ^ ,
1
-6 0 X2
1 , the
0
i
.^2
A2 =-l gives
0 0
1
'1 0 2
1
0
0 0 0
0
i
reduces to
0
1
X2 = ^ 0 and P| = 0 is a basis for the
1
0
Since
0
reduces to
general solution is Xj = s, X2 = 0, X3 =0 or
0 ■
-6 2 X2
2
0
0
0
x.
0
0
0
2
0
eigenvalue must have nonzero geometric
multiplicity, the geometric multiplicity of each
eigenvalue is also 1 and A is diagonalizable.
(A/-A)x =0 is
-4
2
X]
X2 = 0
3
A-4
0
^2
0
0
3
-1
A-3
X3
0
^1
3^3
'A + l
0
125
0
SSM: Elementary Linear Algebra
Chapter 5: Eigenvalues and Eigenvectors
2
-4
2
^1 = 1 gives 3 -3
3
-1
0
-2
X|
0
•^2
0
^3
0
diagonalizes A and
0
^1
p~'ap =
0
0
-4
2
1
0
3 -3
0
reduces to 0
1
_3 -1 -2_
0
0
2
Since
-1 , the
0 0
0
2 0
^3
0 0
0
3
multiplicity 2) and ^2=1^2 = I must also have geometric multiplicity 1.
'^1
is a basis for the
= 5 1 and P| = 1
1
0
19. Since A is triangular with diagonal entries of 0,
0, 1, the eigenvalues are
=0 (with algebraic
0
general solution is x^ =s, X2= s, x^ = s, or
X2
0
0
0
X
0
0
•^3
(X/-A)x = 0 is
0
eigenspace corresponding to I] = I.
3^
2
^2=2 gives 3 -2
0
3
Since
1
3
-4
2
3
-2
0
■^2
0
^3
0
■ 0 0
=0 gives
0 0
-3 0
reduces to
3
0
-3 0 ?^-lJ[x3_
0
-1
0
•^2
1
0
0
1
-I
0
0
0
3
0
0
0
0
0
0
-3
0
-1
Since
0
0
0 ^2
0
-1
0
L-^3j
reduces to
1
0
i
0
0
0 , a
0
0
0
2
the general solution is JC| = — 5, X2= s, 3:3 = 5
general solution is X] =
or x^ = 2t, X2 = 3r, X3 = 3r which is
2
2
X2 = ? 3
and P2 = 3
3
3
X|
•^3
X| =5, X2 = t, X3 = -3i' which is
^3 = 3 gives 3
3
4-4
Since
-4
^2
2
^1
0
-1 0
X2
L^3
0
-1
0
1
3
-1
0
3
-1
0
reduces to
0
1
0
I . Thus X| =0 has
0
0
1
Pi =
0 , P2 =
-3
i
are a basis for the
1
0
eigenspace corresponding to
4
, a
X2 = I gives
0
X2 = —
4
or
Since
X| =t, X2 = 3t, X3 = 4f which is X2
= t
_X3
=0.
■ 1
0
0
Xi
0
0
I
0
X2
0
-3
0
0
^3
0
1
0
0
0
1
0
-3
0
0
reduces to
1
0
0
1
0
0
0
0 , a
0
3
general solution is X| =0, X2 = 0, X3 = ^ or
4
is a basis for the eigenspace
-^1
0
Xj
0
0
and P3 = 0 is a basis for the
•'•'3
4
eigenspace corresponding to ^2=1-
corresponding to ^3 =3.
I
Thus P = [p|
-3
3
general solution is X] = —
4
and P3 = 3
+t
geometric multiplicity 2, and A is diagonalizable.
0 ' -I
0
0
0
= 5
■^3
0
2
I
•^1
is a basis for the
eigenspace corresponding to X2 - 2■4
X2=t, X3 = r or
P2
P3]= 1
1
2
3
3
Thus /’ = [Pl
3
4
126
P2
P3]=
‘ 1
0
0
0
1
0
-3
0
1
SSM: Elementary Linear Algebra
Section 5.2
diagonalizes A and
r>t| 0
p~'ap= 0 ?i| 0
0
1
0
0 0 0
0
1
-1
1
0
0 0
0
0
0
0
0
0
0
0
0
0
0
0 ;l2
1
on the diagonal, the eigenvalues are 3.| = -2 and
X2 = 3, both with algebraic multiplicity 2.
(X/-A)x = 0 is
0
0
0
x.
0
0
1+2
-5
5
X2
0
0
0
A,-3
0
T3
0
0
0
0
3.-3
3.| = -2 gives
X|
0
5
X,
0
0 0-5
0
•»^3
0
-5
X4
0
0
0
0 0-5
0
0
5
0
+t
=S
^3
1
0
X4
0
1
0
-1
are a basis for the
. P4 =
P3 =
0
0
1
1
0
0 0
0
0
0 0-5
0 0
X|
X2
Thus 3.2 = 3 has geometric multiplicity 2, and
0
X4
, a general solution is xj = 0,
X2=s — t, x^= s, X4 = t, or
21. Since A is triangular with entries -2,-2, 3, and 3
~l+2
0 0
0
1
0
eigenspace corresponding to 3.2 = 3.
Since the geometric and algebraic multiplicities
are equal for both eigenvalues, A is
diagonalizable.
1
reduces to
Since
0 0-5
0 0
0 0
1
1
0
0
0
^=[P| P2 P3 P4]=
-5
0
1
1
-1
0 0
1
0
0 0 0
1
0
0 0 0
0
0
0
0 0
0
diagonalizes A, and
, a general solution is X| = s,
0
-1
0 0 0 0
P~‘AP =
X2 =t, X3 = 0, X4 =0 or
0
X,
1
^2
0
+f
=5
•^3
0
L^4j
0
0
0 3,, 0 0
0 0 ^2 0
0 0 0 32
-2
0
0
0 0 0
0-2 0 0
0
Thus 3| = -2 has geometric multiplicity 2, and
0
0
0
0 0
3 0
3
0
3+ 1
0
Pi =
0
- P2 = 0
0
0
-7
1
23. det(3/-A)= 0
3-1
0
0
-15
3+2
are a basis for the
= 3^+23^-3-2
eigenspace corresponding to 3[ = -2.
5
32 = 3 gives
5
0
0
5
0
0 0
•^1
=(3+ 2)(3+ l)(3-l)
0
0
5
5
X2
0
0
0
0 0
•«3
0
Thus, the eigenvalues of A are 3] = -2,
32 = —1, and 3^ = 1.
0
0
0 0
X4
0
(3/-A)x = 0 is
-5
'3+ 1
0 0
-5
5
reduces to
Since
0 0
0 0
0 0
0
-7
1
Xi
0
0
3-1
0
4^2
0
0
-15
3+2
,^3j
-1
-7
0
3, =-2 gives
127
0
1
X]
0
-3 0
X2
0
0 -15 0j[x3
0
0
SSM:Elementary Linear Algebra
Chapter 5: Eigenvalues and Eigenvectors
-1
Since
1
1
-3 0
reduces to 0
-7
0
0
-15
general solution is
1
and D = P~‘AP
0,a
0 0
0
-I
0 -1
‘0 -5
0
=
1
1
1
0
1
1
0 0
1
0
1
5
A-i 0 o'
0 ^2 0
0 0 >.3
1
X2 = i 0 , so Pi = 0 is a basis for the
,^3
7 -1
0
0 1 oJ[ 0 15 -2
=s, X2= 0, X3 = s, or
1
llf-l
4-1
1
-2 0 0]
eigenspace corresponding to A.] = -2.
0-1 0
'0
-7
X2 =-l gives 0
0
0
Since
-7
0
0
xi
0
-2 0
^2
0
X3
0
-15
1
1
-2 0
-15
1
0
0
1
reduces to 0 0
1
0
a" =PD^'p-^=P
0
0
A3 = 1 gives 0
0
Since
1
x^
0
0 0
•^2
0
^3
0
-15
2
-7
1
0
0
0
0
-15
3
3
reduces to
1
I
0
1
-1
0
0
0
5
1
1
1
A-2
1
0
1
A-3
(A/-A)x =0 is
1
'A-3
1
0
Xi
0
1
A-2
1
■^2
0
A-3
X3
0
so
0
5
1
-2
1
0
Xi
0
1
-1
1
X2
0
0
1
-2
,^3
0
A] = 1 gives
is a basis for the eigenspace
5
corresponding to A3 = 1.
A’ = [Pi
P2
1
1
P3]= 0 0
1
1
0
0
The eigenvalues of A are A| =1, A2 = 3, and
A3 =4.
X| =t, X2= t, X3 = 5t which is X2 = r 1
1
0
-2048
=(A-l)(A-3)(A-4)
, a
general solution is X] =-^,
X2=-s,
X3 =5 or
5
5
P3 = 1
1
0 10,245
= A^-8A2+19A-12
0
X3
1
A-3
25. det(A/-/l)=
1
x^
0
-1 10,237 -2047'
eigenspace corresponding to A2 = -1.
-7
1
0-1 0 P~‘
0
0
'2
0
II
-2048 0 0]
=P
X2 =s 0 so P2 = 0 is a basis for the
•*3
P“‘
0
(-1)
0
1
0
0
II
0
1 ,a
0 0 0
1
0
(-2)
general solution is jC] = s, X2= 0, ;C3 = 0, or
^1
1
Since
diagonalizes A,
-2
1
0
1
-1
1
0
1
-2
reduces to
1
0
0
1
0
0
-1
-2 , a
0
general solution is x, = ^, X2 = 2s, X3 = s or
5
1
1
X2 =s 2
and Pi = 2
1
1
Xi
•*3
is a basis for the
eigenspace corresponding to A] = 1.
'0 1 0]f'x,
At = 3 gives
128
0
1
1
1
JC2
0
0
1
0
■^3
0
SSM:Elementary Linear Algebra
Since
0
1
0
1
1
I
0
1
0
reduces to
Section 5.2
I
0
1
0
1
0 , the
31. If A is diagonalizable, then P */lP = £). For the
sameP, P^'a'^P
-I
-1
-1
= P~'A{PP^')A(PP~')A ■ ■ ■(PP“‘)AP
0 0 0
general solution is X] =5, X2 = 0, X3 = -i or
1
X,
with k factors of A
1
1
I
1
={P~'AP)(p-'AP) - iP~‘AP)
X2 = s 0 and P2 = 0 is a basis for the
k factors off 'AP
-1
^3
eigenspace corresponding to X2 = 3.
'1
1
0
X3 = 4 gives 1 2 1
0
Since
1
1
0
1
2
1
0
1
1
1
X,
0
•^2
0
•3^3
0
reduces to
1
0
0
1
,a
X3 = 5, or
is a basis for the
1
1
(b) If A is diagonalizable then each eigenvalue
has the same algebraic and geometric
multiplicity. Thus the dimensions of the
eigenspaces are:
eigenspace corresponding to X3 = 4.
r1
1
1
Thus F =[pi P2 P3]= 2 0
1
X = 1: dimension 1
1
-1
X = 3: dimension 2
X = 4: dimension 3
diagonalizes A and
0
1
p-^AP = D= 0 X2 0
rxi
0
0
0
0
1
1
1
6
3
6
1
0
0
0
(c) Since only X = 4 can have an eigenspace
3 0
with dimension 3, the eigenvalue must be 4.
0 0 4
X3
True/False 5.2
1
(a) True; use P = l.
2
2
i
3
is
dimension is 1, 2, or 3.
X2 = s -1 and P3 = -1
X3
0
1
1
k
33. (a) The dimension of the eigenspace
corresponding to an eigenvalue is the
geometric multiplicity of the eigenvalue,
which must be less than or equal to the
algebraic multiplicity of the eigenvalue.
Thus, for X,= 1, the dimension is 1; for
X == 3, the dimension is 1 or 2; for X = 4, the
-1
0 0
general solution is X| =s, X2 =
X,
k
Since D is a diagonal matrix, so is Z) , so A
also diagonalizable.
_1
3
i
3.
(b) True; since A is similar to B, B = F, * Afj for
-1
A” = PD'^P
some invertible matrix Fj. Since
I
1
6
3
6
i
0
1
1
1
2
0-1
1
1
1"
0
0
3"
0
0
0
0
n
4
C, C = P2^BP2 for some invertible matrix ^2Then C = P2‘'5^2
= P2'(Pf^AP0P2
= iP2'P{-^)AiP^P2)
={P^P2)-'A{P^P2).
1
2
2
i
_i
i
3
3
3
27. Using the result in Exercise 20 of Section 5.1,
one possibility is P =
X,=
-b
-b
a — 'K\1
a — X9
where
2 a + d + yjia-df +4bc
is similar to
Since the product of invertible matrices is
invertible, P = P\P2 is invertible and A is similar
and
to C.
1
(c) True; since 5= P AP, then
^2 = 2 a +d -4ia-df'+4bc
-1
fi-' =(p-‘APr‘ =P“'a“'p so a
are similar.
129
-1
and B
SSM: Elementary Linear Algebra
Chapter 5: Eigenvalues and Eigenvectors
(d) False; a matrix P that diagonalizes A is not
unique—it depends on the order of the
eigenvalues in the diagonal matrix D and the
choice of basis for each eigenspace.
Im(A)=
Im(-5/)
Im(4)
Im(2-0 Im(l + 5()
-5
0
-1
5
det(A)= -5/(1 +5/)-4(2-/)
= -5/+ 25-8+ 4/
= 17-/
I
(e) True; if P AP = D, then
=
=D~\ so A
-I
tr(A) = -5/ + (l +5/)= 1
is also
11.
diagonalizable.
u-v =(/, 2/, 3)-(4, -2/, 1 +/)
= /(4)+2/(^)+ 3(r+/)
=(F“'aP)^ P'^a''{p-'i, so A^
(f) True;
= 4/+ 2/(2/)+ 3(1-/)
= 4/-4+ 3-3/
is also diagonalizable.
= — 1 +/
(g) True; if A is n x n and has n linearly independent
eigenvectors, then the geometric multiplicity of
each eigenvalue must be equal to its algebraic
multiplicity, hence A is diagonalizable.
11 • w =(/, 2/, 3)■(2-/, 2/, 5+ 3/)
=/(2^)+ 2/(^)+3(5+ 3/)
= /(2+/)+ 2/(-2/)+ 3(5-3/)
= 2/-l +4+ 15-9/
= 18-7/
(h) True; if every eigenvalue of A has algebraic
multiplicity 1, then every eigenvalue also has
geometric multiplicity 1. Since the algebraic and
geometric multiplicities are the same for each
eigenvalue, A is diagonalizable.
V • w =(4, -2/, 1 +/)•(2-/, 2/, 5+ 3/)
= 4(2^)-liili)+(1 +/)(5+ 3/)
= 4(2+/)-2/(-2/)+(1 + /)(5-3/)
= 8+4/-4+5+ 2/+3
= 12+6/
Section 5.3
vu = 4(-/)-2/(-2/)+(l +/)(3)
Exercise Set 5.3
= -4/-4+3+ 3/
-i
1.
u =(2-/, 4/, l +/)=(2+/, -4/, 1-/)
=u
V
11 •(V + w)=(/, 2/, 3) ■(6-/, 0, 6+4/)
= /(6+/)+ 2/(0)+ 3(6-4/)
Re(u)=(Re(2 - /), Re(4/), Re( I + /)) =(2, 0, 1)
Im(u)=(Im(2 - /), Im(4/), Im( I +/))=(-1,4, 1)
= 6/-l + 18-l2/
|| = Vl2-/|“+|4if+|l +(f
u
= 17-6/
=(-l +/)+(18-7/)
+1
=u
= ^5+ 16+2
v +u
w
kiu ■ v)= 2/(-l + /)= -2/- 2 = -2- 2/
(ku)■ V =(2/(/, 2/, 3)) ■(4,-2/, 1 + /)
=(-2, -4, 6/)-(4, -2/, 1 +/)
= -2(4)-4(2/)+ 6/(1-/)
= V^
5. If /x-3v = u, then /x = 3v + u, and
= -8-8/+6/+6
X = -(3v + u)= -/(3v + u).
= -2-2/
= k(u ■ V)
X = -/(3(1 +/, 2-/, 4)+(3-4/, 2+ /, -6/))
= -/((3+ 3/, 6-3/, 12)+(3+ 4/, 2-/, 6/))
13.
= -/(6+ 7/, 8-4/, 12+6/)
=(7-6/, -4-8/, 6-12/)
u-v =(/, 2/, 3)-(4, 2/, 1-/)
= /(4)+ 2/(-2/)+ 3(l +/)
= 4/+4+3+ 3/
= 7+ 7/
-5/
7.
4
5/
4
A=
w • u =(u ■ w)= u • w = 18-7/
2-/ 1+5/J L2+' 1-5/
Re(/l)=
Re(-5/)
Re(4)
Re(2-/) Re(l + 5/)
0
4
2
1
(u- v)-w-u = 7+ 7/-(18-7/)
= -11 + 14/
= -11-14/
130
SSM: Elementary Linear Algebra
Section 5.3
15. tr(/l)=4 + 0 = 4
det(A)=4(0)- l(-5)= 5
The characteristic equation of A is
21. Here a= I, b =
The angle from the
positive x-axis to the ray that joins the origin to
n
-1
the point
-4X +5 = 0, which has solutions X = 2± i.
X-4
5
-2-i
|
A =2-; gives
-2-i
Since
-1
-1
W =|l-/V3|= ViT3=2
0 ■
A-9
5
23. tr(A) = -1+7 = 6
det(A)= -l(7)-4(-5)= 13
Xi _ 0
0 ■
2-/J[_X2
1
^ , reduces to
2-i
3
1
X| _ 0
(XI -A)=0 is
is ^=tan
The characteristic equation of A is
-2+;
0
0
, a
-6A,+13=0 which has solutions A = 3 ± 2i.
For A = 3 - 2i, det(A/- Ajx = 0 is
general solution is X[ =(2-i)s, X2 = s or
x.
“4-2;
2-i
2-i
=5
is a basis for the
so X| =
5
-4
Xi _ 0 . Since
0
-4-2;J1_X2_
X2
_
[2+;
X2 = X| =
4-2;
5
-4
-A-2i
1 1 +^; , a
reduces to
eigenspace corresponding to Aj =2-;.
0
is a basis for the eigenspace
1
1.
general solution is x^ = -\-—i s, X2=s or
V
corresponding to X2 =A| = 2+ i.
=t
^1
0
-1
A-3
x-,
0 ■
2
the eigenspace corresponding to A = 3 - 2i.
A^ -8A +17 =0 which has solutions A = 4 ±;.
2
■2l
Re(x) =
-1
2 and Im(x) =
Thus A = PCP
-i
2
A| =4-i gives
■1-;
-;
2
-1
0 ■
JL^2j
reduces to
\-i
l-i
so
X
1 -
1
-1 + ;
0
0
0
and
-2
2
3 ■
A^ -1OA + 34 = 0 which has solutions
A = 5 ± 3;.
ForA = 5-3;, (A/-A)x = 0is
is a basis for the eigenspace
“-3-3;
-6
3
3 - 3;
Since
19. Here a = b = 1. The angle from the positive xaxis to the ray that joins the origin to the point
tan
-1
2
The characteristic equation of A is
is a basis for the
corresponding to X2 =A| =4 + i.
(1, 1) is
-2
25. tr(A) = 8 + 2= 10
det(A) = 8(2) - (-3)(6) = 34
eigenspace corresponding to A| =4-;.
1+;
where P =
, a
X2
X2 = X| =
3
C =
general solution is X| =(1-;)^, X2 =s or
= 5
-1
0 ■
0
^1
l-i
is a basis for
so X =
2
X2
'A-5
-2-i
-2-i
x.
The characteristic equation of A is
(A/-A)x = 0 is
2 J
xj =(-2-i)t, X2 = 2t which is
17. tr(/\) = 5 + 3 = 8
det(/\) = 5(3)- l(-2)= 17
Since
0
\
x, _ 0
0
X2
-3-3;
-6
3
3-3;
■
reduces to
1
1-;
0
0
, a
general solution is X| =(-l + ;)5, X2 =5' or
71
-1
l)
-^'1
4
X2
|A|=|l + l;| = Vl^ = V2
1-;
—;
= 5
, so X =
is a basis for the
-1
eigenspace corresponding to A = 5 - 3;.
-1
Re(x) =
131
and Im(x) =
0 ■
SSM: Elementary Linear Algebra
Chapter 5: Eigenvalues and Eigenvectors
-I
Thus, A = PCP
1
-1
-1
0
Section 5.4
and
where P =
Exercise Set 5.4
C=^3
-3
1. (a) The coefficient matrix for the system is
5
1
4
A=
27. (a) u • V =(2/, /, 3/)•(i, 6i, k)
2 3/
= 2i{-i)+i(-6i)+3i(k)
det(A./- A)=
= 2+6+3ik
-43.-5 =(X-5)(3,+1)
The eigenvalues of A are Xj =5 and
= 8+ 3iT
—
—
8
If u • V = 0, then 3ik =-S or k =
?i2 =-l.
8
=-i
3i
3
?i-l
-4
-2
X-3
(kl -A)\ = 0 is
SO k =-i.
3
4-4 X, _ 0
0
^ 2 X2
X, =5 gives
(b) u v =(3:, A:, l +i) (l,-1, 1-/)
= A:(l)+ ^(-l)+(l +0(1+ 1)
Xi _ 0
.^2j 0
4 -41
Since
2 reduces to
-2
=(l+if
■
■
1
-1
0
0
a
general solution is X] = s, X2 = s, or
= 2i
Thus, there is no complex scalar k for which
X,
u • v:=0.
X2
1
1
is a basis for the
= ^ 1 and P| =
1
eigenspace corresponding to 3.| = 5.
True/False 5.3
3-2 = -1 gives
(a) False; non-real complex eigenvalues occur in
conjugate pairs, so there must be an even
number. Since a real 5x5 matrix will have 5
-2
-4
-2
-4
-2 -4irxi'
0
-2 -4j[x2_
0 ■
1
Since
eigenvalues, at least one must be real.
reduces to
2
0 oj"
general solution is X| = -2s, X2= s or
(b) True; 3.^ -Tr(A)3,+ det(A)=0 is the
^ and p, = ^ is a basis for
characteristic equation of a 2 x 2 matrix A.
X2J L ij
(c) False; if A is a matrix with real entries and
L'
the eigenspace corresponding to 3-2 = -1.
complex eigenvalues and 3: is a real scalar,
1
k^O, 1, then kA has the same eigenvalues as A
with the same algebraic multiplicities but
Thus P =[p^ P2]
tr(3A) = 3dr(A) 5-^ tr(A).
A and p-'AP = D= b
^
(d) True; complex eigenvalues and eigenvectors
^
^1 diagonalizes
1
0
5
0 3-2
0
0
The substitutions y - Pu and y'= Pu' give
occur in conjugate pairs,
the “diagonal system” u'= Du = 0^
(e) False; real symmetric matrices have real
eigenvalues,
or
(f) False; this is only true if the eigenvalues of A
satisfy |^|= 1.
u'f = 5uI
0
u
which has the solution
“2 -““2
5x
u=
c\e
. Then y = Pu gives the
C2e \
solution
5x
y=
1 -2
1
qe
~x
C2e
132
5x
c\e
-2c2e ^
+02^--^
5x
and
SSM: Elementary Linear Algebra
Section 5.4
the solution of the system is
1
X| = — i', X2=s, X3 = 5 or X] = —t.
5x
-2c2e^'‘
>'2 =c,e-'’-'+C2e“''
2
y, =c^e
X,
X2 = 2t, X3 = 2t which is X2 = r
(b) The initial conditions give '
-1
2 is a basis for the eigenspace
and P2 =
which has the solution C| = C2 = 0, so the
solution satisfying the given initial
>1 =0
conditions is
2
corresponding to ^2 = 2-
>’2=0’
■-1
A,3 = 3 gives
3. (a) The coefficient matrix for the system is
'4 0 1'
1
-2
0 .
0
Since
1_
det(A,/-/t) = >t^-6^^+ia-6
1
= a-l)(?.-2)a-3)
The eigenvalues of A are 3-| =1, 3-2 = 2,
1
0
0
-1
Xi
0
2 2
0
•«2
0
2
2
•^3
0
-1
0
-1
2
2
0
2
0
2
0
0
0
0
reduces to
1
-1 , the general solution is x^ =s,
0
and ^3 = 3.
0
-1
X,
0
2
X-l
0
•^2
0
2
0
A,-l
•^3
0
Xj = 1 gives
-3
0
-1
2
0
0
2
0
0
-3
0
-1
2
0
0
2
0
0
-^2
0
-1
corresponding to X^ = 3.
0
1
0
0
0
0
1
0
0
0
0
0
X3 = 0 or X2 = 5 1
and pi = 1
0
0
X3
Thus P = [p|
P2
P3]=
Since
1
0
0
1
0
0
-1
Xi
0
2
1
0
•^1
0
2
0
,^3
0
-2
0
-1
2
1
0
2
0
1
1
2
-1
0
2
-1
0
0
1
0
0
P^^AP = D= 0 X,2
0
0
2
0
the “diagonal system'
[1 0 o'
M, =M|
u' = Du = 0 2 0 u or M2 = 2m2 which
=1.
0
-1
0 0 3
0 0 ?i3
The substitutions y = Pu and y' = Ai' give
IS a
-2
0
diagonalizes A and
basis for the eigenspace corresponding to
3-2=2 gives
and
P3 = -1 is a basis for the eigenspace
0
the general solution is x, =0, X2 =
•^1
-1
■*3
Xi
reduces to
= 5
X2
X2 = -s, X3 = -5 or
~X-4
1
1
Xi
(XI -A)x = 0 is
Since
2
^3
^
c,+C2 =0
A= -2
2
0
0
3
has the solution u =
“3 = ^“3
2x
. Then y = Pu
C2e
reduces to
3x
C3«
gives the solution
I
2
, the general solution is
0
133
J
SSM: Elementary Linear Algebra
Chapter 5: Eigenvalues and Eigenvectors
X
y=
0
-I
1
2
0
2
3 -1 ^1 ^0
A,| = 3 gives -6
2
“0 ■
c\e
3.V
reduces to
Since
-6
3a-
2x
I
3
^3^
3 , the
2
0
0
-Cje ■ +036
2a
= C|e'’^ + 2c2e
1
3a
general solution is A| = —s, X2= s or A| =/,
-C3e
3
3a
2c2e^’^ -C3e
+ c-^e
2a
=C\e^ -vlc^e
.
IS a
3j’ ^° P>=L3
basis for the eigenspace corresponding to
3a
-c^e
=t
^2
3a
V] =
1
3^1
Xt = 3f which is
and the solution of the system is
^,=3.
1
.
3a
y3 =2c2e^^'-C3e
-2 -1 X, ^[0 . Since
i^2 = -2 gives
I
-2
-C2+C3 =-I
C] + 2c9 — C3 = 1
2c2 -C3 =0
2
•^1
X2 = -2t which is
so P2 =
-2
-2
is a basis for the eigenspace corresponding to
^2 = -2. Thus P =[p| P2
3,3 = -2e^-''-f2e^-"
3
-2
diagonalizes A and
5. Let y =/(x) be a solution of y'= ay. Then for
0 _[3 0'
1
p-'AP = D =
we have
-ax
=t
X2
y, =e2-'^-2e
initial conditions is y2
-2<?^'' +2e^’^.
0 X2j~Lo -2_’
-ax
+ f{x){-ae
-ax
= af(x)e
0
solution is X] = —5, X2 = 5 or X] = t.
= -1,
3a
g'(x)= f'(x)e
0
1
C3 = -2, so the solution satisfying the given
-ax
2 , the general
^ reduces to
-6
which has the solution C| =1,
g{x)= f{x)e
0
-6 -3 X2
(b) The initial conditions give
)
The substitutions y = Pu and y'= Pu' give the
-ax
-af{x)e
“diagonal system” u'= Du =
=0
-ax
Hence g(x)= f{x)e
= c where c is a
3
0
0
-2
u or
u[ = 3M|
^2 — —2^2
constant or /(x)= ce""'.
3a
which has the solution u =
7. If j'l = y and y2 = )'">
V2 = y"= y'+6y = y2 +^y\ which yields the
. Then
-2x
C2e
y = Pu gives the solution
system
3’[ = y2
y=
3’2 = 63'! + 3'2
The coefficient matrix of this system is
1
3
2
3a
3a
_
qe
-2a
■^J €26
-2a
c^e" +C2e
-2a
3c|e^'' —2c2e
-f C2e'"^''^ is the solution of
the original differential equation.
1 ■
det(3^/ - A)= A,^-A,-6 =(?i- 3)(:L + 2)
9. Use the substitutions y^ = y, y2 = y'= y'l and
3'3 = y"= y2- This gives the system
The eigenvalues of A are X] = 3 and X2 = -2.
(A./-A)x = 0 is
1
Thus Ji = y =
0 r
6
c^e
X
X|
-6 ^-lj[x2
_ 0
0 ■
134
"0
1
A= 0
0
6
-11
o'
1 .
6
SSM: Elementary Linear Algebra
Section 5.4
del(kl -A)= X^-6X^ + \\X-6
= iX-\)(X-2)(X-3)
The eigenvalues of A are
A.3 =3.
A,| = 1 gives
=1, ^2 = 2, and
0
X
II
1
Since
•^1
0
•^2
0
•^3
0
X-6
1
-1
0
A-,
0
0
1
-1
X2
0
•^3
0
11
-1
0
0
1
-1
-6
11
-5
-5
reduces to
9
1
1
Thus P =[p| P2 P3]= 1 2 3
I
4
9
diagonalizes A and
0
1
0
0
1
0
1
p-^AP = D= 0 ?L2 0
0
0
0
2 0
0 0 3
0 0 A,3
The substitutions y = Pu and y'= Pm' give the
0 0
1
0
0 0
diagonal system” u'= Dm =0 2 0
0
0
u or
3
1
1
x.
9
.^3
the general solution is Xi = s, X2= s, X3 = ^ or
^2
is a basis for the
eigenspace corresponding to ^3 = 3.
-6
-6
1
X2 =t 3 so P3 = 3
0
{XI -A)x = 0 is
1
X,
X
is a basis for the
= S 1 so P| = 1
c\e
M, = M,
2x
U2 = 2u2 which has the solution u = C2e
.•^3j
eigenspace corresponding to
2
-1
0
0
0
2
-1
X2
0
-6
11
-4
X3
0
^2 = 2 gives
-1
0
0
2
-1
1
1
y= 1
2
3
1
4
9
1
0
4
reduces to
Since
1
1
0 0
0
0
3x
2x
3x
C|e'^+C2e‘”' +C3e
2
=
the general solution is X| = —5, X2 = —5,
3x
+2c2e^'^+3c3e
Cie^+4c2e^^+9c3e^^
Thus yi = y = C|e'' +C2e^'^ +036^'*^ is the
X3 = 5 or X| = r, X2 = 2t, X3 = 4r which is
solution of the original differential equation.
1
is a basis for the
= t 2 so P2 = 2
True/False 5.4
4
4
•^3j
eigenspace corresponding to X2 = 2.
■ 3 -1
X^ = 3 gives
0
0
11
-3
x.
0
•^3
0
1
0
3
0
3
-1
-6
11
-3
(a) False;just as not every system Ax = b has a
solution, not every system of differential
equations has a solution,
0
3 -1
-6
Since
2x
C2e
C3e
1
Xo
^3^
Then y = Pm gives the solution
1
1
2
3x
M3 = 3m3
= 1.
0
(b) False; x and y may differ by any constant vector
b.
I
(c) True
9
reduces to
0
1
I
(d) True; if A is a square matrix with distinct real
3
0 0
eigenvalues, it is diagonalizable.
0
1
the general solution is X|
(e) False; consider the case where A is
^2
diagonalizable but not diagonal. Then A is
similar to a diagonal matrix P which can be used
a-3 = X or X| = t, X2 = 3t, X3 = 9t which is
to solve the system y'= Ay, but the systems
y'= Ay and u'= Pu do not have the same
solutions.
135
SSM:Elementary Linear Algebra
Chapter 5: Eigenvalues and Eigenvectors
Chapter 5 Supplementary Exercises
1
X2 = 1 , X3 = 2
X-cosd
sin^
-sin 8
^-cos^
1, (a) det(^/-4)
2
0
1
= A,^-(2cos6')?.+ l
1
Thus P= 0 1
Thus the characteristic equation of A is
1
0
1
-4ac < 0, so the matrix has no real
If
eigenvalues or eigenvectors.
= 0 2 0 , then sf = D.
0
■■■
0
then
dn
1
1
1
1
1
0
0
1
0
1
2
0
2
0
0
1
-1
2
0 0
3
0
0
i
0 0
0
3
-1
opposite direction.
d2 ■■■
9
S = PS,P
1
through
the angle 8. Since 0 < ^< 7i, no vector is
0
0
0 0
0 0
transformed into a vector with the same or
3. (a) If D =
0 0
1
Since 0< 8 <n, cos^ 8 < 1 hence
0
2
P-‘AP = D = 0 4 0
-4ac = 4cos^ 8-4= 4(cos^ ^-1).
0
2 diagonalizes A, and
0 0
-(2cos^)^+ l = 0. For this equation
(b) The matrix rotates vectors in
1
1
1
0
2
1
0 0
3
-1
2
2j
0
2
S = A, as required.
0
S=
-
0
0 7^ •••
0
0
0
X-3
9. det(3./-A)=:
■■■
-6
= X^-5X
X-2
The characteristic equation of A is X^ -5X = 0,
n
hence A^ -5A =0 or A^ = 5A.
(b) If A is a diagonalizable matrix with
nonnegative eigenvalues, there is a matrix P
3 6
1
such that P^^AP = D and
D=
0
0
0
X2
0
0
0
•••
=5a2 =5
Is a diagonal
=5A^ =5
Xn
matrix with nonnegative entries on the main
A^ =5A‘^ =5
diagonal. By part (a) there is a matrix S|
such that si = D. Let S = PS^P~\ then
2
30
5
10
15
30
75
150
5
10
25
50
75
150
■375
150
25
50
125
250
375
750
1875
3750
125
250
625
1250
11. Suppose that Ax = X\ for some nonzero
C,Xi+C2X2+--- + C„X„
= psfp-'
= PDP
15
=5A=5
X =
-1
^2
. Since Ax =
= A.
C,Xi+C2X2+--- + C„X„
C]X^+C2X2+■■■ + C„X,
then either Xj = X2 = • • • = x„,
(c) A has eigenvalues 3,| = 1, 3-2 = 4, 3,3 = 9
3. = C| +C2 H
1
i-c„ = tr(A) or not all the x,- are
equal and X must be 0.
with corresponding eigenvectors X| = 0
0
136
SSM: Elementary Linear Algebra
Chapter 5Supplementary Exercises
13. If A is a ^ X ^ nilpotent matrix where A" = 0,
then det(^/- A")= det(X/)=
characteristic polynomial
17. If X is an eigenvalue of A, the
and A" has
eigenvalue of
= 0, so the
A'^Xj =^^X|. Since A^ = A, the eigenvalues
must satisfy X^ =X and the only possibilities are
eigenvalue of y4, then A” is an eigenvalue of
-1,0, and 1.
A", so all the eigenvalues of A must also be 0.
0
1
0
1
-1
1
-1
1
1
0 0
-1
and P
AP = D = 0
1
will diagonalize A
0
0 , thus
0 0-1
-I
A = PDP
0
1
1
-1
1
0
1
0 0
0
0
0
0 0-1
1
0
0
-1
i
_1
2
2
1
i
-1
2
2
1
1
I
2
2
0
0
i
2
corresponding to the same
eigenvector, i.e., if Ax] =^Xi, then
eigenvalues of A” are all 0. But if A, is an
15. The matrix P =
will be an
1
137
Chapter 6
Inner Product Spaces
Section 6.1
(d) (/oi, v)=((-12,8),(4,5))
= -12(4)+ 8(5)
Exercise Set 6.1
= -8
k{u, v)= -4(2)= -8
(u. Icy)={{3,-2),(-16,-20))
1. (a) (u, v)= l -3+l-2 = 5
(b) {k\, w)=(3v, w)= 9-0+ 6-(-l)= -6
= 3(-16)+(-2)(-20)
(c) (u + V, w)=(u, w)+(V, w)
(e) (0, v)=((0, 0),(4, 5))= 0(4)+0(5)=0
(V, 0)=((4, 5),(0, 0))= 4(0)+5(0)= 0
= l-0+l-(-l)+ 3-0+ 2-(-l)
= -3
(d)
I v||=
=^3^^= Vi3
2
1
5. Let A =
(e) <i(ii, v)= |u-v|
. Then
a’'a = a- =
5
2
1
1
1
=
= /I so
3
3 2 ■
= V(-2)2+(-|)2
(a) (u, v)=
= ^5
5
=[-l
(f) ||u-/tv||= i(l, l)-3(3,2)
l|(-8.-5)||
1]
3
2
3 2j[l
=[-2
7(-8)2+(-5)2
= -4-1
= -5
3. (a) (u, v)= 3(4)+(-2)(5)= 2
(v,i.)= 4(3)+5(-2)= 2
(b) (v,
w)= yp'A^ Ay
0
=[0 -1]
(b) (u + V, w)=((7, 3),(-1, 6))
5
3
3
2
=[-3 -2]
= 7(-l)+3(6)
= 3-2
)+(v, w)
=1
u, w
=(3(-l)+(-2)(6))+(4(-l)+5(6))
= -15+ 26
(c) (u + V, w)=
A(u + v)
[0 -1]
5 3]rr
3
(c) (u, v + w)=((3,-2),(3, 11))
=[-3
= 3(3)+(-2)(l l)
= -13
-2^ 2
=-3-4
(u, v)+(u, w)
= -7
=(3(4)+(-2)(5))+(3(-1)+(-2)(6))
= 2-15
= -13
138
2
2
SSM: Elementary Linear Algebra
Section 6.1
(b) (u, v)= 9(-3)(l)+ 4(2)(7)= -27+56 = 29
(cl) (v, v)=
=[-l
I]
5
3
3
2
11. (a) The matrix is A =
\I3
, since
[-2 -1]
A^A = A^ =
= 2-1
llv|| =
3 0
0
v) =Vl =I
5
2
(b) The matrix is A =
0
-I
(e) V - w =
o'
0 >/5
-1
-1
A = A^ =
2
I
4
0
0 ^^6
, since
0
0 6 ■
(v-w, v-w)=(v-w)^A^A(v-w)
=[-l
2]
5 3ir-r
3
13. (a) ||a||= V('4, a)
2
2
= V(-2)^+5^+3^+6^
= ylT4
=[1
= -1 +2
(b) ||a|| =0 by inspection.
d{\, w)=||v-w||
15. (a) A-B=[8t
.^(v-w, v-w)
-1
-2_
d{A, B)=\\a-B\\
=vr
= 7(A^fi, A-g)
= ^6^+(-1)2+8^+(-2)2
(f) ||v-w||2=(c/(v, W))2 =l2=l
= n/K)5
7. (a) u2^v =
3
4
-2
8
-1
1
13
10
2
3
(b) A-B
(u, v)= tr(u‘^ v)= 1 + 2= 3
1
-3
4
6
(b) u^v= 2
5
0
8
3
0
0
2
3
-2
c^(A, B)=||A-B||
4
= 7(A-B, A-B)
= ^32+32+(-5)2+(-2)2
-18
52
= 747
(u, v)= tr(u2'v)= 4+52= 56
9. (a) A^ =
3
-5
= A, so A^’a = a2 =
9
17. (p.q)
0
0 4 ■
= /7(-l)c/(-l)+ /7(0)^(0)+ Kl)g(l)+ /7(2),j(2)
=(-2)(2)+(0)(l)+(2)(2)+(10)(5)
For u =
11,
and V =
>'1
= -4+0+4+50
V2
“2
= 50
(u, v)= A^All
=[V|
1p|| = x/(p. p)
9
0
V2] 0
4
=[9v| 4v2]
u
1
= V[p(-l)f +[B(0)f +[B0)f +[p(2)]2
= V(-2)2+02+2^+102
.«2
M|
=vrH
= 673
«2
= 9k|V| +4w2V2
139
SSM: Elementary Linear Algebra
Chapter 6: Inner Product Spaces
19. u-v =(-3,-3)
(a) c((u, v)=||u-v||
u- V, u
-v)
= yji-3f+i-3f
= ^/^8
= 3>/2
(b) dju, v)=||u-v||
= ^{ii-v, u-v)
= V3(-3)2+2(-3)2
= >/45
= 3V5
(c)
^
3
1
2
2
-1
-1
3
-1
13
(u-v, u-v)=(u-v)^/\^/4(u-v)
=[-3
=[-3
-3l[, ^
-1
-1
-3
13
-3
-36]r“’
-3
= 9+ 108
= 117
^^(u, V)=||u_v||
= yJ{u-\, u-v)
= 3Vl3
1/2
21. (a) For u =(x, y), ||u|| =(u, u)
1
1 ..2 + —
—X
4
2
y
and the unit circle is |-x^+ —
16
4
is the ellipse shown.
A
+
+
-2
2
140
16
=1 or — + — = 1
4
16
which
SSM: Elementary Linear Algebra
Section 6.1
(b) For u = (x, y), ||u|| = (u, u)'^^ =
I
I
and the unit circle is ^J2x^ + y~ =] or
^
- 1 which is the
2
ellipse shown.
I
+
25.
27.
U+ V
n
|2
vl- = (u + V, u + v) + (u - V, u - v)
= (u, u+v) + (v, u + v) + (u, u-v)-(v, u-v)
= (u, u) + (u, v) + (v, u) + (v, v) + (u, u)-(u, v)-(v, u) + (v, v)
= 2(u, u) + 2(v, v)
2
ii2
= 2||u|| +2||v|f
+ u-
Let V =
0
1
-1
0
, then
{V,V) = 0(0) + (1)(-1) + (-1)(1) + (0)(0)
= -2<0
so Axiom 4 fails.
29. (a) (p, q)= _fj(l-x + x^+5x-'’)(x-3.r2)filx
= J^^(x-4x“ +4x^ +2x"^ -15x^)<ix
1
Jx^_^ + x'^ + 2x-‘^
2
5
3
5x^
2
-1
29
15
15
28
15
(b) (p, q)= J^^(x-5x^)(2 + 8x^)Jx
= f_^(2x-2x^ -40x^)clx
6 3
1
X
2
37
6
f
3
-1
37
6
=0
True/False 6.1
(a) True; with
= W2 = 1, the weighted Euclidean inner product is the dot product on R^.
141
■“
SSM: Elementary Linear Algebra
Chapter 6: Inner Product Spaces
'y
(u, v)=(-l)(2)+5(4)+ 2(-9)=0
(b) False; for example, using the dot product on R"
with u =(0,-1) and v =(0, 5),
0
(u, v)= 0(0)-l(5)= -5.
llullllvll
(c) True; first apply Axiom 1, then Axiom 2.
(cl)
(d) True; applying Axiom 3, Axiom 1, then Axiom 3
and Axiom 1 again gives
=0
cos6* =
lull = 74^+12+82 =7^ =9
Ivll =
k\)= ^'(u, k\)
= k {kv, u)
= m{y,u)
=
v).
M
(ii, v)= 4(l)+ l(0)+8(-3)=-20
20
(». v)
cos^ =
llullllvll
(e) False; for example, using the dot product on
(e)
with u =(1, 0) and v =(0, 1),
(u, v)= 1(0)+0(1)= 0, i.e., u and V are
9VlO
||u|| = Vl^+02+l2+02 =V2
llvll = V(-3)“+(-3)2+(-3)2+(-3)2
orthogonal,
=6
(u, v)= l(-3)+0(-3)+1(-3)+0(-3)= -6
(f) True; if 0 =||v|p =(v, v), then v = 0 by Axiom
4.
cos0=
(u, v) _ -6 _
(g) False; the matrix A must be invertible, otherwise
there is a nontrivial solution Xq to the system
(f)
4x = 0 and (xq, Xq)= Axq ■ Axq =0-0= 0, but
^2^ +1^ + 72+(-1)2 =7^
|u,
+0“ +0“ +0“ = 4
|v||=
Xo^tO.
(u, v)= 2(4)+ 1(0)+ 7(0)+(-0(0)= 8
Section 6.2
2
COS0 =
Exercise Set 6.2
llullllvll
1. (a) Hull = x/l^+(-3)2 =Vk)
llvll = V22+42 =Vm
(u, v)= l(2)+(-3)(4)= -I0
-10
(u. v)
cos 6'=
llullllvll
(b)
VioV^
3. (a) \\a\\ = J{aT)
= 722+52 + 12+(-3)2
= x/^
= 5V2
10
IIbII = 7(fi, 5)= x/32 + 22 +12 + o2 = Vl4
V2
{A, B)= 2(3)+6(2)+1(1)+(-3)(0)= 19
llu|| = 7(-l)^+0^ =1
llvll = x/32 +82 =V^
cos6'=
{A,B)_
19
lUllllfill 5V2V14
(u, v)=(-l)(3)+0(8)= -3
COS0 =
(b) \\a\\ = 4{a7a)
(u, v) _ -3
= 732+42+(-1)2+32
llullllvll
= yl3d
(c)
||u|| = 7(-')^+5^+2“
I|5|| = 7(^> B)
llvll = 72-+42+(-9)2 = Viol
= 7(-3)^+1^+42+22
=7^
142
19
10V7
Section 6.2
SSM: Elementary Linear Algebra
{A, B)= 2(-3)+4(1)+(-1)(4)+3(2)=0
(A
0
cos^ =
5. (p,q)= l(0)+(-l)(2)+ 2(l)=0
7. 0=(u, w)= 2(l)+/t(2)+6(3)= 20+ 2L'
Thus /t = -10 and u =(2,-10, 6).
0=(v, w)= /(l)+5(2)+3(3)= / + 19
Thus,/ =-19 and v =(-19, 5,3).
(u, v)= 2(-19)-10(5)+ 6(3)= -70 9^0
Thus, there are no scalars such that the vectors are mutually orthogonal.
9. (a) (u, v)= 2(l)+ l(7)+3L'=9+ 3L'
u and V are orthogonal for k =-3.
(b) (u, v)= k(k)+ ki5)+ \i6)= k^+5k +6
k~ +5k +6 = ik + 2){k + 3)
u and V are orthogonal for k = -2,-3.
11. (a) (u, v) =|3(4)+ 2(-l)|=|l0|= 10
u
I I v||= 732+22^42+(-1)2
= ^/^3Vi7
= 14.9
(b) (u,v) =1-3(2)+!(-!)+0(3)1= -7 =7
u
l llvll = V(-3)2 +12+02^22+(-1)2+32
= ^/^40
= 1 1.8
(c) (u, v) =|-4(8)+ 2(-4)+1(-2)|= -42 = 42
u
l llvll = V(-4)2 + 22 + i2 ^82 +(-4)2 +(-2)2
= 42
143
SSM: Elementary Linear Algebra
Chapter 6: Inner Product Spaces
(d) |(u, v)|=|0(-1)+(-2)(-1)+ 2(1)+1(1)1
=|5|
=5
llullllvll = yjo^ +{-2f +2^+127(-1)2 +(-1)2 +,2 +,2
= ^^^/4
=6
13. (u, W|)=0 by inspection,
(u, W2)= -1(1)+1(-1)+0(3)+ 2(0)= -2
Since u is not orthogonal to W2, it is not orthogonal to the subspace.
15. (a) W'^ will have dimension 1. A normal to the plane is u =(1,-2,-3), so W'^ will consist of all scalar
multiples of u or m =(r, -2r,-30 so parametric equations for W'-*’ are x = t,y = -2t,z = -3f.
J.
(b) The line contains the vector u =(2,-5,4), thus W is the plane through the origin with normal u, or
2x-5y + 4z = 0.
(c) The intersection of the planes is the line x + z = 0 in the xz plane (y = 0) which can be parametrized as
(x, y, z)=(r, 0,-0 or r(l, 0,-1). W-^ is the plane with normal (1, 0,-1)orx-z = 0.
17. |u-
viP =(u- V, u- v)
=(u, u-v)-(v, u-v)
=(u, u)-(u, v)-(v, u)+(v, v)
Ip-O-O+Ilvlp
u
=2
Thus ||u-v||= ^/2.
19. span{U], U2,.... u^} = ^|Ui +^2^2
where k\, k2,
are arbitrary scalars. Let
ve span{u,, U2, ..., u^}.
(w, v)=(W, ^,U,+^2U2 +•••+^r“r)
= A:,(w, u,)+l:2(w, U2)+ --- + k,{w,u^)
=0+0+ --- +0
=0
Thus if w is orthogonal to each vector Uj, U2, ..., u^, then w must be orthogonal to every vector in
span{u,, U2,.... u^}.
21. Suppose that v is orthogonal to every basis vector. Then, as in exercise 19, v is orthogonal to the span of the set of
basis vectors, which is all of W, hence v is in W'^. If v is not orthogonal to every basis vector, then v clearly
cannot be in W'^. Thus IT"*" consists of all vectors orthogonal to every basis vector.
144
SSM: Elementary Linear Algebra
Section 6.3
Section 6.3
27. Using the figure in the text, AB = v + u and
Exercise Set 6.3
BC = v-u. Use the Euclidean inner product on
1. (a) ((0, 1),(2, 0))=0+0=0
and note that ||u|| =l|v||-
The set is orthogonal.
{ab, bc)~(
v + u, v-u)
=(v, v)-(v, u)+(u, v)-(u, u)
i2
V|
1
1
1
— +-=0
2 2
J 1_"
^f2’ V2j’lV2’ V2.
(b)
|2
|u|
The set is orthogonal.
=0
Thus AB and BC are orthogonal, so the angle
\
1
(c)
at 5 is a right angle.
/
1
VI’ VIj’lVI’VI
1
1
2
2
= -l
31. (a) W'"'" is the line through the origin which is
;t0
perpendicular to y = x. Thus W'^ is the line
The set is not orthogonal.
y = -x.
(d) ((0, 0),(0, 1))=0+0=0
(b)
is the plane through the origin which is
The set is orthogonal.
perpendicular to the y-axis. Thus, W'^ is the
3. (a)
(c) W'^ is the line through the origin which is
1
1
1
xz-plane.
1
1
0
IVI’ ’Vlj’U’VI’ s
1
1
-^+0-^
V6
V6
perpendicular to the yz-plane. Thus W'^ is
=0
the x-axis.
1 X /
1
0
0
True/False 6.2
vi”^’vi;i x/i’
VI
1
--+0+2
2
(a) False; if u is orthogonal to every vector of a
subspace W,then u is in W'^.
=0
1
1
(b) True; ITnVT-" ={0).
1
1 ^
1
1
0
VI’VI’ VIj’l VI’^’VI
1
(c) True; for any vector w in W,
+0
(u + V, w)=(u, w)+(V, w)= 0, so u + V is in
VI
S
1
IT".
VI
(d) True; for any vector w in W,
The set is not orthogonal.
(All, w)= A:(u, w)= A:(0)=0, so ku is in W-^.
(b)
(e) False; if u and v are orthogonal (u, v) =|ol = 0.
2 _2
f2 I _2
4_2_2
3’ 3’3j’l3’3’ 3
9
9
9
=0
2 _2
(f) False; if u and v are orthogonal.
n 2 2'
3’ 3’3j’l3’3’3j
llu +vlP =llu|p +||v|p thus
2
|u + v|
_2^ fl 2 2"
3’!’ 3j’l3’3’3.
The set is orthogonal.
145
2
4
9
9
2
+-=0
9
2 2_4 =0
9’^9 9
SSM: Elementary Linear Algebra
Chapter 6: Inner Product Spaces
\f2'] 2(2
(c)
(I, 0, 0), 0,
+-
=0+0+0=0
3^3
V2’ 42)
_2 2_4
((1, 0, 0),(0, 0, 1))=0+0+0=0
1
”9^9 9
1
=0
1
0,
,(0, 0, 0 =0+0+
4~2 4i
The set is an orthonormal set.
4~2
^0
(b) \\p^{x)\\ =^
The set is not orthogonal.
a , a2
1
||/^2W||=
2
(d)
\
+
r2
/
^/6’^/6’ ^/6JA^/2’ V2
,0
1
2'''2
1
+0
= 41
4n 4n
=0
The set is orthogonal.
=1
(piW,;22W)= 1(0)+0(1)+0(1)=0
(/j,(x), p3(x))= l(0)+0(0)+0(l)=0
n.2 +( 24 +(1 n2
5- (a) ||aiW||= J 73
4
3
4
3
1
(P2W. /23(x))=0(0)+J=(0)+J=(1)
'-+-+9
9
9
= 41
7^0
4i
=1
n2
/ ,\2
A2W||=
+
/
+
The set is not an orthonormal set.
2f
7. (a) ((-1,2),(6, 3))= -6+6 =0
3
fi+1+1
9
9
(-1,
9
2)\\ = yl{-\4+2^ =^/5
2^
(-1, 2) _ (-1, 2)
=^/^
4~5’r5
ll(-^2)|^ 4~5
/.,\2
2
fo
|(6, 3)||= V6-+3- =4^= 345
+
3
0
4
(6, 3) _ (6, 3)
4
1(6,3)|| 3V5
9’^9’^9
6
= 41
3
3V5’ 341
2
2(2^ 2(\
(/r|(x), P2{x))=3U
2O
3U
3
3
20
4_2_2
9
=0
9
The vectors
and
45'45
9
A A
A’45
are an orthonormal set.
, ,, ,
2f\^ 2(2^+1(2
(Pl(x), /?3(x))=- -7 \3
3L3
(b) ((1,0,-0,(2, 0,2))= 2+0-2=0
((1, 0, -1),(0, 5, 0))=0+0+0=0
((2, 0, 2),(0, 5, 0))=0+0+0= 0
3U
2_4 2
9
1 0
l^/5’ S)
+-
9
9
=0
1(1,0, - oil = ViAoA(-o^ = 42
146
SSM: Elementary Linear Algebra
(L0,-1)
(1,0,-1)
|(L 0,-1)
V2
Section 6.3
1
I
. 0,-
I
1
2)
2 2
|(2,0, 2)|| =
+0
+
-,-,o
IV2
2J
1
= 2V2
2
(2, 0, 2) _ (2, 0, 2)
1
2 "1
0
2V2’ ’ 2V2
2
yi
1
1
0
2’ 2
1
0
"U’ ’V5j
1
1
,0
V^’V2
||(0, 5, 0)||= Vo“+5'+0“ = 5
(0. 5. 0)
(0, 5, 0)
11(0, 5, 0)11
5
The vectors
=(0, 1, 0)
/
11
if + if r 2f
3’ 3’ 3.
3
3J 1 3
'6
r 1
9
1 ^
1
0
, and (0, 1,0) are an
3
orthonormal set.
_ Lfl 1 _2
I
V
(c)
/
1
3’ 3’
1
1
-,-,o
5’ 5’ sj’
I
10
2 2
I) ^^/6l3’ 3’ 3
h0
1
10
5’ 5’ sj’b’ 3’
1
2
1
2
1
r 1
1
2’ 2
/
V
1
1
,0 and
1 I _2''
4i’4~2
--+-+0
6 6
-,o , -
3’3’ 3.
1
V^’V5’V3j’
=0
y /
1
r 1 \2
/11 \2
s’s’sj Vis
nf
5)
y 3f + Mf +0^
9.
Vs
5j
3
9
5)
16
25"^25
25
= V1
5
I
Mil 5 r
(i.i.i) J5l5'5'5
r4f
x2
3
+
+0^
'll 1.
1
25"^25
IV3’ VI’ V3
= VI
V3|= VoMoMM
=1
(v,,
V2)=-—+
^
25 —+
25 0=0
147
2
are
an orthonormal set.
+
+
1
Vo’ 46’ 46
=0
111 =
46
The vectors
IS’^IS 15
3
2^
^-y/6 \f6
=0
1
1
SSM: Elementary Linear Algebra
Chapter 6: Inner Product Spaces
linearly independent. Since S contains 4
linearly independent orthogonal vectors in
(v], V3)=0+0+0=0
(v2, V3)=0+0+0=0
it must be an orthogonal basis for R^.
(a) u =(l,-l,2)
' ''
5 5
(b) ||v,f =l^+(-2f+3^+(-4)^ =30
||v2f= 2^ + 12+(-4)2+(-3)2 =30
IKsf =(-3)^+4^+l^+(-2)^=30
5
(u, v2)=---+0=i5
'^^ 5 5
(u, V3)=0+0+2= 2
V4
(u, v,)=-l-4+9-28 = -24
(u, V2)=-2+2-12-21=-33
(u, v3)= 3+8+3-14 =0
(u, V4)=-4+6+6+7 = 15
Thus, (1, -1, 2)=-|V|+^ V2 + 2V3.
(b) u =(3,-7,4)
37
(u,v,)=-^-^
5 5 +0= -.5
'
/
\
12 21
-24
u=
9
30
5
-33
Vi +
4
30
11
— Vi
0
37
15
^2+—V3+ —V4
30
n
30
1
Vt +OV4 +—V4
5 ' 10 ^
(u, V3)=0+0+4=4
3 2 ^
9
Thus, (3,-7, 4)=-
T''>--V2+4V3.
1
(c) u= -
2 =42+32+22+1^ =30
13.
_3 5
,^ /
\
10
4 „ 10
(a) (w,u,)=-+0+ y
14
3
|
(w,U2)= +0-^
7’ 1'1
“3)=1
^+0-^3
11--1
(u, v,)=-A-li+o='
35 35
35“ 7
14
1
(u,v2)
'
2/ =^-^-^0
35 35 =-A
35
8
3
1
3
8
1
Thus, w =—uI “
3
y
3U2-3U3.
(U, V3)=0+0+^=^
Thus,
1
3 5^
7’
3
1
5
-.yJ =-yV,--V2+-V3.
,
1
2
2
5
-v/b ^/6 Vb
I
11. (a) (vi, V2)= 2-2-12+ 12 =0
(vi, V3)=-3-8+3+8 =0
(vj, V4)= 4-6+6-4 =0
(v2, V3)= -6+4-4+6=0
(v2, V4)= 8+3-8-3=0
(v3, V4)= -12+12+ 2-2=0
\
1
4
-v/b
14
11
11
Thus, w = Ouj H—j
Thus, since 5 = {v,, V2, V3, V4} is a set of
nonzero orthogonal vectors in
V
(W, U2)=-^+-pr+-= =-=
5 is
148
“3
SSM: Elementary Linear Algebra
15. (a)
Section 6.3
Vl
l|v2f= 1^ + 1^+(-1)2+(-1)^=4
||v3f =i2+(-1)2+i2+(_i)2=4
(x, v,)= l + 2+0-2= l
(x, V2)= 1 +2+0+2=5
(x, V3)= 1-2+0+2= 1
1
5
Pro)ivx =-Vi+-V2+-V3
\
ill ll+fl 1 -1 -1
4’4’4’4J U’4’ 4’ 4
/
/
1 _1 1 _1
V
4’ 4’4’ 4
7 5 _3 _5"l
U’4’ 4’ 4
(b) |v,f =:02+l2+(^)U(-l)2 =18
||v2f =32+52+12+1^ =36
||v3||2 =i2+o2 + i2+(-4)2 =18
(x, v,)=0+ 2+0+2= 4
(x, V2)= 3+ 10+0-2= 11
(x, V3)= 1 +0+0+8 = 9
11
4
9
projwX =-v,+-V2+-V3
2
11
1
=-V,+
V3 +—V3
9
36
2
2
= 0,
,
9
9
2^+
9/
11,11,11,11U1
0,12 -2
36 36 36 367 U
\
n 7 _j_
U2’4’ 12’ 12
1
3
17. (a) (x, Vi)=0+ Vis
^ +0+ V18
x/lS
(x, V2)=l+ —+0-16 = 2
' 2/ 2 6
1
4 _ 5
''3)= Vis +0+0+ Vis ~ Vis
prpiwx =
3
^
5
j= V] +2v2 +
Vis
Vis
3
= 0, —
I 18
'^3
5 1
12
is’
+ 1,, , ,
IS /
23 11
17
18’ 6 ’ 18’
18
V
n
5 ^5
+ —.0, —,
3 3 3j US
149
IS
20
ISj
SSM: Elementary Linear Algebra
Chapter 6: Inner Product Spaces
1
(L -3)
vi
(b) (x,
v,)
=-+l+0—=1
\
1/ 2
2
qi =
(x, v2)=-+I+0
+=2
2
2
||v,||
3
Vio Ivio’ yfloj
V2
q2 =
V2
(x, v3)=--I+0+-=0
(f-l)
proju^x = 1 V| + 2v2 + OV3
4^/^0
5
5
+(1, 1,-1,-1)
2 2 2 2
f\2 4
4Vh)I 5 ’ 5
3 3
1
2’2’ 2’
2
I
3
vio’ ^/^o
19. (a) From Exercise 14(a),
3 3
1 , so
vv| =projn,x =
2’ 2’
1
W2 = x-pro)^^^x = -
1
’
‘I2
1
2’ 2’
,v
-I
1,-1 •
--I
(b) From Exercise 15(a),
r? 5
W| = prq]vi,x =
3
5
U’4’ 4’ 4
3 3 3
\V2 = X - projw X =
(b) First, transform the given basis into an
, so
orthogonal basis {V|, V2).
3
V, =u, =(1, 0)
4’4’4’ 4;
^1
= Vl^T^ = l
21. (a) First, transform the given basis into an
V2 =U2-
orthogonal basis {V|, V2).
V| =U| =(1, -3)
3+0
=(3,-5)
3/12+(-3)2
''1
=(3,-5)-(3, 0)
=(0,-5)
(»2^
V, =U2 -
Vi
||v2|| = 7o2+(-5)2 :=5
2-6
=(2, 2)-
(1. -3)
The orthonormal basis is
10
2 6
=(2,2)- -
vi
(1, 0)
Vl
1
qi =
5’ 5
12 4
=(1, 0)
(0, -5)
=(0,-1).
V2
92 =
5’5
v.
^(1,0)
5
V2
ri2f +r4f
5 J ys)
1
160
'll
25
-1
4Vk)
12
-I
5
The orthonormal basis is
150
I
A'
SSM: Elementary Linear Algebra
Section 6.3
23. First, transform the given basis into an orthogonal basis {V|, V2, V3, V4}.
V| =u, =(0, 2, 1, 0)
= v/o2+2^+i2+0^ =a/5
Vi
Vt =Ut -
0-2+0+0
=(l,-1, 0, 0)-
(0, 2, 1, 0)
5
4 2
=(1,-1, 0,0)+ 0,-,-,0
5 5
1
2
= 1, —,-,0
5 5
l2+
2f +0^ =
+
=^|
5j
5J
^3 = U3 ''2
^1
0+4+0+0
(0, 2, 1, 0)-
=(1, 2, 0, -1)-
1 2
i-|+o+oa ,—,-,o
6
5
5
1
8 4
5 5
1
=(1,2, 0,-1)- 0,-,-,0 V
1
.2’ lO’S’*^.
5 5
1
l2’2’ ’
1
\2
^2
+(-1)2+(-1)2
+
2j
2J
5
2
Vio
2
(U4. ^^3)
(U4, V2)
-V2-
V4 =U4-
^3
O+O+O+O
=(1, 0, 0, 1)-
l+O+O+O
V|-
5
2
1, —
6
5
1
,
=(1, 0, 0, l)-[6
/4
4
15’ 15’
'^4
8
1
1
1
-,-,o -
I 10’ 10’ 5’ 5
6 3
4
15’ 5
(4 v2 +/ 4
=
15
VU5
?
/
8
\2
/.n2
4
+
+
15
5j
240
225
4
^5
151
^+0+0-ln 1
.0 -
5’ 5
5
2
2’ 2’
-1, -1
SSM: Elementary Linear Algebra
Chapter 6: Inner Product Spaces
The orthonormal basis is qj =
V,
ri
q2 =
2
5
I|V2||
M
5
l^/30’
^3
Vio
~I^Vio’Vio’ ^/^o’ n/IoJ’
2
1^- U5’ 15’
^
7,0 ,
1
f_l
15’ 5j
2
1
J[
(± ± -± 4\
1
" ^5 i’V5’V5’7
(!’-i5’ 25’ 0)
l2__
q3 =
^4 =
(0,2, 1,0) r
2
1
2
^
4
''4II
25. First, transform the given basis into an orthogonal basis {V(, V2, V3).
V] =Ui =(1, 1, 1)
= Vl +2+3
= V^
(»2. V|)^.
V2 =U2-
1
=(1, 1,0)-
1(1)+ 2(1)(1)+3(0)(1)
(1, 1,1)
6
1
=(1, 1,0)--(1,1, 1)
1
1
2’ 2’
2
1^21 = V(V2, V2)
1
\2
\2
+2 -
2J
2)
+3 --
2)
=.6 ■
4;
2
V3=U3
|2
Vl
‘
(»3. V2)^2
I|v2f
i+o+ori 1
=(1,0,0)-
4
=(i, 0,0)-
,6’6’6j U’6’
1
2’ 2’ 2
2
6;
0
3
3
.
152
SSM: Elementary Linear Algebra
V3
Section 6.3
= 7(^3- V3)
W| =prq)n,w
_(w, Vi)
f2V + 2 --if + 3(0)2
3
4
(w, V2)
-''2
rv,+
''I
3
^2
5_2_J2
1 + 2+3
2
9’^9
42
3
^V2
9
9 rs
14 1,3’ 3’ 3
The orthonormal basis is
91 =
Vb
tVb a/6 Vbj
1
2)
1
M
W2 = W-W|
Vb J
13 31 20
2
=(1,2,3)- —
/2
1
,a/6
Vb
A
I|V3
3
14’ 7
1
,a/6 ^^6
A
V3
93 =
13
6^
15
14’
14’ 14’ 7
(1 i _1)
l2_- I2’ 2’
V2II
=(2, 2, 2)
1
vi _ (1, 1, 1)
^1
92 =
4^
= 2(1, 1, 1)- —
3
14’ 14’ 7 J
3
,0 .
14’
3
27. Let W =span{u,, U2).
14’ 7
29. (a) A= 2
Vj =u, =(1, 1, 1)
n
3
Sincedet(A)= 3 + 2 = 55^0,A is invertible,
so it has a 2/?-decomposition.
V2 =U2-
v.i
Let U] =(1, 2) and U2 =(-1, 3).
2+0-1
=(2, 0,-1)-
Apply the Gram-Schmidt process with
l^+l^ + ijd, 1, 1)
1
=(2, 0,-1)- -
normalization to U| and U2.
\_ I
V, =U| =(1, 2)
l3’ 3’3
5
4^
V3’ 3’ 3
||v,|| = Vf+ 2^ =V5
V2 =U2-
{v,, V2} is an orthogonal basis for W.
''I
||v,l| = Vf+f+f =V3
''2t|=
rsf +
3
\2
/
+
3
=(-l,3)-^(l,2)
4f
=(-l, 3)-(l, 2)
3
=(-2J)
'42
V2
=
=a/5
9
V,
a/42
9i =
V|
3
92 =
153
(1,2)
2^
A [A'A
V2 _(-2,i)_r 2 _i_
IIV2II
A [ A’A
SSM: Elementary Linear Algebra
Chapter 6: Inner Product Spaces
The Q/?-decomposition of A is
2 ~3 =[qi ^2]
1
(u|,q|) (u2,q,)
0
(U2, q2)_
2
-M^5 S
2
1
0
I
By inspection, the column vectors of A are
linearly independent, so A has a
(2^-decomposition.
i
Us
1
(c) A= -2 \
Vs '
Let U| =(1, -2, 2) and U2 =(1, I, 1).
Apply the Gram-Schmidt process with
I
2
(h) A= 0
1
normalization to U| and U2.
V| =U| =(1, -2, 2)
1 4_
Vl-+(-2)-+2^ =3
By inspection, the column vectors of A are
linearly independent, so A has a
2^-decomposition. Let U| =(1, 0, 1) and
V2 =U2-
U2=(2, 1, 4).
Apply the Gram-Schmidt process with
1-2-I-2
normalization to U| and 112.
=(1,1,1)-^-|21^(1,-2, 2)
vi=u, =(1, 0, 1)
=(1, 1, 1)- 4-,--,-
f]
2 2
-
V9
||v,|| = Vl^+0^-)-l^ =V2
8 22 2
9’ 9’9j
(“2- ''1)
V2 =U2-
— V,
^1
+
V9
2+0+4
=(2, 1, 4)-
(I, 0, 1)
81
V2M
||v2||= V(-l)^ + l^+l^ =V3
''2
9
1
(1, 0, 1)
IV2’ ’V2
41
(ui-9i)=
''1
3
1
2^
3’ ~"3’3
92 =
''2
V3’ Vs’ Vs
(8 LL 7\
l9’ 9 ’ 9/
= 4~2
-l-O-t-
(1,-2, 2)
V2
1
(-L I, 1)
92 =
Vl
91
0
V2
''1
9)
234
=(2, 1, 4)-(3, 0, 3)
=(-l, 1, 1)
91 =
+
9j
\9
2
''1
9 9
^/234
V^
9
7 ^
11
(-2.<i,)=;|+o+^=37J
V^’ V^’ v^
1
/
\
1
4
4
,
1
2
^
(ui, qi)= —I 1— = 3
^ ' '^ 3 3 3
The 2/?-decomposition of A is
1
0
1
2
1 =[q, q2]
4
(ui,9l) (u2,qi)
0
(112,92)
i
i
^/2
S
0
,
(U9, q,)=
'^
I
1
1
VI
VI
/
' 3 3
\
2
1
+-=-
8
3
3
11
7
V^'^V2M^V^
26
V2 3V2
0 Vs
sV^
V^
3
154
SSM: Elementary Linear Algebra
Section 6.3
The ^/^-decomposition of A is
-2 1 =[qi
q2]
(ui,q,) (u2,q|)
0
I
8
3
yJlM
(U2, q2)_
I
3
3
2
1
(d) A= 0
1
3
Jim
2
7
3
J2M
0
2
1
1
0
3
2 0
Since det(/\) = -4 0, A is invertible, so it has a (2/?-decomposition. Let Uj =(1, 0, 1), U2 =(0, 1, 2), and
U3=(2, 1, 0).
Apply the Gram-Schmidt process with normalization to Uj, U2, and U3.
V, =u, =(1, 0, 1)
\\v^ \ = ^\^ +0^ +\^ =42
(U2> V|)
(1, 0, 1)=(0, 1, 2)-(l,0, 1)=(-1, 1, 1)
V2 =U2-
^2
= V(-1)^+1^ + 1^ =V3
(u3- '^2)
fclZilv
V3=U3-
2
Vi
^2
■2 + 1-t-O
2 + O-rO
= (2, 1, 0)-
(I, 0, 1) —
2
f I
= (2, 1, 0)-(l, 0, 1)-!- —
(-1, L 1)
3
\
V 3’3’3j
2 4 _2
3’ 3’
/ « \2
3
/
4f
^ _ 2^
2Y
9 ~ 3
3
=
1
v, _ (1,0, 1)
0
V2
q2 =
IV2’ ’72
1
1
V2 _(-l, Ll)_f
^/3’ 73’ 73
7^
;'2
2
''3
q3 =
^3
2J6
^76 To
3
1
To
1
(ui-qi) = —j=
+ 0 + —^ = 75
75
75
(u2, q,) = 0-l-0-r-^ = 72
155
SSM:Elementary Linear Algebra
Chapter 6: Inner Product Spaces
/
\
/
\
2
1
2
+0=-
2
^
4
The (2/?-ciecomposition of A is
'1
0 2'
0 1
1
1 =[q, q2 qs]
(ui,qi) (U2,qi) (u3, q,)
0
(“2-q2) (u3-q2)
2 0
0
i
i
i
^/2
n/3
^/6
i
2
0
0
V2 V2
0
n/3
0
0
(u3- qs).
V2’
^/6
1
(e) A= 1
0
2
1
1
1
3
1
1
J_
L
>/2
73
n/6
^
76
Since det(A)= -1 0, A is invertible, so it has a ^/^-decomposition.
Let U| =(1, 1, 0), U2 =(2, 1, 3), and U3 =(1, 1, 1).
Apply the Gram-Schmidt process with normalization to uj, 03, and U3.
vj =u, =(1, 1, 0)
||v,||=7i2+i2+o2=V2
^V,
V2 = U2 -
Vl
=(2,1, 3)-^i^(l,1,0)
=(2,1,3)-
3 3
i0
I2 2
1
2’“2’^.
n2
if +
1
2J
2)
4-3^ =
19
2
156
SSM: Elementary Linear Algebra
(U3, V|)_
=»3-“il
Section 6.3
(U3, V2)_
li liT"” 2
I| V2
^ >
Vl
-,-2,3
=a L1)-
U
2
r3
19’
1 18
=(1, 1, 1)-0, 1,0)- —
3
2 1
19’ 19
3
19’r9’ 19j
3f +
3
^2
r 1
1 _ 1
~Vi9“Vi9
19
1
(1, 1, 0)
Vl
91 =
V2
IV2’ n/2
''3
/_J_
_L
I 19’ 19’ 19
2
)
1
(ui>9i)=
\
^
3
1_
Vi9’ V19’ ^/i9
1
IIV3II
2
1
1
1
V? 3^
,2^’ 2Vi9’a/i9
;''2
19
/
,0
' V2
^2
92 =
93 =
1
+0= V2
3
+0=
VI
(u3,9i)=
+o = VI
/
V2
\ 2V2
9V2 _ Vl9
^““’‘*'^"2VI^"2VIV^Vi^“ VI
VI VI ^ 3V2 _ 3V2
(u33 92)= 2Vi9
2V19 Vl9 Vil
K<13)--^+^+jL=^
The Q/?-decomposition of A is
“1
2 r
1
1
1 =[9i
0
3
1
92
(ui,9i) (u2,q|) (uj, q,)
0
93]
(u2>92) (“3-92)
0
0
(“3-93).
3
1
V2
2Vl9
J_
V2
3
2^^
x/Tg
n/2
0
n/I?
3x/2
n/I?
VI -f
V2 VI
0
a/i?
^/2
0
0
3V2
x/i9
1
719
157
SSM: Elementary Linear Algebra
Chapter 6: Inner Product Spaces
1
(f)
A=
0
1
1
1
0
1
1
= [C|
C2
ll''2| = 12 iS
C3]
■\dx--x^
(P3.v,)=J[,
=
3
0
By inspection, C3 -Cj = 2c2, so the column
(P3-V2)=|;3c'
vectors of A are not linearly independent
and A does not have a 2/?-decomposition.
31. The diagonal entries of
are
i= 1 ,2, ..., n, where q,- =
V;/
— + x dx
. 2 j
fT —1 X 2 + X 3'! dx
.
=
for
3
K 2
r 1 3
I 6
is the
V/
1
1 3
4
0
normalization of a vector v,; that is the result of
12
applying the Gram-Schmidt process to
(P3. V,)
^3=P?i--
{u|, U2, ..., u,j}. Thus, V,- is u,- minusalinear
combination of the vectors V|, V2,
V,_|, so
u,- = V,-+A:|V|+^2''2 +—
Thus,
(P3. V2)
-2
^V2
Vl
V2
i
1
—+x
(u,., v,.) = (v,., V,) and
2
12
V/
V;
U/'
I
= x
■
1
2
1
V;
X
3
/
Since each vector v,- is nonzero, each diagonal
2
— x+x
2
6
entry of R is nonzero.
V3f =(V3, V3)
33. First transform the basis
= {Ph P2’ P3I = {!>
■■''to
f'f >
orthogonal
2
basis {V|, V2, V3}.
£ L36
v, =p, =1
=(^i’ ''i)=
vi
(P2> V,)= Jj
1
3
1
-X
.36'^ 6
1
x-\dx = — x^
2
0
''I
2
4 3
1
9
2
+-X-
1
''311 =
^ V, =x-|(l) = --2 + X
bVs
91 =
n2
'
180
The orthonormal basis is
''2f =(V2. V2)
—+X
V9
dx
92 =
2
■^2
f'f
>
2
L --X+X
1
2
T+x
dx
JoU
i
1
9
1
3
— X — x" + —x‘
4
2
180
i
(P2/ ''1)
Jo
4 2 -2x
.., 3 +x 4 1 dx
/
1
I
''
1
3
— x + —X
1
=1
=V£ = i
V2=P2-
dx
2
3
= 2V3f-+x
V 2
0
1
= V3(-l + 2x)
12
158
X
4
1
<5
+-X-
5
0
SSM: Elementary Linear Algebra
Section 6.4
''3
2-1
''3
3
1
2
-1
4
5
1
2
4
03 =
(b) A =
2
I
-T-X + X
6
6^/5
2
= 6^/5f--x + x^
0
3
A^A= -1
U
V5(l-6x+6x^)
1
0
True/False 6.3
-1
4 2
2
2
-1
3
1
0
2
4
5
2
4
1
5
4
15
-1
5
-1
22
30
5
30
45
2
3
(a) False; for example, the vectors (1,2) and (-1,3)
-1
in
are linearly independent but not
orthogonal.
A^b= -1
1
0
2
-1
-1
1
0
4 2
9
1
5
13
4
2
(b) False; the vectors must be nonzero for this to be
The associated normal system is
true.
'15 -1
5
x.
-1
22
30
•^2
9
5
30
45
,^3
13
3
(c) True; a nontrivial subspace of R' will have a
basis, which can be transformed into an
orthonormal basis with respect to the Euclidean
inner product.
1
3. (a) A^A =
(d) True; a nonzero finite-dimensional inner product
1
1
-1
3
-2
2
-2
6
1
-1
space will have finite basis which can be
transformed into an orthonormal basis with
2
7
respect to the inner product via the GramSchmidt process with normalization.
A^b =
1
-1
14
-1
0
>
1
2J
L-7_
The associated normal system is
(e) False; proJiyX is a vector in W.
3 -2 X, ^[14
(f) True; every invertible n x n matrix has a
(2^-decomposition.
-2
3 -2
-2
Section 6.4
Exercise Set 6.4
-7 ■
6 X2
1
14
6
reduces to
0
0
-7
5
1 1
the least squares solution is Xj = 5, X2 =^.
1. (a) A = 2
3
4
1
5
1
(b) A^A =
2
4
_
2
21
25
25
35
2
1
1
0
2
1
1
0
1
1
1
-1
-2
0
-1
-2
0
1
3
; 5
7
4-6
4
3
-3
-6
-3
6
1
2
1
1
0
1
1
1
2
1
2
4
20
-1
3
5
20
A^b =
6
5
The associated normal system is
A^b =
18
0
12
9
'21 25]fx,
25
, so
35
X,
20
-1
-2 0
20
The associated normal system is
159
-9
SSM: Elementary Linear Algebra
Chapter 6: Inner Product Spaces
7
18
4-6
4
3
-3
X2
12
-6
-3
6
^3
-9
7
4
-6
18
4
3
-3
12
-6
-3
6
-9
7. (a)
0
A=
1
2
24
8
8
6
1
1
3
12
2 4-2
A^b =
reduces to
2
1
2
8
1
1
The associated normal system is
"24 8]rx|1_[l2'
1 0 -3 so the least squares
1
1
2
-2
1 0 0 12'
0 0
2
4
2 4-2
8 ■
8 6 X2
9
solution is Xj = 12, X2 = -3, X3 = 9.
24
8
1
12
i
0
10
reduces to
8
6
0
5. (a) e = b-/4x
5
7
1
1
0
-1
1
1
-7
-1
2
2
1
5
the least squares solution is Xj = —
10’
6
X2 = — or X =
5
JI
7
2
2
-7
i
10
6
5
_9
0
6 ’
1
The error vector is
-4
e = b-y4x
3
2
3
2
9
2
4
2
-2 1 Ls
1
-3
1
2 10
6
7
3
T
1
-1
-1
1
1
2
2
A'e =
0
9
2
0
3
5
2
14
5
, SO e is
1
1
-3
8
orthogonal to the column space of A.
5
4
5
6
1
0
-1
0
12
(b) e = b - Ax =
0
2
1
-2
9
1
1
0
-3
so the least squares error is
9
3
1
6
3
0
3
9
9
1
„ „ IfsV r 4f
-1
+
3
V25 5
V 5j
(b) A^A =
0
1
-2
1
3
-2
-6
3
3-6 9
3
3
14
42
42
126
9
1
-3
A"b =
0
1
-2
3
4
0
3
-6
9
12
3
The associated normal system is
3
T
A‘e =
1
2
1
1
0
1
1
1
-1
-2
0
-1
0
"14 42irx,l^r 4"
-3
0
0 , so e IS
42 126j[x2j~Ll2_'
0
3
“14
42
4
42
126
12
1
3 1
reduces to
orthogonal to the column space of A.
160
, so
0 0
0
SSM:Elementary Linear Algebra
Section 6.4
2
1
the least squares solutions are Xj = — 3t,
7
7 +t
6
X3 = f or X =
2
7
X2 = f or X =
^1 for f a real
+f
0
-1
6
for t a real
-1
1
0
number.
number. The error vector is
1] r 1
e = b-/4x = 0
The error vector is
e = b-Ax
3' rr2
2
7
-6
-3
+f
0
ij [ 3 9
1
rr 2i '
1
3
2
0
2
1
3
rr 7
6
7 +t
-7
0
1
1
0
\_
+t
7
1
-1
6
0
7
4
0
7
VL
0
f-
0
6
14
7
rr 7
5
0
3
7
0
+r
0
6
-7
0
7
7
6
4
2
7
I
3
7
7
6
The least squares error is
49
6
rsf + 4f +^ if
|e|
The least squares error is
1)
1
1
^ 7f +r 7f +r
42
vU
V49
6;
V6
4
7
-1
3
2
2
1
3
2 3 lj_ 0
1
1
-1
2 0
3
A=
1
1
= ZV6.
2
2
1
-2
1
0
-1
10
1
1
0
14
1
1
-1
' 5 -1
4
9. (a) Let 4 =
-1
11
4
-1
4^b =
10
2 0
3 1
1
7
-7
0
14
_ 2 3 IJL-7
7
4^4 =
2
1
1
1
1
0
1
1
5
-2
-1
0
X2
14
4
3
-3
•^3
7
-6
-3
6
11
10
14
11
10
14
4
10
14
7
0
1
I
>
0 0 0
reduces to
1
-1
4 are linearly independent and the column
4
space of 4, VL, is the subspace of R
spanned by V|, V2, and V3.
7
" 3 -2
X , SO the least squares
0
(4^4)-' = -2
0
2
7
solutions are X| =
0
-1
6
0
0
det(4^4)= 3 0, so the column vectors of
4-7
-1
1
-2
4-6
-7
10
5-1
7
•«1
4
1
1
4
-1
2
1
The associated normal system is
1
^2
294
=lV42.
(c)
49
6
2
2 -1
5
3
7
1, X2 = — t,
6
161
SSM: Elementary Linear Algebra
Chapter 6: Inner Product Spaces
projjyU = A(A^A)'A^u
[2
1 -2lr
1
0
3 -2
-2
6
2
2
2
1
1
0
1
I
3
-1
9
0
2
-1
-2
-1
0
3
1
6
7
2
9
5
prqiiyu =(7, 2, 9, 5)
(b) Let A =
I
-2
3
-2
-3
0
3
-2
3
-3
0
-1
T
A' /l = -2
1
-2
1
-2
-3
0
11
-9
-9
10
8
-1
8
20
I
3
-1
T
det(A A)= 10 0 so the column vectors of A are linearly independent and the column space of A, W, is the
subspace of
spanned by V[,
M
M
5
5
M
(A^A)-'
Hi
5
_I2
5
10
10
11
_V9
29
10
10
5
and V3.
-If
proJiyU = A(A^A)“'A^u
'
1 -2 -3
68
5
I
3
-2
0
1
3
31
5
5
86
5
219
10
79
31
5
79
10
29
10
-2
I
3
0
0
-2
-1
-2
1
-3
-1
I
3
2
4
10
12
5
4
5
12
5
16
5
projiyii =
_4 _I2 26
5’
5’ 5 ’ 5
a
162
SSM: Elementary Linear Algebra
-1
11. (a)
A=
2
0
-1
3
2
(b) W = the yz-plane is two-dimensional and
2
1
3
one basis for Wis {wj, W2) where
0
1
1
3
2
3
1
5
= -1
11
4
10
Section 6.4
W| =(0, 1, 0) and W2 =(0, 0, 1).
4
0 0
10
Thus A= \ 0 and A^ =
14
0
det(A^ A)= 0, so A does not have linearly
-1
3
1
1
0 -5
1
-2
-1
0
-2
4
-5
3
0
1
0
0
0
1
1
1
20
-22
27
-17
20
-17
23
1
0
1
0
0
1
0 0
1
1
0
1
0
0 0
1
0
1
det(A^ A)= 0, so A does not have linearly
0 0 0
0
13. (a) W = the xz-plane is two-dimensional and
one basis for Wis {w,, W2) where
0
0 0
0
1
W| =(1, 0, -5) and W2 =(0, 1, 3) are a
0
1
1
0 0
15. (a) If X = ^ and y = t, then a point on the plane is
(5, f, -55 + 30 = 5(1, 0,-5)-H r(0, 1, 3).
W] =(1, 0, 0), W2 =(0, 0, 1).
0
0
basis for W.
1
1
1
1
1
A^A =
0
0 0
independent column vectors.
1
1
1
1
0
Thus A = 0 0 and A^ =
0
0
0
-22
1
0
0
0 0
3
21
1 '
[P]= A(A^Ar‘A^
4
(b) A^ A= -1 1
3
2
0
0
1
0
0
1
0 0
0 0
A^A =
independent column vectors.
2
0
0
0
1
0
0
(b) A =
0 0
0 0
0
1
0
1
0
0 1 , A^ =
-5
1
0
0
1
1
0
0
1
-5
3’
3
1
[P]= A(A^A)"'a^
1
T
0
0
0
1
0
1
0
1
0 0
1
0
0
0 0
1
1
26
-15
-15
10
-5
A' A =
3
-5
0 0
3
0 0
0
1
10 15
det(A^A)= 35, (A^A)'= _l_
35 15 26
1
0
0
[7>]= A(A^A)“'a^
0
0
1
0
0 0
0
1
0
0 0
-5
0
0
1
3
J_ 10 15
1
35 15 26
0
10
15
15
26
1
35
35
163
-5
J
0
-5
3JLO 1
3
10
15 -5'
15
26
3
-5
3
34
0 -5
1
3
SSM:Elementary Linear Algebra
Chapter 6: Inner Product Spaces
10
•^'0
15
-5
^0
15 26
3
3^0
3
34
1
(C) [/>] Jo
L^oJ
L^oJ
j r 10xo +15yo-5zo
=:^ 15xo +26yo+3zo
'^L-5xo+3yo+34zo_
10xo+15y(|-5Z|)
35
15xo+26y,|+3Z||
35
-5-«^i+3>’o+34z(|
35
The orthogonal projection is
2xo -3yo -zq ISxq + 26yQ +3zq -Sxq+3}'o+ 34zo ^
7
35
35
(d) The orthogonal projection of /’qCL “2, 4) on W is
2(l)-3(-2)-4 15(l)+ 26(-2)+ 3(4) -5(1)+ 3(-2) + 34(4)
7
35
_ 5
7’ 7’ 7 /
35
8
^2
The distance between PqCL -2,4) and W\s d =
r
TJJ
Using Theorem 3.3.4, the distance is
5^f +
+ -2- -
IJ)
25
n2
4—2^
3^^
77
1
|5(l)-3(-2)+4| _ 15 _3V^
752+(-3)2+i2
7
17. By inspection, when t = 1, the point (r, f, r) =(1, 1, 1) is on line/. When ^ = 1, the point (s, 2^- 1, 1)=(1, 1, l)is
on line m. Thus since ||P-Q||S 0, these are the values of y and t that minimize the distance between the lines.
19. If A has linearly independent column vectors, then
A is invertible and the least squares solution of Ax = b is
the solution of A^Ax = A^b, but since b is orthogonal to the column space of A, A^b = 0, so x is a solution of
T
T
A Ax = 0. Thus, X = 0 since A A is invertible.
21. A^ will have linearly independent column vectors, and the column space of A^ is the row space of A. Thus, the
standard matrix for the orthogonal projection of /?" onto the row space of A is
m = A^[(A")^A"r'(A")^=A^(AA^)-'A.
True/False 6.4
(a) True; A^A is an n x n matrix,
(b) False; only square matrices have inverses, but A^A can be invertible when A is not a square matrix,
(c) True; if A is invertible, so is A^, so the product A^A is also invertible.
(d) True
(e) False; the system A^Ax = A^b may be consistent.
(f) True
164
SSM: Elementary Linear Algebra
Section 6.5
(g) False; the least squares solution may involve a
V* =
a
1
=(A/^A/r‘Ay^y = 5
-3
.«2
(h) True; if A has linearly independent column
vectors, then
2
%
parameter,
The desired quadratic is y = 2+5x-3x^.
A is invertible, so
A^Ax = A^h has a unique solution.
5. The two column vectors of M are linearly
independent if and only neither is a nonzero
multiple of the other. Since all the entries in the
first column are equal, the columns are linearly
independent if and only if the second column has
at least two different entries, i.e., if and only if at
Section 6.5
Exercise Set 6.5
1
1.
M=
0
1
1 , A/^ =
1
1
1
0
1
1
least two of the numbers X], X2, •..,
2’
2
I
m'^m =
1
0
1
0
1
1
3
1
1
1
3
11. With the substitution X =—, the problem
3 5’
2
1
are
distinct.
X
2
becomes to find a line of the form y = a + b ■ X
1
5
-3
J_ 5 -3
I5-9L-3 3
6L-3
1
3
that best fits the data points (1,7), -,3 ,
3
v3
U’ J
0
=ir 5 -3iri 1 1 2
6[-3 3JL0 1 2 7
1
1
-1
7
. y= 3
M =
2
7
2
1
1
7
The desired line is y =
M^M =
1—x.
2
1
2
J- -L
3
3
1
3
2
1 II
I
6.
1
2
36
6
1
Xi
xf
2
1
3.
2
4
3
9
X-,
X2
1
^3
4
1
5
25
1
6
36
M=
X4
-I
(M^M)
J_
41 -54
42 -54
108
5
a
2
1
1
v* =
b
X4
=(Af^Af)“'A/^y =
21
48
7
0
The line in terms of 2f is y = —+—A", so the
-10
y=
21
-48
5
48
required curve is y= — H
-76
^
1
1
1
2
4
1
3
9
.y
1
m'^M= 2
3
5
6
4
9
25
36
4
16
74
16
74
376
74
376
2018
1989
-1
1
1
5
25
1
6
36
-1116
to
X
135
1
90
21 7x
-1 116
649
-80
135
-80
10
10
165
7
SSM: Elementary Linear Algebra
Chapter 6: Inner Product Spaces
3. (a) The space W of continuous functions of the
form a + be^ over [0, 1]is spanned by
True/False 6.5
(a) False; there is only a unique least squares
V| =1 and V2 =e''.
straight line fit if the data points do not lie on a
vertical line.
Let gi = V| =1.
(b) True; if the points are not collinear, there is no
solution to the system,
~=t\^dx
=\
Jo
8l
1
'...r
e^dx =
(c) False; the sum df+d^-^
= e-\
\-d}^ is minimized,
(^2>g|)gl
not the individual terms d-,.
I
82 = ^2-
ig.f
(d) True
e-1
= e^-
(1)
I
Section 6.6
= l-e+ e"^
Exercise Set 6.6
82
1. With/(A-)= I+x:
= \{\-e+e^fdx
•J
1
2^
An
[ {\ + x)dx
X
%=K
2
= JQ''o-2e+ e^ + 2e^-2ee^+e'^^)dx
= 2+ 2k
x+
7t
0
1
/
1
1
r2;r
aI, =- I
= x-2ex + e^x+2e''-26^^*^ + — e2x
(\ + x)coskxax
2
rr
*
cos kx dx + —
0
3 ^
1 2
— + 2e —
2
2
X cos kx dx
n
In
I
kn
= --(l-e)(3-<?)
+0
sin/:x
0
gl and g2 are an orthogonal basis for W.
=0
The least squares approximation of x is the
orthogonal projection of x on W,
1 r2;r
I
h=
(l + x)sinLxc/x
n
1
An
g|)
1 r2;r
sinLrc/xH—
K
xs,\nkxdx
g|
1
cos Lx
kn:
0
IT 82
82
2- 2
1
/
^
\
J
^ 2
^
(x,g,)=_(^xr/x =-x ^=-
2
k
g2)= lx{\-e+e^)dx
3
(a) aQ=2+ 2n:, a^=a2=0, 6, =-2, 62=-!
1
—e
2
Thus /(x)= 2(2+ 2.ir)-2sinx-sin2x
i
projH/X = ^(l)+
(b) a, = £12 = «„ =0
2
— so bi. sin Lx =
L
2
= 2(3-.)
=(1 +;ir)-2sin x-sin 2x.
bk =
(--82)
T"gl +
prqiH/X =
K
{(3-e)
Xsin Lx'
^
(1-.+.-^)
k
sin 2x
f(x)~ (\ + n)-2 sinx +
sm/;x
+ •■• +
2
n
2
I
\-e
2
1-e
1-e
1
2
166
(l-e+ e"')
-i(l-.)(3-e)
e-1
1-e
SSM: Elementary Linear Algebra
Section 6.5
(b) Let f| =x and f2 =projiyf|, then the mean
square error is ||f| -f2|f • Recall that
f| -projjyfi =f| -f2 and {2 are orthogonal.
=0
-1
fi “f2|f -(f|-h’fi “fa)
(V3.gl)„
(V3,g2)g2
81-
=(fl.f|-f2)-(f2.fl-f2)
=(f|.fl-f2)
=(fl-f.)-(f|-f2)
g3 = ''3 -
l|8.f
llgaf
2
= ^2
= llf.f-(fl-f2)
-|(l)-0
9
— + x~
Now decompose f2 in terms of g| and g2
3
from part (a).
{g|. g2> 83} 'S an orthogonal basis for W.
The least squares approximation of
2 (f|.8i)
f, = sin nx is the orthogonal projection of
f| on VL,
(fi.82)
—(fl>g2)
^(fi'gl)-
I
82
2 (fh8l)" (fl.82)^
1
ft.8,).
i s.r ^,ft.82)^^,ft^^^
feif
proJwfi =
83
|2
|82
I
||flf = Jjx2r/x =
1
(fl,gi)= ^ s\n7!:xdx =
3'
A'
1
cos;^'jc
n
=0
-1
3
3 0
rl
2
(f], 82)= I .^sin Kxdx = —
Thus,
(i)' (io-.)f
ft.83)=t(-3+A^ ?,\n7txdx =0
-l(l-e)(3-e)
2
3-e
1
3
proJu^f, =0gi+^A +0g3 =
3 4’^2(l-e)
(b) Let f2 = projiyf|, then in a clear extension
1+e
of the process in Exercise 3(b),
“ 12^ 2(1-e)
i2 (f|.8i)^ (fi.82)^ (fi’g3)^
5. (a) The space W of continuous functions of the
form flQ
+
1
over[-l, 1] is
2
“ l|8.f " ||82|f
spanned by v, = 1, V2 = a, and V3 =a .
I
Let gi = V, = 1.
hf = t^"dx =
sin^ Kxdx = 1
Thus, the mean square error is
I
A|_i = 2
2)2
6
-0=1-
0
=0
xdx = —x''
1
2
(^2.81)g| = V2 = A-
9. Let /(a)=
llsif
I
'f
2
3
l|82f = J-1 x~dx = —3 x
;r2
3
-1
82 = ''a -
9
||83f
•L|
I
(v23 8i)=
7t
3
l 2-2g + l + e
”12
2(l-e)
13
—X
ao=7t
-I
3
167
1, 0 < A < ;r
0, n<x<'ln'
dx = \
SSM: Elementary Linear Algebra
Chapter 6: Inner Product Spaces
at =—
fix)cos kxdx
n -O
'^coskxdx
— -^5x2 = ^2 4~5.
||x|| —
=0
n
1
If llxll = 1, then JC2 -±
, so
^/5
n
1
x =± 0,
'f
sm kxdx
n
kK
3. Recall that\f U =
So the Fourier series is
-+5
—(l-(-l)'=)sinfa.
2
UiI
u
2 and V =
L“3 “4
V,
V2
V3
y4
then (f/, V)= MiVi+U2''2 ■*'“3''3
(a) If 1/ is a diagonal matrix, then U2 =1/3=0
True/False 6.6
and (f/, V) =/<iV|+M4V4.
(a)
False; the area between the graphs is the error,
not the mean square error.
(b)
True
(c)
True
For V to be in the orthogonal complement of
the subspace of all diagonal matrices, then it
must be the case that V] = V4 = 0 and V
must have zeros on the main diagonal.
(b) If f/ is a symmetric matrix, then «2 = “3
(d) False; ||i|| = (l, l) =
and (l/, v) = //,v,+m2(v'2+V3) + //4V4.
= 2k^\
Since //j and
(e)
can take on any values, for
V to be in the orthogonal complement of the
subspace of all symmetric matrices, it must
True
be the case that V| = V4 = 0 and V2 = ~''3-
Chapter 6 Supplementary Exercises
thus V must be skew-symmetric.
1. (a) Let v = (v|, V2, V3, V4).
(v,U,) = V,, (v,U2) = V2, (v,U3) = V3,
(v, U4) = V4
If (v, u,) = ^v, U4) = 0, then Vj = V4 = 0
5. Let u =
and
1
. By the Cauchy-Schwarz
and V = (0, V2, V3, 0). Since the angle 6
between u and v satisfies cos 0 =
...,
Inequality, (u ■ v)^ = (1 -1- ■ ■ ■ •+• 1)^ < ||u|P ||v|P
(u, v)
or
n terms
Iu|
n
then V making equal angles with U2 and U3
means that V2 = V3. In order for the angle
^ < (ai H
l-fli)
1
1
a
«l
n
1. Let X = (xj, X2, X3).
between v and U3 to be defined ||v|| ^ 0.
(x, U,) = X,+X2-X3
(x, U2 ) = -2xi - X2 + 2x3
(x, U3) = -x,-hX3
(x|, U3) = 0 => -X| -h X3 =0, so X| = X3. Then
(x, Uj) = X2 and (x, U2) = -X2, so X2 = 0 and
Thus, V = (0, a, a, 0) with a^O.
(b) As in part (a), since (x, U|) = (x, U4) = 0,
xj = X4 = 0.
Since ||u2|| = ||u3|| = 1 and we want ||x|| = 1,
then the cosine of the angle between x and
x = (X|, 0, X|). Then
U2 is cos ^2
“2) = ^2
similarly,
cos^3 = (x, 03) = X3, so we want X2 = 2x3,
and x = (0, X2, 2x2, O)-
llxll = yjxf+xf =
168
= X,
SSM:Elementary Linear Algebra
Section 6.6
n2
1
If llxll = 1 then X| = ±
and the vectors are
cos^ cr, =
U
1
V;
/
1
2
± -p,0, -p. .
lV2
V2 j
a./
(h||= l)
Vllul
2
9. For u =(«|, U2)> v =(V[,V2) in
a:
let
I
1'
af +132 +**‘+ a„
(u, v)= aM|Vj +bu2V2 be a weighted inner
Therefore
product. If u =(1, 2) and v =(3, -1)form an
2
orthonormal set, then
cos^
iP =ai\f+b{2f =a +4b = l,
\\yf =aOf+bi-\f-^9a +b = l, and
= a\
+--‘+ cos^
+ <32 "f
U
15. To show that
(u, v)= a(l)(3)+ b(2){~ 1)= 3a -2b = 0.
'l
4
9
1
1
Since
4
1
reduces to
0 0
0
Theorem, that v = W|+ W2 where w, is in IF
1
system is inconsistent and there is no such
and W2 is in IF"*". By definition,
(v, W2)=(w|, W2)= 0. But
(v. W2)=(W|+W2, W2)
=(w,, W2)+( 2> 2)
=(W2, W2)
so that (w2, W2)= 0. Hence W2 =0 and
weighted inner product.
11. (a) Let U| =(^,0,0, ..., 0),
U2 =(0, k, 0,..., 0), .... U2 =(0, 0, 0,.... k)
be the edges of the ‘cube’ in
w
w
and
u =(k, k, k, ..., k) be the diagonal.
Then ||u,-1 = k, ||u|| = kyfn, and
therefore v = Wj, so that v is in IF. Thus
, so
(IF-^)-^ cIF.
k^
1
cos^ =
-1
1
k{kyfn) 'In
u,
I
= W, we first show that
Since v is in F, we have, by the Projection
0 , the
1
0 0
3 -2 0
n
Thus IFc(IF'^)'^.
To show that (IF'*')'*' c IF, let v be in (IF'*')'*'.
0
1
1
9
<3^
to every vector in IF"*", so that w is in (IF'*')'*'.
1
3
2
W c(IF'*")'*'. If w is in IF, then w is orthogonal
1
a
This leads to the system
2
+a2+---+ a,,
17.
A= 2
3 ,
4
(b) As n approaches 0°,
4~n
approaches 0, so 6
A^A =
2
-1
4
3 5 ’
5
21
25
25
35 ’
K
approaches —.
=
1
2
4
-1
3
5
A^b =
4^+3
1
5i + 2
s
The associated normal system is
13. Recall that u can be expressed as the linear
combination u = aiVi+—
'21 25irx,1^r45+3"
where
_25 35j[x2j L5‘y'^2_'
a,- =(u, V,)for( = 1, ..., n. Thus
If the least squares solution is X| = 1 and X2 = 2,
21
then
25
1
25 35j[2
71
4^+ 3'
95
5s + 2 ■
The resulting equations have solutions ^ = 17
and s = 18.6, respectively, so no such value of s
exists.
169
Chapter 7
Diagonalization and Quadratic Forms
A is orthogonal with inverse
Section 7.1
I
I
0
71
Exercise Set 7.1
1. (b) Since A is orthogonal,
4
9_
il
5
25
25
0
1
1
5
5
1
_J1
16
5
25
25
= 21^ =
2_
n/6
^/6
_L
_L
_L
^/3
^/3
J_
(e) For the given matrix A,
A^A
3. (a) For the given matrix A,
2l^A = 1 oin 0l_
0 1 0 I "
1
n/6
1
0
0
1
■
T
A is orthogonal with inverse 2\
1
0
0
1
I
I
I
I
1
1
1
1
9
2
9
2
2
2
2
2
I
5
I
I
1
5
I
I
9
6
6
2
6
6
6
1
1
1
_5
1
1
1
_1
2
6
6
6
2
6
6
6
I
5
I
I
I
5
I
6
6
6
2
6
6
6
i
.2
■
1 0 0 0'
0
(b) For the given matrix A,
J_ J-lf-L
a’’’A = V2 n/2 n/2
x/2
1
I
I
I
42
42
42
42
1
1
0
1
2l is orthogonal with inverse
1 ■
A is orthogonal with inverse
A^ =
0 0
0 0 0
0
0
1
0 0
I
i
i
1
2
2
2
-2
1
_1
1
1
2
6
6
6
i
I
1
_5
6
6
i
1
2
4i
4i
1
2
i
I
4i
4i
3
—
2
U2)
1 so the
n
3
X
a4 A
1J
~T2
1
I
1
i
41
4i
4~6
4~2
2
I
4^
45
I
7
46
46
46
i
i
I
I
i
45
45
45
4E
45
. 45
0
1
1
0
1
3
2
2
2
-2
^/5
1
2
2
6
-I + 3V3
n^+3
Thus, (/,/)=(_i+373, 3+ x/3)-
0 0"
0 0
n
sm— =
2
1
(d) For the given matrix 2l,
0
.
7. (a) cos—= —,
matrix is not orthogonal,
I
2J
7
— ?il
12
The matrix is not orthogonal.
n2
V
1
6.
if
<n
V
0^+1- +
6
1-5
6
6
Note that the given matrix is the standard
matrix for a rotation of 45°.
I
0
1
170
SSM: Elementary Linear Algebra
Section 7.1
(b) Since a rotation matrix is orthogonal,
Equation (2) gives
I
9
^3
i
2
2
. Thus the transition matrix
0
[u3]b =
COS(9
f-V5
5
2
sin6>
COS0
2
from B' to 5 is P =
X
i.e.,
sin6>
0
cos^
0
-sin^
Thus, (x, y)= — - V3, —73
+1 .
2
0
0
X
>- =P y' . Then
9. If 5 = {U|, U2, U3) is the standard basis for R^,
and B'= {uj, u^, U3) is the rotated basis, then
cosd
-1
A =P
the transition matrix from B' to B is
"cos| 0 sinI
P=
0
0
sin6>
I
0
0
I
0 -sin^
0
0
cos^
2
0
S
-siny 0 cosj
0
1
1
0
(b) With the same notation, [U]]g = 0
2
0
-1
y' =P
y
z
z
[u^]^ = cosd , and [u^]^ = -sin^ , so
cosd
sin^
s
1 0 -2
6 1 0
the transition matrix from B' to B is
-1
2
0
0
p= 0
cos^
-sin (9
0
sin^
cosd
# 0 1 jL 5
1
2
0
0
X
X
(a)
0
0
cos(9
sin6>
2^'
A= 0
0 -sin 6’
Thus
a+b
b-a
a-b
b+a
cosd
. Then
13. Let A =
(x, y, z)=
Li_2V3.2,-iTS+n
2
2
2 2
T
A' A =
i
X
X
2
0 2 r I
0
1
2(0^+b^)
0
0
2(0^+b'^)
9
(b)
and
y =P / =
2
0
-4 0 i
6
, so a and b
9
1
must satisfy a +b = —.
2
-3
17. The row vectors of an orthogonal matrix are an
orthonormal set.
6
n2
-1734J
Thus (x, y, z)=
1-1S.6.-2S-3\
2 2
2
2j
1 f = a^ +\
+
I*'!
Thus a = 0.
(I
n2
1
+
T2
= b^+-
re
11. (a) If 5 = {uj, U2, U3} is the standard basis for
6
2
Thus b =±
and B'= {u,, U2, U3}, then
cosB
K]b =
0
-sinB
re
0
1
, [u,]^ = 1 , and
r
0
171
2
2
= c +-
+
r
3
SSM: Elementary Linear Algebra
Chapter 7: Diagonalization and Quadratic Forms
Section 7.2
1
Thus c = ±
-S'
Exercise Set 7.2
The column vectors of an orthogonal matrix are
an orthonormal set, from which it is clear that b
1. (a)
and c must have opposite signs. Thus the only
2
X-l
-2
-2
X-4
= a2-5A
1
possibilities are a = 0, b =—^, c = —^ or
s
The characteristic equation is
-5^ =0
and the eigenvalues are A. = 0 and A = 5.
s
1
a = 0, ^ =-4
S’ "^S'
Both eigenspaces are one-dimensional.
21. (a) Rotations about the origin, reflections about
any line through the origin, and any
combination of these are rigid operators.
(b)
A-1
4
4
A-1
-2
2
-2
2 =A^-27A-54
A+2
=(A-6)(A +3)2
(b) Rotations about the origin, dilations,
contractions, reflections about lines through
the origin, and combinations of these are
angle preserving.
The characteristic equation is
A^ -27A-54 =0 and the eigenvalues are
A = 6 and A = -3. The eigenspace for A = 6
is one-dimensional; the eigenspace for
A = -3 is two-dimensional.
'y
(c) All rigid operators on R are angle
preserving. Dilations and contractions are
angle preserving operators that are not rigid.
A-1
(c) -1
True/False 7.1
-1
-1
A-1
-1
-1
-1 =A^-3A^ =A^(A-3)
A-1
(a) False; only square matrices can be orthogonal.
The characteristic equation is A^ -3A^ -0
(b) False; the row and column vectors are not unit
and the eigenvalues are A = 3 and A = 0. The
eigenspace for A = 3 is one-dimensional; the
eigenspace for A = 0 is two-dimensional.
vectors.
(c) False; only square matrices can be orthogonal.
(d)
(d) False; the column vectors must form an
orthonormal set.
A-4
-2
-2
A-4
-2
-2
-2
-2 =A^-12A2-h36A-32
A-4
=(A-8)(A-2)2
(e) True; since A^A = / for an orthogonal matrix A,
The characteristic equation is
A must be invertible,
A^-12A^^-36A-32=0 and the
eigenvalues are A = 8 and A = 2. The
eigenspace for A = 8 is one-dimensional; the
eigenspace for A = 2 is two-dimensional.
(f) True; a product of orthogonal matrices is
orthogonal, so A^ is orthogonal, hence
det(A2)=(detA)2 =(±1)2 =i.
A-4
(g) True; since IIAx||= l|x|| for an orthogonal matrix.
-4
(e)
\
0
0
(h) True; for any nonzero vector x, Tj-r is a unit
-4
0 0
"o' A
0
0
A
The characteristic equation is A'^ -8A‘^ =0
and the eigenvalues are A = 0 and A = 8. The
eigenspace for A = 0 is three-dimensional;
the eigenspace for A = 8 is one-dimensional.
vector, so A^-jr =1. It follows that
1
IIAxil = 1 and ||Ax||=||x||, so A is orthogonal
by Theorem 7.1.3.
172
SSM:Elementary Linear Algebra
1-2
(f)
1
0
Section 7.2
0
3
4
I
X-2
0
0
0
0
X-2
1
0
0
1
X-2
=
3
is an orthonormal
X3 = 0
1
5
0 1
+ 22%} -2AX +9
basis. Thus P =
The characteristic equation is
0
1
0
orthogonally
. 10 f.
-%X^ + 22X^ -24X +9=0 and the
eigenvalues are A,= 1 and A = 3. Both
diagonalizes A and
eigenspaces are two-dimensional.
P~'AP =
-I
A,
0
0
0
0
0
At
0
0-3
0
A3
0
0
3. The characteristic equation of A is
A^-13A +30 =(A-3)(A-10)= 0.
0
25
0 -50
7. The characteristic equation of A is
A| = 3: A basis for the eigenspace is
A^-6A^-r9A = A(A-3)^ =0.
2
2
X
1 -
1
S . Vl =
_
73
77
X|
1
IS an
A, = 0: A basis for the eigenspace is x, = 1
orthonormal basis.
i
A2 = 10: A basis for the eigenspace is
X2=
2
^1
[Til
[77
•
77
Vo =
Ihi
1
vi =
is an orthonormal basis.
1
IS an
2
77
2
77
77
77
orthonormal basis. Thus P =
A2 = 3: A basis for the eigenspace is X2 =
77
0
2
77
orthogonally diagonalizes A and
p~'ap =
I
77
^3 =
0 . The Gram-Schmidt process gives the
1
Ai 0]^r3 o'
0 A2j“L0 10 ■
1
1
7^
"77
orthonormal basis V2 =
5. The characteristic equation of A is
1
77
A^+28A^-1175A-3750
I
. V3 =
2
0
77J
=(A-25)(A-t-3)(A-F50)
= 0.
A| = 25: A basis for the eigenspace is
V
3
xi =
0
Thus, P =
4
4
1^1
1
76
_
j_
L
77
77
76
0
is an orthonormal
3
orthogonally
2
S
73
diagonalizes A and
5
A,
basis.
0
0
0 0 0
P~'aP= 0 X2 0
X2 = -3: A basis for the eigenspace is
0
0
X2 =
1
77
5
0
''1 =
I
77
1 = V2, since ||x2|| = 1.
0
0
3 0
0 0
A2
3
9. The characteristic equation of A is
0
A'^ -1250A^ -(- 390,625 =(A -1- 25)^(A-25)^
A3 = -50: A basis for the eigenspace is
= 0.
A| = -25: A basis for the eigenspace is
173
SSM: Elementary Linear Algebra
Chapter 7; Diagonalization and Quadratic Forms
4
15. No, a non-symmetric matrix A can have
eigenvalues that are real numbers. For instance,
0
3
0
1
X| =
, X2 =
4
0
. The Gram-Schmidt
the eigenvalues of the matrix 8^
3
0
are 3 and
0
-1.
4
5
17. If A is an orthogonal matrix then its eigenvalues
3
process gives the orthonormal basis V| =
5
have absolute value 1, but may be complex.
0
Since the eigenvalues of a symmetric matrix
must be real numbers, the only possible
eigenvalues for an orthogonal symmetric matrix
0
0
are 1 and -1.
0
4
V2 =
19. Yes; to diagonalize A,follow the steps on page
399. The Gram-Schmidt process will ensure that
columns of P corresponding to the same
eigenvalue are an orthonormal set. Since
eigenvectors from distinct eigenvalues are
orthogonal, this means that P will be an
orthogonal matrix. Then since A is orthogonally
diagonalizable, it must be symmetric.
5
3
.S
^2 = 25: A basis for the eigenspace is
2
0
4
0
, X4 =
^3 =
3
0
4
0
1
. The Gram-Schmidt
True/False 7.2
3
.S
0
True; for any square matrix A, both AA^ and
A^ A are symmetric, hence orthogonally
0
diagonalizable.
4
process gives the orthonormal basis V3 =
(a)
.s
0
(b) True; since V[ and V2 are from distinct
0
''4 =
3
eigenspaces of a symmetric matrix, they are
. Thus,
orthogonal, so
.5
4
i^'l +''2f =(''1 + V2, Vi -I-V2)
5
P =[V|
V, V3
V4]=
0
0
=(v|, V|)+ 2(v,, V2)+(v2, V2)
0
0
= lhf+0-.||v2f.
0
4
.S
3
5
0
3
4
.S
5
4
3
5
.S
5
5
0
0
(C) False; an orthogonal matrix is not necessarily
symmetric.
-1
(d) True; (P^'APr' = P“‘a~'p since P“‘
and
both exist and since P~^AP is diagonal, so
is (P"'APr‘.
orthogonally diagonalizes A and
0
p‘‘ap = 0
0
0
0
0
'
>.2 0 0
0 ?i| 0
0 0 ^2
-25
0
0
O'
0
25
0
0
0
0 -25
0
0
0
(e)
True; since ||Ax||=||x|| for an orthogonal matrix.
(f) True; if A is an n x n orthogonally
O '
diagonalizable matrix, then A has an
orthonormal set of n eigenvectors, which form a
25
basis for R”.
174
SSM: Elementary Linear Algebra
Section 7.3
(g) True; if A is orthogonally diagonalizable, then A
must be symmetric, so the eigenvalues of A will
2
3
all be real numbers.
2
bases for the eigenspaces are ^ = 1
3
i
Section 7.3
L
2
Exercise Set 7.3
i
3
1
X = 4:
1. (a) 3x| +7x3 =[X|
x^]
3
0
3
a:,
2
0 7j[a:2
3
3
2
■1 = 1:
3
_2
-11
(b) 4xf-9x2-6x|X2 =[x,
_3
-3
•^1
-9
X,
For P =
3
3
2
i
2
3
3
3
= [X|
X2
X3]
-4
3 -1
-4
Ax = y^
3.
AP)y
1
= [>’1
2 -2
12
3
0
0
3'!
^3] 0 4
0
>'2
0
7
>"3
x.
1
4
, let X = Py, then
11-1
(c) 9X|^ - x| + 4x1 + 6x, X2 - 8X| X3 + X2X3
3
1
3
3
9
3j
12
X3
0
and
Q = y1 +Ayl+lyl.
-3
3. [X
y]
X
sJLr
= 2x^ +5y^ -6xy
r o
5. (2 = x^Ax = [X| X2]
-1
9. (a) 2x^+x>' + x-6y+ 2 = 0 can be written as
2x^+0y^+xy + (x-6y) + 2 = 0 or
X,
2
2JU2.
— r
" +[1 -6]
i, ^0 L3^J
y]
The characteristic equation of A is
X
+ 2 = 0.
y.
x2-4;^ + 3 = (X-3)(X-1) = 0, so the
eigenvalues are ^ - 3, 1 . Orthonormal bases for
(b)
1
^2
the eigenspaces are /\, = 3;
;
1
+7x-8>’-5 = 0 can be written as
Ox^+ y^+0xy + (7x-8y)-5 = 0 or
I:
Tl
[X
0
y]
0 lj[y
1
^2
+ [7
-8]
11. (a) 2x2+5y^ =20 is —
+
10
I
1
75
75
L
L
-5 = 0.
y
9
I
For F =
Olfx
2
y
= 1 which is the
4
equation of an ellipse.
, let X = Py , then
75 75j
(b) x2 -y ^-8 = 0 is x^ - y^ = 8 or
3
x’' Ax = y''{p'^ AP)y =[y^ y2]
0
x2
y,
0 lj[y2_
y2
— = 1 which is the equation of a
8
8
hyperbola.
and 2 = 3y|^ + y|.
3
7. g = x^Ax = [x, X2 X3] 2
2
0
9
4-2 X2
[0 -2
7
9
(c) 7y -2x = 0 is x = —y which is the
^1
equation of a parabola,
5JL^3
The characteristic equation of A is
(d) x^ + y^-25 = 0 is x^+y^ =25 which is
I? -12?i^ + 391 - 28 = (1 -1)(1 - 4)(1 - 7) = 0,
the equation of a circle.
so the eigenvalues are 1 = 1,4, 7. Orthonormal
175
SSM: Elementary Linear Algebra
Chapter 7: Diagonalization and Quadratic Forms
13. The equation can be written in matrix form as
x^Ax = -8 where A =
i
eigenvalues of A are
= 3 and X2 = -2, with
-
corresponding eigenvectors v, =
-1
0
0
-2
(b) The eigenvalues of
, are ^ = -1
and X = -2, so the matrix is negative
definite.
and
-1
-1
0
0
2
(c) The eigenvalues of
are X = -1 and
1
X = 2, so the matrix is indefinite.
V2 = 2 respectively. Thus the matrix
2
1
1
(d) The eigenvalues of
P=
2 orthogonally diagonalizes A.
1
^/5
rs
0
(e) The eigenvalues of
.ry'-coordinate system is
Qjx' = -8
-4/.
19. We have Q = xf+xj>0 for (x,,X2)^(0, 0);
thus Q is positive definite.
1
-1
= -26.6°.
2j
21. We have g =(X]-X2)^ > 0 for X] ^ X2 and
g 0 for X| = X2; thus g is positive
15. The equation can be written in matrix form as
11
T
X Ax = 15 where A =
12
semidefinite.
12
. • The
4_
23. We have xf -x|> 0 for X] 5^ 0, X2 =0 and
eigenvalues of A are A-j = 20 and X2 = -5, with
corresponding eigenvectors
^
g < 0 for X, = 0, X2 A 0; thus g is indefinite.
and
25.
(a) The eigenvalues of the matrix
-3
A=
^ respectively. Thus the matrix
i
5
L5
5 -2l
are A = 3 and A,= 7; thus A is
positive definite. Since |5| == 5 and
_3
^ orthogonally diagonalizes A. Note
5
-2
2 ^ = 21 are positive, we reach the
5_
that det(P)= 1, so P is a rotation matrix. The
equation of the conic in the rotated
same conclusion using Theorem 7.3.4.
x'y'-coordinate system is
[x' y']
20
2 are A.= 0 and
semidefinite.
The angle through which the axes have been
rotated is P = tan
0
0
A,= -2, so the matrix is negative
which can be written
as 2y'^ -3x'^ = 8; thus the conic is a hyperbola.
V2 =
are A,= 1 and
X-Q,so the matrix is positive semidefinite.
Note that det(f’)= 1, so P is a rotation matrix.
The equation of the conic in the rotated
r , ,J3
0
0 0
2-1
Olfx'
0 -5j[y'
(b) The eigenvalues of A = -1
= 15 which we can write
0
are
5
definite. The determinants of the principal
The angle through which the axes have been
-1
2 0
0
X= 1, A = 3, and A = 5; thus A is positive
as 4x'^ - y'^ = 3; thus the conic is a hyperbola.
rotated is P = tan
0
2
3^
—1
submatrices are |2|= 2, ^
4 =36.9°.
3, and
4j
2-1
17. (a) The eigenvalues of
1
0
0
2
-1
are A = 1 and
0
0
2 0=15; thus we reach the same
0
5
conclusion using Theorem 7.3.4.
A = 2, so the matrix is positive definite.
176
SSM: Elementary Linear Algebra
^
Section 7.3
^
0
T'
27. The quadratic form Q = 5X| + X2 + kx^ + 4xiX2 -2X|X3 -2x2X3 can be expressed in matrix notation as Q = x Ax
5
where A =
1
5
2
1 -1 . The determinants of the principal submatrices of A are |5| = 5’
2
5
2
2
1
= 1, and
k
-1
2
1 -1=^-2. Thus Q is positive definite if and only if ^ > 2.
2
-1
-1
k
31. (a) For each / = 1,
n we have
(x;-x)^ = xf -2xjX + x^
^2
I f"
= X!
I
T
n
ry
n
y=i
n
1
n
'*
^
n
"
=
7
n
n
;=l
7=1
,.U±
n-l
II
n
1
—,
and the coefficient of XjXj for i itjis
n
n
Z
n
i
n
that si =x^Ax where A =
i
«(«-!)
i
n(/!-l)
1
(b) We have
)
[(X]-x)^ +(X2-x)^ +•• •+(x„ - x)^] the coefficient of xf is
Thus in the quadratic form s], =
1
j=\k=j+\
=
2
2
2
2
. It follows
n
n-\
n
-r7
n(n-])
1
i
/i(n-l)
n(n-\)
1
i
»(«-!)
It
1
i
n(;i-l)
n
[(Xi - x)^ +(X2 -x)^ +• ••+(x„ -x)^]> 0, and
=0 if and only if x, = x,
n-1
X2 =X,
X„ = X,
i.e., if and only if X| = X2 = • • • = x„. Thus
is a positive semidefinite form.
33. The eigenvalues of A must be positive and equal to each other. That is, A must have a positive eigenvalue of
multiplicity 2.
True/False 7.3
(a) True
(b) False; because of the term 4X1X2X3.
(c) True; (xj -3x2)^ = xf -6x|X2 +9x|.
(d) True; none of the eigenvalues will be 0.
(e) False; a symmetric matrix can be positive semidefinite or negative semidefinite.
(f) True
(g) True; x • x = X|2 + x|H
1- x,^
177
SSM: Elementary Linear Algebra
Chapter 7: Diagonalization and Quadratic Forms
(h) True
0
V2 = 1 , V3 = 0 . Thus the constrained
(i) False; this is only true for symmetric matrices.
0
maximum is 9 occun'ing at (x, y, z)=(±1,0, 0),
and the constrained minimum is 3 occurring at
(X, y, z)=(0, 0,±1).
G) True
(k) False; there will be no cross product terms if
a-.;
u
0
= -Oji for all( j.
7. Rewrite the constraint as
T
2
rxf +
3’
= 1,
U2
then let x = 2x| and y ='J2y^. The problem is
now to maximize z = xy = I'JlX]y| subject to
X|“ + yj^ = 1. Write
2)
(I) False; if c < 0, x Ax = c has no graph.
Section 7.4
Exercise Set 7.4
1. The quadratic form 5x“ -y^ can be written in
Z = x^Ax =[x, y,]
0
matrix notation as x^Ax where A = ^
0 V2 [x
I
x/2 0 [>'1.
0
Since
The eigenvalues of A are X, = 5 and I2 ="I
-V2
the largest eigenvalue of A is V2, with
and
with corresponding eigenvectors v, =
-2=(X+ V2)U-V2).
1.
0
I
0
V2 =
. Thus the constrained maximum is 5
corresponding positive unit eigenvector
occurring at {x, y)=(±1, 0), and the constrained
Thus the maximum value is
minimum is -1 occurring at (x, y)=(0, ± 1).
\/
1
z = 2V2
3
0
0
7 ■
occurs when
I72JL72
X = 2x| = V2, y = ^/2y| =1 or x = -72,
3. The quadratic form 3x^ +7y^ can be written in
matrix notation as iJAx where A =
1
s/5
y = -1. The smallest eigenvalue of A is -75,
1
The eigenvalues of A are
=7 and ^2 =3,
with corresponding unit eigenvectors ±
0
with corresponding unit eigenvectors V| =
s/2
Thus the minimum value is z =-75 which
1
and V2 =
0 ■
Thus the constrained m aximum is
occurs at (x, y)=(-75, l) and (75,-l).
7 occurring at (x, y)=(0, ±1), and the
constrained minimum is 3 occurring at
9.
57-7=5
(X, y)=(±l,0).
5
,v
2
0
0
0
..2
7+7=175
5. The quadratic form 9x +4y +3z" can be
A-
["9 0 o'
expressed as x^ Ax where A = 0 4 0 . The
(0,-1
1,0)
5
0 0 3_
eigenvalues of A are X, =9, ^2 = 4, ^3 = 3,
1
13. The first partial derivatives of/are
with corresponding eigenvectors V| = 0 ,
/.(x, y)= 3x^-3y and /^.(x, y)= -3x-?>/.
0
To find the critical points we set /j. and fy
178
SSM: Elementary Linear Algebra
Section 7.5
17. The problem is to maximize z = 4xy subject to
equal to zero. This yields the equations y = x^
\2
and X = -y^. From this we conclude that
x^+25y^ =25, or
y = y"* and so y = 0 or y = 1. The corresponding
maximize z=20x|)'i subject to ||(xi, y|)||= l.
(-1, 1).
Write z=
The Hessian matrix is
/xr(^> y) fxy(^’ y)
6x
-6}’.'
0
-3
-3
0
Ax =[X| y,]
0 lOlfxi'
10
-3
fyx(x^ >’) fyy(x,
The eigenvalues of H{0, 0) =
= 1. Let
U
X = 5X| and y = y\, so that the problem is to
values of x are x = 0 and x = -1 respectively.
Thus there are two critical points;(0, 0)and
H{x, y) =
Is.
+1
-10
K
The largest eigenvalue of A is A- = 10 which has
are
1
positive unit eigenvector
X = +3; this matrix is indefinite and so/has a
. Thus the
i
saddle point at (0, 0). The eigenvalues of
-6
Hi-l 1) =
^ are A. = -3 and
1
maximum value of z = 20
■6
-3
= 10
[yflAyflj
A, = -9; this matrix is negative definite and so/
has a relative maximum at (-1, 1).
which occurs when x = Sxj =
15. The first partial derivatives of/are
1
3' = 3'| =
Aix, y) = 2x-2xy and fy{x, y) = 4y-x“. To
find the critical points we set /^ and fy equal
then q{x) = x^Ax = x^(A.x) = A.(x^x) = A.(l) = X.
X = 0 or y = 1. Thus there are three critical
points: (0, 0), (2, 1), and (-2, 1).
True/False 7.4
The Hessian matrix is
(a)
fxxA y) fxyA y)
fyxi^’ y) fyyA, y)
False; if the only critical point of the quadratic
form is a saddle point, then it will have neither a
maximum nor a minimum value.
2-2y -2x'
4
(b)
True
(c)
True
(d)
False; the Hessian form of the second derivative
■
The eigenvalues of the matrix N(0, 0) =
2
0
0
4
test is inconclusive.
are A. = 2 and A, = 4; this matrix is positive
definite and so/has a relative minimum at (0, 0).
0
The eigenvalues of H(2, 1) =
(e)
-4
are
-4
4
True; if det(A) < 0, then A will have negative
eigenvalues.
A, = 2 ± 275. One of these is positive and one is
Section 7.5
negative; thus this matrix is indefinite and /has a
saddle point at (2, 1). Similarly, the eigenvalues
Exercise Set 7.5
of H(-2, 1) =
0
4'
4
4
are
, which are the corner points of the
21. If X is a unit eigenvector corresponding to A-,
and y =-x^.
From the first, we conclude that
4
-2x
75
rectangle.
to zero. This yields the equations 2x( 1 - y) = 0
H{x, y) =
ir and
75
/ = 2 + 275; thus/
1.
has a saddle point at (-2, 1).
179
A* =
-2/
4
5-(
l+(
3-(
0
SSM: Elementary Linear Algebra
Chapter 7: Diagonalization and Quadratic Forms
3.
1
i
2-3/
-/
-3
1
2+ 3/
1
2
A=
5. (a) A is not Hermitian since a^i
(b) A is not Hermitian since c/22
9. The following computations show that the row vectors of A are orthonormal:
3^ 4.2
- + -/
5
25
5
hlh i
25
25
5
25
\r
3.
r3Y —
43+[4
—/
—/
l5jl 5) {5 I 5
_\2 n
r, •r2 = —
5
5
=0
Thus A is unitary, and A
-1
= A* =
3
4
5
5
4;
3;
5
5 J
11. The following computations show that the column vectors of A are orthonormal:
2
2
1
1
(v^+/) +
=
2^I2
(l+/>/^)
2V2
Z+f
8
8
=1
2
hll=
2
V2V2
-+-
8
8
=1
1
C, C2 =
1
(v^+/)
2V2
2V2
1
(l+/>/3)+
2yf2
1
(l+/V^)
1
-I
Thus A is unitary, and A
2^/2(^-/)
= A* =
(-/-V^)= o
2^
I
2^
ii-isy
13. The characteristic polynomial of A is Y -93.+18 =(3^-3)(A,-6); thus the eigenvalues of A are A,= 3 and A = 6.
-1
The augmented matrix of the system (3/- A)x = 0 is
-/
-1 +/
-2
I
I
I
0
0
, which reduces to
1 I-/ IO
0
. Thus
0 II 0
-1+;
-1+/
Vl =
1
is a basis for the eigenspace corresponding to A = 3, and P[ =
1
V3
180
is a unit eigenvector. A
SSM: Elementary Linear Algebra
Section 7.5
i-(
similar computation shows that P2 = , is a unit eigenvector corresponding to ^ = 6. The vectors P| and P2
V6
are orthogonal, and the unitary matrix P =[pi
-I-/
P*AP =
-1+/
1
73
V3
4
1-i
73
i±i
^
\+i
5
J_
76
76
pi ] diagonalizes the matrix A:
I-/'
76
_ 73
76,
3 0
0 6
-103.+ 16 =(A,-2)(3,-8); thus the eigenvalues of A are A, = 2 and 3,= 8.
15. The characteristic polynomial of A is
-4
The augmented matrix of the system (2/- A)x = 0 is
-2-2/ I O',
I
-2+ 2/
-2
I
0
which reduces to
'2 1+i lO’
0
0 II 0 ■
-l-i
Thus V I
2 * is a basis for the eigenspace corresponding to A,= 2, and p| =
-
7^
2
is a unit eigenvector. A
76
1+1
similar computation shows that P2 =
S
i
is a unit eigenvector corresponding to ^ = 8. The vectors pj and P2
s
are orthogonal, and the unitary matrix P =[pj P2] diagonalizes the matrix A:
P*AP =
-1-1
1+/'
-l+i
2
76
76
6
2+2/
76
73
l-<
1
2-2/
4
_2_
J_
76
73
2 0
0
8
17. The characteristic polynomial of A is {X — 5)(^“ + A,-2)=(A,+ 2)(A -1)(A-5); thus the eigenvalues of A are
[-7 0
A, = -2, A2 = 1, and A3 = 5. The augmented matrix of the system (-21- A)x = 0 is 0
0
'1
0
0
I0
0 I o'
l-( 10 ,
1 +/
-2 I 0_
0
which can be reduced to 0 1 -1 + /|0 . Thus v, = 1-/ is a basis for the eigenspace corresponding to
0 0 0 I oj
1_ 1 _
0
0
I-/
-i+i
is a unit eigenvector
Ai = -2, and p| = 7^ is a unit eigenvector. Similar computations show that P2 =
1
2
75
ISl
corresponding to A2 = 1, and P3 = 0 is a unit eigenvector corresponding to A3 = 5. The vectors {p,, P2, P3)
0
form an orthogonal set, and the unitary matrix P =[pi P2 P3] diagonalizes the matrix A:
181
SSM: Elementary Linear Algebra
Chapter 7: Diagonalization and Quadratic Forms
0
P*AP = 0
-2
=
0
1+/
1
^/5
^
5
-l-(
^ :
2
0
S
n/6
0
0
0
0
+(
0 -l-(
0
0
0
1-1
-1+/
^/3
^/6
_L
_2_
x/3
Ve
0
0
0 0
1
0
_0 0 5
Note: The matrix P is not unique. It depends on the particular choice of the unit eigenvectors. This is just one
possibility.
19.
A=
0
i
i
0
-2-3/
-1
2-3/
4/
21. (a) 021 =-i^i = -a^2 and £231 ^^-^13
(b) |
fl|
^-a||, 031
-a,3, and 0^2 ^-«23
0
23. The characteristic polynomial of A =
+/
\ +i
is
i
-iX + 2 =(X-2i){X + i); thus the eigenvalues of A are
X - 2/ and X = -/.
I
1
-le
1 +1
27. We have A*A =
e
/6i
■W
V2 /le
le
i0
:.-ie
-le
2
1+1
1
0
0
1
-1
; thus A* = A
and A is unitary.
I
29. (a) If fi =-(A + A*), then B* = -{A + A*)*
= ^(A*+A**)
= -(A*+A)
2
= B.
Similarly, C* = C.
1
1
2
2
(b) We have B + iC = -{A + A*) + -{A-A*) = A and B-iC = -{A + A*) — (A-A*) = A*.
2
(c)
2
AA* = (B + iC)(B-iC)
= B^-iBC + iCB + C~
and A*A = B^ + iBC - iCB +
. Thus AA* = A*A if and only if -iBC + iCB = iBC - iCB, or 2iCB = 2iBC.
Thus A is normal if and only if B and C commute i.e., CB = BC.
31. If A is unitary, then A“‘=A* and so (A*)“' = (A“')* = (A*)*; thus A* is also unitary.
33. A unitary matrix A has the property that ||Ax|| = ||x|| for all x in C". Thus if A is unitary and Ax = A,x where x ^ 0,
we must have U| ||x|| = ||Ax|| = ||x|| and so Ul = l.
182
Chapter 7 Supplementary Exercises
SSM: Elementary Linear Algebra
Chapter 7 Supplementary Exercises
35. If// = /-2uu*, then
H* =(/- 2uu*)* = /* - 2u**u*= I- 2uu* = H;
3
thus H is Hermitian.
_4
53 , A^A =
1. (a) For A = ^
HH* = iI-2uu*){I-2uu*)
= I -2uu *-2uu * +4uu * uu *
5
=/
and so H is unitary.
37.
3
4
5
5
4
5
3
5
I 0
i
L
1
, so
3
/
A=
0
5
= /-4uu*+4u|lu|Pu*
I
1
0
is both Hermitian and unitary.
5
(b) For A = -J25
4
5
25
11
3
16
. 25
5
25
39. If P = uu*, then
_i2
1 0 O'
T
Px =(uu*)x =(uu )x = u(u x)=(x-u)u. Thus
AM = 0
1
0 , so
0 0
multiplication of x by P corresponds to ||u||“
1
12
4
times the orthogonal projection of x onto
W = span{u). If Hull = 1, then multiplication of x
A->=A^ =
5
25
25
0
4
5
3
5
1
_I2
16
5
25
25
hy H = I- 2uu* corresponds to reflection of x
about the hyperplane u"*".
True/False 7.5
5. The characteristic equation of A is
?.^-3?t^+2A,= X(>i-2)(?L-l). so the
(a) False; (VI.
eigenvalues are X = 0, 2, 1. Orthogonal bases for
I
I
(b) False; for r,
V2 ^ ^/3
and
0
the eigenspaces are A, = 0:
/
/
r, =[0
i
I
v/6 VsJ’
I
I
(0)+
riT2=-
I
I
I
1
. \2
0
^2
^/6l VbJ
=0+ 4
S)
0
• \2
(
; A= 1;
i
1
1—
6
; X = 2:
I
3
Thus P =
6
0
1
I
Vi
Vi
0
0
0
1
i
Vi
Vi
orthogonally
0
Thus the row vectors do not form an
0 0
orthonormal set and the matrix is not unitary.
0
diagonalizes A, and P^AP= 0 2 0
0 0
1
(c) True; if A is unitary, so A '= A*, then
=A =(A*)*.
7. In matrix form, the quadratic form is
r
r >
3
9
X'l . The
(d) False; normal matrices that are not Hermitian are
also unitarily diagonalizable.
X Ax =[x| X,]
(e) False; if A is skew-Hermitian, then
characteristic equation of A is
(a4* =(A*)(A*)=(-A)(-A)=
'-i
-A^.
183
4
X2
-5A + — = 0
4
Chapter 7: Diagonalization and Quadratic Forms
which has solutions X =
SSM: Elementary Linear Algebra
5+3V2
or
2
X ~ 4.62, 0.38. Since both eigenvalues of A are
positive, the quadratic form is positive definite.
9. (a)
(b)
y-x^ =0 or y = x^ represents a parabola.
3a:-11/ =0 or
11 2
x=—
y
represents a
parabola.
184
Chapter 8
Linear Transformations
Section 8.1
Exercise Set 8.1
1. r(-u)= |-u,
Hull = T{u)^ -T{u), so the function is not a linear transformation.
3. T{kA)= {kA)B = k(AB)= kT(A)
T(/\| + /42)=( + A)B
= AyB A^B
= T{A0+ T{A2)
Thus 7is a linear transformation.
5. F{kA)={kA)'' = kA'' = kF{A)
F{A + B)={A + Bf =A^ +b'^ = F{A)+ F{B)
Thus 7is a linear transformation.
7. Let p{x)= aQ+a\X + a2X and q{x)= bQ+b^x + b2X''.
(a) T(kpix))= koQ + ka^{x +\)+ ka2ix + \)^
= kT{p{x))
T{p{x)+ q{x))= aQ+bQ+(a^+b^){x+\)+{a2+b2)ix+\f
= Gq + d](x+1)+ G2(-*+ bfj + b^{x+ \)+ b2{x+\)^
= T{p(x))+ T{qix))
Thus 7is a linear transformation,
(b) T(kp{x))= T{kaQ+ka\X+ ka2X^)
={kaQ +1)+{kuy + l)x +(^^2 +
^kTipix))
7is not a linear transformation.
9. For X =(Xj, X2)= qv]+C2V2, we have (q, X2)= qfl, l)+ C2(l, 0)=(q+C2, q) or
q+C2 =x,
Cl
= X2
which has the solution q = X2, C2 = X| -X2.
(X|, X2)= X2(1, 1)+(X|- X2)(1, 0)
= X2V,+(X|-X2)V2
and
7(X,, X2)= X27(v,)+(x, -X2)7(V2)
= X2(1,-2)+(x,-X2)(-4, 1)
=(-4X| +5x2, “3x2)
7(5,-3)=(-20- 15, 5 + 9)=(-35, 14).
185
SSM: Elementary Linear Algebra
Chapter 8: Linear Transformations
(b),(c) Neither 1 +j:nor 3-x^ can be
11. For X =(a-|, ^2, X3)= C|V[+C2''2+<^3''3’
have
expressed as xp(x) with p(x) in P2, so
(X,, X2, a-3)= c,(1, 1, 1)+ C2(1, 1, 0)+ C3(l. 0, 0)
=(C| +C2+C3, q +C2, q)
neither are in R{T).
21. (a) Since -8x + 4y = -4(2x - y) and 2x-y can
C| + C2 + C3 = X|
= x^ which has the solution
or c, + C2
^1
assume any real value, R(T) is the set of all
vectors (x,-4x) or the line y = -4x. The
= ■^3
vector (1,-4)is a basis for this space.
q=A3, C2=A2-A3,
(b) T can be expressed as a matrix operator
C3 = A, -(a-2 -A3)-A3 = a-| -a-2.
from R‘^ to R^ with the matrix
(A|, A2, A3) = A3V, +(A2 -A3)V2 +(A| -A2)V3
7(A|, A2, A3)
A =
= A37(v,) + (A2 -A3)r(V2) + (A, -A2)7(V3)
= A3(2,-1,4) + (a2-A3)(3, 0, 1)
+ (a,-A2)(-1, 5, 1)
= (-A| + 4a2 - A3, 5A| - 5 A2 - A3, A| + 3A3)
4
1
-2
2
1
1
6
0
-9
-3'
-4 . A basis for R{T) is a
9
basis for the column space of A. A row
I
2
T(2, 4, -l) = (-2+16+l, 10-20 + 1, 2-3)
= (15,-9,-1)
-L2 -2I
echelon form of A is S = 0
1
4-5 -
0
0
0
1
The columns of A corresponding to the
columns of B containing leading Is are
13. 7(2v,-3V2+4V3)
= 27(v,)-3r(v2) + 4r(v3)
4
1
-3
= (2,-2, 4)-(0, 9, 6) + (-12, 4, 8)
2
1
-4
= (-10, -7, 6)
6
0
, and
. Thus, (4, 2, 6),
9
(1, 1,0), and (-3, -4, 9) form a basis for
R{T).
15. (a) 7(5, 10) = (10- 10,-40+ 40) = (0,0)
so (5, 10) is in ker(7).
(c) R{T) is the set of all polynomials in
(b) 7(3, 2) = (6-2,-24+ 8) = (4,-16) so
with
constant term 0. Thus, {a, a“, a^) is a basis
(3, 2) is not in ker(7).
for R(T).
(c) 7(1, 1) = (2-1,-8+ 4) = (1,-4) so
(1, 1) is not in ker(7).
1
23. (a) A row echelon form of A is
17. (a) 7(3,-8, 2,0)
= (12-8-4,6-8 + 2, 18- 18)
= (0, 0, 0)
so (3, -8, 2, 0) is in ker(7).
3
0
1
0
0
thus columns 1 and 2 of A,
(b) 7(0, 0,0, 1)
= (0 + 0 + 0-3, 0 + 0 + 0-4, 0-0 + 9)
0
5
and
6
4
form a basis for the column space of A,
which is the range of 7.
= (-3, -4, 9)
so (0, 0, 0, 1) is not in ker(7).
1
7(0,-4, 1,0) = (-4-2,-4+1,-9)
= (-6,-3,-9)
(b) A reduces to
0
0
0
M
1 1
' -if
0
, so a general
0
so (0, -4, 1,0) is not in ker(73.
14
solution of Ax = 0 is Ai =
19. (a) Since a + a~
I I
1
7
(c)
19
19
s, At = — s,
‘
11
^ 11
A3 = s or A] = -14r, A2 =197 +3=117 so
a(1 + a), a +a^ is R{T).
186
Section 8.1
SSM: Elementary Linear Algebra
(b) dim(P4)= 5, so nullity(T)= 5 - rank(T)= 4
-14
19 is a basis for the null space of A,
(c) Since R{T)= R^, 7hasrank3.
II
which is ker(7).
nullity(7) = 6- rank(7)= 3
(c) R{T) is two-dimensional, so rank(T)= 2.
ker(70 is one-dimensional, so nullity (7)= I.
(d) nullity(7) = 4- rank(7)= I
31. (a) nullity(/4) = 7 - rank(A)= 3, so the solution
space of Ax = 0 has dimension 3.
(d) The column space of A is two-dimensional,
so rank(A)= 2. The null space of A is one
dimensional, so nullity(A) = 1.
(b) No, since rank(A)= 4 while R^ has
dimension 5.
25. (a) A row echelon form of A is
[1 2 3
0
0l
2 , thus columns 1 and 2 of
1
33. /?(7) must be a subspace of R^, thus the
7
possibilities are a line through the origin, a plane
through the origin, the origin only, or all of R'.
A. ^1 and 2 form a basis for the column
35. (b) No; F(kx, ky)
space of A, which is the range of T.
= {a^k'^x^ +b^k^y'^, a2k^x^ +b2k'^y^)
1
0
1
0
1
1
4
(b) A reduces to
= k^F(x, y)
2 , so a general
^kF{x, y)
7
4
37. Let w = CV| +€2^2
solution ofAx = 0is X| =-5—t,
1
be any vector in
V. Then
2
7’(w)= C|7'(V|)-Hc27(v2)+"-+ c„7'(v„)
X2=-s+ — t, X2=s, x^=tor
= C|V,+C2V2+'--+ C„V„
X\=-s-Ar, X2=s+ 2r, ^3=5,
= w
Since w was an arbitrary vector in V, T must be
the identity operator.
-4
-1
a'4 = Ir, so
and
1
0
?
form a basis
0
41. If p'(x)= 0, then p(x) is a constant, so ker(D)
7
consists of all constant polynomials.
for the null space of A, which is ker(70.
43. (a) The fourth derivative of any polynomial of
(c) Both R{T) and ker(7) are two-dimensional,
so rank(70 = nullity(7)= 2.
degree 3 or less is 0, so T{f(x))=
has ker(7')= /’3.
(d) Both the column space and the null space of
A are two-dimensional, so
(«-H)
(b) By similar reasoning, r(/(x))=/
rank(A)= nuIlity(A)= 2.
has ker(7)=
27. (a) The kernel is the y-axis, the range is the
entire xz-plane.
(X)
.
True/False 8.1
(b) The kernel is the x-axis, the range is the
entire yz-plane.
(a) True; q = k, C2 =0 gives the homogeneity
property and q = C2 = 1 gives the additivity
(c) The kernel is the line through the origin
perpendicular to the plane y = x, the range is
the entire plane y = x.
property,
(b) False; every linear transformation will have
r(-v)=-7’(v).
29. (a) nullity(7) = 5 - rank(77 = 2
187
Chapter 8: Linear Transformations
SSM:Elementary Linear Algebra
(c) True; only the zero transformation has this
1
property.
0
(c) A reduces to 0
1 , so nullity(A)= 0 or
0 0
(d) False; TfO)= Vq +0= Vq 5^ 0, so T is not a
Ax = 0 has only the trivial solution, so
multiplication by A is one-to-one.
linear transformation.
(e) True
5. (a) All points on the line y = -x are mapped to
0, so ker(7)= {yt(-l, 1)}.
(f) True
(b) Since ker(7) {0}, T is not one-to-one.
(g) False; T does not necessarily have rank 4.
(h) False; det(A +
7. (a) Since nullity(7) = 0, T is one-to-one.
det(A) + det(/?) in general,
(b) nullity(7) = n -(n - 1)= 1, so 7" is not one-
(i) False; nullity(7)= rank(7)= 2
to-one.
Section 8.2
(c) Since n<m,T is not one-to-one.
Exercise Set 8.2
(d) Since R(T)= R'\ rank(7)= n, so
1. (a) By inspection, ker(7)= {0}, so Tis one-to-
nullity (T)= 0. Thus T is one-to-one.
one.
9. If there were such a transformation T, then it
(b) T(x, y)= 0 if 2x + 3>’ = 0 or x =--y
would have nullity 0(because it would be oneto-one), and have rank no greater than the
dimension of W(because its range is a subspace
VF). Then by the dimension Theorem,
dim(V)= rank(7^ -t- nullity(7) < dim(VV) which
so
3
ker(7’)= U
1
and T is not one-to-
^ 2
one.
contradicts that dim(W) < dim(V). Thus there is
no such transformation.
(c) (x, y) is in ker(7) only if x -i- y = 0 and
X - y = 0, so ker(7)= {0} and T is one-to-
a
one.
b
a
b
c
c
(d) By inspection, ker(7) = {0}, so Tis one-to-
11. (a) T
b
d
e
c
e fJy
d
one.
VI-
e
(e) (x, y) is in ker(7) if x - y = 0 or x = y, so
ker(T)= {A:(l, 1)} and Tis not one-to-one.
,/
a
(f) (x, y, z) is in ker(r) if both x -i- y + z = 0 and
X - y -z = 0, which is x = 0 and y + z = 0.
yr
b
c
d
b
=
(b) T,
VL
Thus, ker(r) = {^(0, 1, -1)} and Tis not
-|\
a
and
c
J/
d
one-to-one.
a
1
3. (a) By inspection, A reduces to 0
zr
-2
a
b
c
c
d
b
Tl
0 , so
VL
0
0
d
nullity(A) = 1 and multiplication by A is not
one-to-one.
(c) If p(x) is a polynomial and p(0)= 0, then
p{x) has constant term 0.
(b) A can be viewed as a mapping from R^ to
R^, thus multiplication by A is not one-to-
a
T(ax^ + bx^ +cx)= b
c
one.
188
SSM: Elementary Linear Algebra
Section 8.3
Section 8.3
a
(d) 7’(a +/7sin(x)+ ccos(x))=
b
Exercise Set 8.3
c
1. (a) (T2oT0(x, y)= T2(2x,3y)
13. T(kf)= kfiO)+ 2kf'(0)+ 3lrf'(\)= kT(f)
T(f + g)= /(O)+ ^(0)+ 2(/'(0)+ g'(0))
+3(/'(l)+ g'(l))
=(2x-3y, 2x + 3y)
(b) (T2°Ti)(x,y)
= T2ix-3y, 0)
= r(f)+ 7(g)
Thus 7is a linear transformation.
=(4(x-3>')-5(0), 3(x-3>>)-6(0))
=(4x-12>', 3x-9y)
Let f = /(x)= x^(x-l)^, then
/'(x)= 2x(x-1)(2x-1) soj[0)= 0, /'(0)= 0,
and /'(I)= 0.
7(f)= /(0)+ 2/'(0)+ 3/'(l)= 0, so
(c) (72 o7,)(x, y)= 72(2x,-3y, x+y)
=(2x + 3y, -3y + x + y)
=(2x-i-3y, x-2y)
ker(7) (0) and 7is not one-to-one.
(d) (72o7|)(x, y)
= 72(x-y, y, x)
15. Yes; if T{a, b, c)= 0, then a = b = c = 0, so
ker(7)={0}.
=(0,(x-y)+ y-(-x)
17. No; 7is not one-to-one because ker(7)
7(a)= a X a = 0.
=(0, 2x)
{0} as
rr
3. (a) (7|o72)(A)= 7,(A^)= tr
19. Yes, since (7(u), 7(v))=(u, v), so
||7(u)||= 7(7(u), 7(u))= 7(u, u)= II
u
a
c
= a+d
b
\l-
d J/
= I and
(b) {T2°T\){A) does not exist because 7;(A) is
7(u) is orthogonal to 7(v) if u is orthogonal to v.
not a 2 X 2 matrix.
True/False 8.2
1
5. 72(v)= —V, then
(a) False; dim(/?^)= 2 while dim(P2)= 3.
(1
(7,o7,)(v)= 7| -V =
(b) True; if ker(7)= {0} is one-to-one rank(7)= 4
V4
so 7is one-to-one and onto.
— V
= V and
4 ;
1
(72o7|)(v)= 72(4v)=-(4v)= v.
(c) False; dim(yW33)= 9 while dim(79)= 10.
9. By inspection, 7(x, y, z)= (x, y, 0).
(d) True; for instance, if V consists of all matrices of
a
b
0
c
d
0
the form
Then
7(7(x, y, z))= 7(x, y, 0)= (x, y, 0)= 7(x, y, z) or
, then
7o7=7.
a
a
b 0'
c
d
7
0-J/
11. (a) Since det(A)-0,Tdoes not have an inverse.
^ is an isomorphism 7:
d
V
(e) False; 7(/)= 7- 7 = 0 so ker(73
0.
(f) False; if there were such a transformation 7, then
dim(ker(7))= dim(7(7)), but since
dim(ker(7))-i-dim(7(7))= dim(74)= 5, there
can be no such transformation.
189
SSM: Elementary Linear Algebra
Chapter 8: Linear Transformations
15. (a) Since 7|(/?(x))= xp(x),
(b) det(A) = -8 5* 0, so 7has an inverse.
' 1 1 _3'
I
4
i 1
A-'
71 (p{x))=-p(x).
1
X
so
4
11
Since 7’2(/?(x))= pix+\),
1
8
4
T2\p(x))= p{x-\).
(72 o7ir'(p(x))=(7f' oT2'){pix))
= Tf'{p(x-\))
X,
-1
=A
7
-I
L-^3j
-1/
= ~p(x-l)
i^l+i^2+i-^3 •
-|X|+|X2 +^X3
17. (a) 7(1-2x)=(l-2(0), 1-2(1))=(1,-1)
(c) Let p(x)= aQ+aiX, then
(c) det(A) = -2
0, so 7 has an inverse.
I
1
I
1
1
1
1
±
i
2
2
1
i
_i
. 2
2
2
A-‘
/-
T(p(x)) =(0, 0), then aQ=a, =0 and p is
the zero polynomial, so ker(71i = (0).
so
(d) Since 7(/j(x))=(ao> <30+
then
-IN
X'l
X|
-1
7
7(p(x))=(ao, ao+«|) so if
X,
=A
7“'(2, 3) has rtg = 2 and aQ+a^ =3 or
a, =1. Thus, 7“'(2, 3)= 2+ x.
-I
^1
X3
VL X3 - /
1
I
I
-X,--X2 + 2X3
-i-^'l+i^2+T^3
^X,+jx2 -jxy
X
(d) det(A) = 1 0, so 7 has an inverse.
■ 3
3 -I
A
-I
-2
-2
1
-4
-5
2
21. (a) 7|(x, y)=(x, -y) and 72(x, y)=(-x, y)
(7,o72)(x, y)= 7|(-x, y)=(-x,-y)
-IN
/r-
X,
-1
7
so
X|
(72 o 7i )(x, y)= 72(X, - y)=(-x,- y)
-1
X,
=A
^3 jy
L-^3j
3X] +3x2“^3
= -2X|-2x2 +^3 ■
-Axj-5x2 + ^^3
^2
7,o72=72o7,
(b) Consider (x, y)=(0, 1):
(72 °7,)(0, 1)= 72(0, 0)=(0, 0) but
(7, °72)(0, l)= 7,(-sin6», cos^*)
=(-sin6', 0)
13. (a) For 7to have an inverse, all the
Thus 7| o 72 7^:72 o7|.
a,-, i- 1, 2, ..., n must be nonzero.
I
(b) 7 (X|, X2, .... x„)
1
— X,,—X2,...,—x„
a
|
fl
a2
n
190
Section 8.4
SSM: Elementary Linear Algebra
(c) T^{x, y, z)=(kx, ky, kz) and T2{x, y, z)=(xcos0-ysind, xsin6'+>’cos(9, z)
(7| °T2){x, y, z)={k{xco59-y?,in9), L'(A-sin0+}'sin6’), kz)
{T2°T\){x, y, z)={kxco59-ky?,m9, kxs\r\9+ kycos9, kz)
7j 0T2 =T2 oT
True/False 8.3
(a) True
(b) False; the linear operators in Exercise 21(b) are an example.
(c) False; a linear transformation need not have an inverse.
(d) True; for T to have an inverse, it must be one-to-one.
(e) False; T
I
does not exist.
(f) True; if Ti is not one-to-one then there is some nonzero vector V| with 7j(vi)= 0, then (7’2 ori)(V|)= 72(0)=0
and ker(72 o 7,)?* {0}.
Section 8.4
Exercise Set 8.4
1. (a) 7(U|)= 7(l)= x= V2
T(u2)= T{x)= x- =^2
.s
7(u3)= 7(x2)
=X
= ''4
0 0 0
0
0
1
0
0 0
1
Thus the matrix for 7relative to B and B' is
0
0 0
(b) [T]i2',bWb =
0
0
0
1
0
0 0
0
0
1
<^0
Co
Cl
0 .C2
Cl
C2
0
[Tix)]/}' =[CoX + C^X^ +C2X^]b' =
Co
Cl
Co
3. (a) 7(1)= 1
7(x)=x - 1 =-l + x
7(x^)=(x-l)^ = 1-2x4Thus the matrix for 7relative to B is
0
1
-2
0
0
1
191
Chapter 8: Linear Transformations
(b) [T]b[x]5 =
1
-1
1
0
1
-2
a.
0
0
1
^2
SSM: Elementary Linear Algebra
9. (a) Since A is the matrix for T relative to B,
%
^ =[[^(vi)b I [nv2)]B]. That is,
1
3
[T(v,)]b=
aQ-a^+ a2
and [T(y2)]B = 5
ay -2a2
1
«2
(b) Since [r(v,)]g =
For x = aQ+ayX+ a2X^,
T{x)= aQ+ai{x-l)+a2(x-\f
= a^-ay +^2 +(<2]-2a2)x+ a2X^,
r(v,)= lv,-2v2 =
1
-2
3
3
8
-5
Similarly,
'ao-ay+a2'
so [7’(x)]g =
-2 ’
3
7(v2)= 3v, +5v2 =
ay -2a2
-5
-2
20
29
1
-1
+
9
«2
1
/r. n\
\l-
-1/
-1 = 0v I
= CiVi+C2V2=Ci ^ +C2 4 ,
X2
8
1
1
5. (a) T{yiy)= T ^ =
^1
(c) If
-V2^-V2
Xy =Cy-C2
then
0
X2 ~ 3ci +4c2
/r
-21^
T(u2)= T
6
VL
J/
4
4
2 = OVr + Vt H
4
3
0
. Solving for q and C2
3
1
1
gives Cy=-Xy+-X2, C2=--Xy+-X2,
''3
SO
0
0
i
\T]b\ b =
/r
1
a^2 J/
2
4
3
-\
^1
T
= Ci7'(v,)+ C27'(V2)
3
4
I? ' 7 ^
7. (a) 7(1)= 1
^ -^i+y^2
7(x)^ =(2x+1)2 =1+4x+4x^
1
107
, 24
1
{T]b^ 0 2 4
0
0
18
I
7
7
^1
24
a:,
107
4
7
7 JL z
2
(b) The matrix for 2-3x+4x^ is -3
(d) 7
4
2
-5
/L
r 18 V + I V
7(x)= 2x + 1 = 1 + 2x
1
-2
3
1
= —Xy+ — X2
1
1
1
2
1
\i- -J/
18
1
7
7
107
7
24
7
19
1
7
1
M
7
3
[7]g -3 = 0 2 4 -3 = 10
11. (a) Since 4 is the matrix for 7relative to B, the
[ 4J [0 0 4j[ 4J [l6_
columns of4 are [7(V|)]g, [7(v2)]b, and
[7(v3)]g, respectively. That is.
Thus 7(2-3x+4x^)= 3+10x +16x^
1
3
[7(v,)]5 = 2 , [7(v2)]fi =
(c) 7(2-3x+4x^)= 2-3(2x +1)+4(2x +1)^
6
= 2-6x-3+16x^+16x+4
= 3+10x+16x^
-1
[T(v3)]b =
5
4
192
0 , and
-2
Section 8.4
SSM: Elementary Linear Algebra
1
(d) In 1 + x^, aQ=\, a^=0, 02 =1.
(b) Since [r(v,)]g= 2 ,
Ti\ + x^)
6
239+ 289
201 + 247
24
8
61+107 2
x+
nv,)
= V| +2v2 +6V3
= 3x +3x^-2+6x +4x^+18+42x+12x^
12
= 22+56x+14x^
= 16+51x+19x^.
Similarly, 7’(v2)= 3V|-2V3
= 9x +9x^-6-14x-4x^
13. (a) (72o7,)(1)= 7'2(2)= 6x
(72o7,)(x)= 7’2(-3x)= -9x2
--6-5x +5x^,
Thus, [T2 o7|]g' g -
and 7’(v3)=-Vj+5v2+4V3
= -3x-3x2-5+15x +10x^+12
+ 28x +8x^
7’,(1) = 2
= 7+40x+15x^.
7i(x)= -3x
(c) If aQ+a^x +a2X^
Thus {T\
0
O'
6
0
0
-9
0
0
2
0
g- 0
-3
0
0
= C|V, +C2V2+C3V3
72(1)= 3x
= C|(3x +3x^)+ C2(-1 +3x + 2x^)
+ C3(3+7x+2x^),
72(x)= 3x2
72(x2)= 3x^
^0 =-C2+3c3
Then a^ = 3c| +3c2 +7c3 .
02 = 3c| + 2c2 + 2C3
0 0 0
Thus [T2]b' h’ —
Solving for q, C2, and C3 gives
3
0 0
0
3 0
0 0
1
3
^1 =-(^0-^1+2^2)’
(b) [T2 °7,]g' g - lT2]g' g’[T\ Ig' g
C2 =-^(-500+30,-302),
C3=-^(«0+«l-«2), so
0 0 0 r,
3
(c)
0
6
0
0 -3
0
3
0
0 0
3
0 -9
0
7(oo +a, x + a2X^)
0
0
0 0
0
0
0
= C|r(V|)+ C27(v2)+ C3r(v3)
15. If 7is a contraction or dilation of V, then 7 maps
=^(oo -o,+2a2)(16+51x+19x^)
any basis 5=[v,, V2,
v„) of V to
{k\], 7v2, .■■, ky„} where ^ is a nonzero
+(-5oo + 3o,-3o2)(-6-5x +5x^)
8
constant. Thus the matrix for 7 relative to B is
+-(oo+a,
-a2)(7+40x+15x^)
8
239^0 -16 lai + 289a2
~k
0
0
0
1:
0
0
0
k
0
0
0
o'
•••
0
0
24
201oo-llla,+24702
8
6I0 0
-31o,+107o2
^x
2
12
193
•
A:
Chapter 8: Linear Transformations
SSM: Elementary Linear Algebra
17. The matrix for T relative to B is the matrix
21. (a) [7'2o7',V,b=[7'2]b',zj'[7',V,s
whose columns are the transforms of the basis
vectors in B in terms of the standard basis. Since
(b) [7'3o7-2or,V,B
B is the standard basis for R'\ this matrix is the
standard matrix for T. Also, since B' is the
standard basis for R'", the resulting
True/False 8.4
transformation will give vector components
relative to the standard basis.
(a) False; the conclusion would only be true if T:
V
V were a linear operator, i.e., if V=W.
19. (a) T)(f|) = Z?(l)=0
(b) False; the conclusion would only be true if T:
V
V were a linear operator, i.e., if V=W.
D(f2)= D(sin x)= cosA'
D{{^)= D(cos jc) =-si n X
The matrix for D relative to this basis is
'0 0
(c) True; since the matrix for T is invertible, ker(7)
= {0).
O'
0 0-1 .
0
1
(d) False; the matrix of 5°7 relative to B is
0
(b) D(f,)= D(l)=0
(e) True
7)(f2)= D(e")= e"'
Section 8.5
2x
D{f^)= D{e^’^)= 2e
The matrix for D relative to this basis is
Exercise Set 8.5
'0 0 O'
0
1
-2
0 .
0 0
1. Tfu,)
2
and 7(u2)=
0
1
\T]b =
2x
(c) D(^^)= D{e^-'‘)= 2e
2a:
D{t2)= D{xe'^^)=:e^^ +2xe
D(f3)= D{x^e^^)= 2xe^^ + 2x^e^^
1
0'
0
2
2 .
0
0
2
, so
-2
0
By inspection, V| =2u|+U2 and
V2 = -3ii| +4u2, so the transition matrix from
The matrix for D relative to this basis is
'2
-1
B' to 6 is P =
2 -3'
. Thus
4
±
-I
Pb^b'- P
(d) D(4e^-'^ +6xe^^ -Wx~e^^)
II
±
2.
11
l l_
[7]^' = P
Pb'^B
A
2
I
0
4
0
2
2
6
__L
0
0
2
-10
. II
_
II
3
A
II
1
^
0
l lJ
56'
14
1 1
-8
2
3
I I
II
-20
lx
Thus, D(4e^-'‘ +6xe -lOx^e^-'-)
= 14e^-''-8xe2x -2L)x^e^\
194
J,
I I
1 1
-2
2
-3
1
4
Section 8.5
SSM: Elementary Linear Algebra
9
2
3. Since B is the standard basis for R",
7
p=
1
-sin45°]_ ^
cos 45°
1
^ . Thus
3
6
1
and {T]g - Pb^b'^P^bPb'-^b
'^^e same as
= 4
2
2
I
3
9
9
9
1 i
JL2
4
1
I
3
3
6
2
1
[2
in Exercise 1, so
IPh' ~ PB^B'iPh PB'-4B
_
1
I
I
n/2
n/2
2 -3
J_
_L
1
- 1 1 luL^/2
n/2
" 4L
i
II
It
L
r
2.
13
1 ■
0
4
11. (a) The matrix for T relative to the standard
1
25
_ I In/2
basis B is {T]g =
I In/2
5
9
I In/2
I In/2
0
0
1
0
. The eigenvalues
and
and 7'(u3)= 0 , so
0
0
0
4
corresponding eigenvectors
0
5. 7’(U|)= 0 , 7'(u2)= 1
1
2
of [Tig are 3. = 2 and X. = 3 with
0
0
2
1
2
"l 1\\2.
matrices for Pg^n' and Pb'<^b
4
=P
. The
1
cos 45°
sin 45°
1 1
-I
-2 ■
1
[Th
Then for P =
-1
-2
, we have
0 0 0
I
1
B' to 5 is P = 0
1
1
0 0
1
and p-'[7’]gF =
-1
V3 =U|+U2 +U3, so the transition matrix from
1
2
P~'
By inspection, V] =U|, V2=U|+U2, and
2
0
0 3 ■
Since P represents the transition matrix from
the basis B' to the standard basis B, then
1
B’=
is a basis for which
ij’ L-2_
0
-I
Thus P^^g = P
[7]g' is diagonal.
and
0
0
0
1
(b) The matrix for T relative to the standard
[T]b' - Pb^b'\-P^bPb'^b
1
-1
0
1
0
1
-1
0
0
0
1
I
4
0 0
1
0
0 0 0
basis B is
1
0
=
1 ■
-3
1
1
0 0
1
5+ V^
The eigenvalues of [Tig are X =
0 o'
0
1
and X =
0 0 0
5-V^
^
with corresponding
~-3-n/2T1
eigenvectors
2
6
[-3+^/2!
and
'-3-N/2T
2
2
so [T\b = ]
2
2
Then for P =
9
3
4
n/^
3
n/^
3.
1
7
fll =-gP|+3P2 and q2=-P|--P2> so the
transition matrix from B' to B is
195
-3+N/2f
6
6
1
1
-3+N/2T'
2n/2T
3+V2I
2n/2T
6
1
1
7. r(p|)= 6+ 3(jc+1)= 9+ 33t = —P|+ —P2, and
7’(p2)= 10+2(x+1)= 12+2x =-^P,+^P2-
2
and
, we have
Chapter 8: Linear Transformations
5+721
P~'[I]bP=
^
SSM: Elementary Linear Algebra
True/False 8.5
0
. Since P
0
5-72F
(a) False; every matrix is similar to itself since
2
A^r'AI.
represents the transition matrix from the
basis B' to the standard basis B, then
1
B'= <
6
• is a basis for
6
1
(b) True; if A = P BP and B = Q CQ, then
r-3-x/2Tl [-3+7^1
A = P~'{Q~'CQ)P =(QP)-'C(QP).
1
which [Tig' is diagonal.
(c) True; invertibility is a similarity invariant.
(d)
13. (a) r(l)= 5+ x^
1
True; if A = P~'BP, then
=(P"'bP)“' =P"*B“‘p.
T{x)= 6-x
T(x^)= 2-Sx-2x^
Thus the matrix for Trelative to the standard
■5
6
2
basis B is [Tig = 0 -1
-8
. The
L1 0-2
characteristic equation for [7]g is
A,^-2^^-15X + 36 = (X + 4)(X-3)^ =0, so
(e)
True
(0
False; for example, let Tbe the zero operator.
(g)
True
(h)
False; if B and B' are different, let [7]g be
given by the matrix /g_>g'. Then
the eigenvalues of T are ^ = -4 and A, = 3.
\-'P]b\ B -
= Pb-^B'Pb'^B = ^n-
(b) A basis for the eigenspace corresponding to
Chapter 8 Supplementary Exercises
-2
A, = -4 is
, so the polynomial
1. No; TCx,+X2) = A(X|+X2) + B
^>t(/4X[ +B) + (Ax2 +B)
= r(X|) + 7(x2),
1
2 + -x + x^ is a basis in P^. A basis for
and if
1, then
T{cTi) = cAx + B 5* c(Ax + B) = cT{\).
3
the eigenspace corresponding to ^ = 3 is
5
3. Tik\) = kv\\k\ = k k y \ ^ kT{\) if k=;t \.
-2 , so the polynomial so the polynomial
1
5 - 2j: +
5. (a) The matrix for T relative to the standard
is a basis in P^.
15. If V is an eigenvector of T corresponding to X,
then V is a nonzero vector which satisfies the
'1
0
1
basis is A == 2
1
3
r
1 . A basis for the
1
0
0
1
range of T is a basis for the column space of
equation 7(v) = X\ or (XI - 7)(v) = 0. Thus v is
in the kernel of XI - T.
A. A reduces to
17. Since C[x]g = D[x]g for all x in V, then we can
let X = V,- for each of the basis vectors
r1 0 0
0
1
0
0
1]
0-1 .
0
Since row operations don’t change the
dependency relations among columns, the
V|, ..., v„ of V. Since [v,]g =e,- for each i
reduced form of A indicates that 7’(e3) and
any two of Tfci), 7’(e2), 7(64) forma
where (cj, ..., e„) is the standard basis for B",
this yields Ce, = De, for 1 = 1,2, ..., n. But Ccj
basis for the range.
and De, are the ith columns of C and D,
The reduced form of A shows that the
respectively. Since corresponding columns of C
and D are all equal, C = D.
general solution of Ax = 0 is x = -s, X2 = s,
196
SSM: Elementary Linear Algebra
Chapter 8 Supplementary Exercises
one-dimensional. Hence, nullityCT) = 1 and
X3 = 0, X4 = 5 so a basis for the null space
rank(r)= dim(A/22)- nullityCT")= 3.
-1
1
of A, which is the kernel of T is
1
0
13. The standard basis for M 22
IS u
1 -
1
0
u
(b) Since R{T) is three-dimensional and ker(7)
is one-dimensional, rank(7)= 3 and
1
0 0
0 0
0 ’ U4 = 0
0 0 ’ U3 = 1
2-
0
0 0’
. Since
1
L(U|)= Ui, L(u2)= U3, L(u3)= U2, and
nullity(r) = 1.
L(u4)= U4, the matrix of L relative to the
7. (a) The matrix for T relative to B is
1
1
2
-1
I
-I
-4
6
1
2
5
-6 ■
3
2
3
-2
1
0-1
me-
0
[7"]^ reduces to
1
1 0 0 0'
0 0
1
0
standard basis is
0
1
0
3
1
I
1
P= 0
1
1 , thus
-4
which
0
0
0
0
0
0
0
0 0
1
15. The transition matrix from B' to B is
2
0
0 0"
0 0
1
^0
has rank 2 and nullity 2. Thus, rank(7)= 2
and nullity(7) = 2.
1
1
[Th'= p~'msP^
0
9
0 -2
1
1
(b) Since rank(7) < dim(V), T is not one-to-one.
0
1
I
9. (a) Since A = P BP, then
17. T
0
0
rroi
T
0
0 , T
1
1
0
0
1
, and
1
0 , thus [Tig = 0
•1
Thus, A^ and B^ are similar.
-1
1
1
-1
1
1
0
0-1
Note that this can also be read directly from
(b) A~^ =iP~'BPr^ =P~'b-'p
Thus /4~' and 6“’ are similar.
a
b
c
d
19. (a) D(f + g)={f{x)+ gix))'= rix)+ g\x)
and Dikf)={kfix))’= kf\x).
11. For X =
(b) If f is in ker(£)), then f has the form
a+c
b+d
b
0
0
^d d
T(X)=
b
f = fix)= aQ+ a^x, so a basis for ker(D) is
fix)= 1, gix)= .r.
a + b +c
2b+ d
d
d
(c) D(f)= f if and only if f = fix)- ae^ +be~^
TiX)= 0 gives the equations
for arbitrary constants a and b. Thus,
a+b+c=0
fix)= e^ and gix)=^e~^ form a basis for
2b +d=0
this subspace.
d=0.
Thus c = -a and X is in ker(7) if it has the form
k
-k
0
0’
so ker(7)= k
1
0
-1
0
which is
197
SSM: Elementary Linear Algebra
Chapter 8: Linear Transformations
25.
21. (c) Note that a|/i(x)+ a2/2(-^)+'33^3(-r)
y(l)= x
0
evaluated at x^, X2, and x^ gives the values
J{x)=
X
2
a^, 02, and a^, respectively, since
I 3
J{X^)=
Pi{Xi)= ] and Pi(Xj)=0 for i ^j.
—X
3
Thus
a\
7’(a|/^(x)+ a2^2W + <^3^3W)=
.n+\
y(x")=
n+1
a-j , or
This gives the (« + 2) x (n + 1) matrix
«3
0
/r
0
0
0
0
0
0
0
0
i
0
a
I
-I
T
a.
= a^P^ix)+ a2P2{x)+ P^ix).
0
1
2
VL «3 J/
0
0
3
(d) From the computations in part (c), the points
lie on the graph.
0
0
0
i
n+]
23.
D(1)=0
D(x)= 1
D(x^)=
D{x")=
2x
«-l
nx
This gives the matrix shown.
198
Chapter 9
Numerical Methods
Section 9.1
-1
-1
I
-1
-1
0-2
2
0
1
-1
0
1
0
1
Exercise Set 9.1
3
3 0][l -2"l['x|]_[0'
-2
5 ,^2 ^
-2
1
0
.^2
4
1
-1
-1
1
-1
-1
1
0
1
-1
0
1
0
0
5
0
0
1. In matrix form, the system is
—>
4
-1 = U
1
1
1 -2
which reduces to
X| _ y,
0
x.
The multipliers used were
and
1
1, —,-4, and
2
2
>’2
2
3 0 y, ^ 0
-2 lJL3'2j [l.
5
3y,=0, which has
-2yi + y2 = I
The second system is
-1
0 o'
0-2 0
4
1
-2
X| _ 0
0
or
1
1JLx2_
X]-2x2 =0
X2 = 1
5
1
-1
-1
0
1
-1
0
0
1
0 0
3'!
-4
0-2 0
3'2
-2
3'3
6
2
which gives the solution to the original system:
X| = 2, X2 = 1.
4
5
4
1
0
3
0
4
-1
-1
-4
^2
-2
L-^sJ
6
IS
-y,+4y2+5y3=6
3'i=-2, y2=l> 3'3=0-
->
1
^1
= -2 which has the solution
-2^2
1
4
5
= -4
23'!
3. Reduce the coefficient matrix.
2
4
system is
'2
solution V] =0, y2 =1, so the first system
becomes
0 0
0-2 0 , so the
-, which leads to L =
=U
The multipliers used were —, 1, and -, which
2
3
2 0
leads to L =
-1
-1
Xi
-2
0
1
-1
^2
1
0
0
1
•^3
0
IS
Xi - X2 -X3 = -2
X2-X3=l
X3 =0
which gives the solution to the original system:
so the system is
3’
1
1
1
X| =-l, X2 =1, X3 =0.
' 2 0 1 4 X, ^ I-2
-2 ■
-1 3 0 1 X2 ^
2 0
-2
y,
.
IS
-1 ^hl.
-2
7. Reduce the coefficient matrix.
2y, =-2
-3'i +3y2 =-2
which
'5
5
-8
-7
0
4
has the solution y|=-l, y2=-l.
1 4
0
1
X| _ -1
IS
1
X[ +4x2
X2
X2 =
which
X2 =-l.
5. Reduce the coefficient matrix.
-2
-2
0-2
2 ->
-1
2
5
1
-1
0-2
2
5
0
26
2
-+
1
1
2’
0
1
7 =1/
0
0
4
26
1
1
2'
0
1
7
0 0
-2
1
1
-1
-1
2
-7 -9
0 4 26_
gives the solution to the original system: X| = 3,
2
1
7
0
1
10"
-9 ^ -8
1
The multipliers used were —, 8,-4, and —
2
2
199
Chapter 9: Numerical Methods
SSM:Elementary Linear Algebra
5
0
0
which leads to L = -8
1
0
-1
5
0
0
1
1
2
0
-8
1
0
0
1
7
4^2
1
0 0
1
^3
4
0 4-2
5
0
0
>'1
0
^2
1
0 4 -2j[y3
4
1
1
2
2
•^2
-1
0-1
2
0
0 0
1
1
^3
3
0
0
1
4
0 0
0
1
.-^4
7
-1
0
0
0
3 0 0
>'1
5
3'2
-1
2
0
}'3
3
1
4
,•>’4
7
0
IS
=5
= -l
2yi+3y2
, which has the
=3
-y2+2y3
>-3+4^4 =7
which has the solution
>'3 = 0-
solution yi = -5, y2=3, y3 = 3, y4 = 1.
0
X|+ 4:2 + 2x3 =0
0 1 7 X2
1 IS
•*2+7x3 = 1
0 0 1 4^3
0
X3 =0
which gives the solution to the original system:
X] =-l, X2 =1, X3 =0.
*1
5
0
-y\
5)'1
=1
-8yi + y2
4>'2-2>'3 =4
y, =0, y2 =
^1
1
0
=0
0
0
2
IS
0-1
0
0-1
0
1
0
2
0 4-2
IS
0 0 0
3
, so the system
1
0-1 0
^1
-5
1
0
2
X2
3
0 0
1
1
X3
3
0 0
0
1
X4
1
0
X,
= -5
-X3
+ 2x4 =3
.. . .
,
^ , which gives the solution
X3 + X4 = 3
X4 = 1
X2
9. Reduce the coefficient matrix.
IS
■-1
0
1
0
1
2
3
-2
6
2
0-1
2
0
0-1
2
0
to the original system: X] = -3, X2 =1, X3 == 2,
0
1
5
0
1
5
X4 = 1.
0
0
-1
0
3-2
6
0
1
0-1
0
1
0-1
0
0
3
0
6
0
1
0
2
0-1
2
0
0-1
2
0
2
1
0
1
5
0
1
5
-2
2
1
0
0-1
0
0
1
0-1
0
0
1
0
2
0
1
0
2
0
0
2
2
0
0
1
1
0
0
1
5
00
1
5
1
0-1
0
1
0-1
0
0
1
0
2
0
1
0
2
0
0
1
1
0
0
1
1
0
0
0
4
0
0
0
1
—>
The multipliers used were-1,-2,
1
and —, which leads to L =
4
11. (a) Reduce A to upper triangular form.
-1
2
-2
-1
2
1
0
2
1
0
i
> i -i
0
0
1
0
0
1
2
1
0
0
0
1
=u
1
The multipliers used were —, 2, and -2,
2
=U
1,
-1,
0
0
2
3
0
0
0-1
2
0
1
4
0
1
2
1
1 i -2
■-1
0
1
2
-1
2
0
0
which leads to L = -2
1
0
2
0
1
where the
I’s on the diagonal reflect that no
multiplication was required on the 2nd and
3rd diagonal entries.
0
, so
(b) To change the 2 on the diagonal of L to a 1,
the first column of L is divided by 2 and the
diagonal matrix has a 2 as the 1, 1 entry.
the system is
200
Section 9.1
SSM: Elementary Linear Algebra
A = L^DU^
1
0
0
2 0 0
-1
1
0
0
1
0
1
1
' i -2
1
0
0
0
1
0 0
1
0
0
1
1
2
2
X|
0
1
4
^2
2
0
0
17
•^3
12
Xi + 2x2 ^''‘3 “*
X2 + 4X3 = 2
17x3=12
IS
which gives the solution of the original system:
21
is the LDf/-decomposition of4.
12
14
. ^2=-
•*3 = —
17’
17
(c) Let U2 - DU, and 1^=1^, then
17. If rows 2 and 3 of4 are interchanged, then the
resulting matrix has an Lt/-decomposition. For
1 0 0
r3 -1 0"
1 0 0]r2 1 -1
A = LoU2= -1
1
1 0 0 0
1
0
1
1
0 0
p= 0
0
0
1
1 , P4= 0
2
3
-1
0
1 .
1
13. Reduce 4 to upper triangular form.
■3 -12 6]
0
2
_6
0
-28
^
13
-4
2
0
2
0
3
-1
0
6
-28
13
0
2
1
3
-1
1
'l
-4
2'
1
-4
2
^0
2
0
0
1
0
0
-4
1
0
-4
1
1
-4
0
1
0
0
Reduce P4 to upper triangular form,
1
0 =u
0
2
1
3
1
4 0
1
2'
-i 0"
1
—>
1
-i 0
0
2
1
0
1 i
0
0
1
0
0
= u
1
1
The multipliers used were -, -3, and —, so
2
1
The multipliers used were - , -6, —, a nd 4,
2
■3
0
which leads to Lj = 0
2
L =
O’
0 . Since
3
0
0
0
2
0
3
0
1
[e -4 1_
1
0
0
3
0
0‘
L,= 0
1
0
0
2
0, then
2
-2
1
0
0
1
0
0
1
4 =
A = PLU =
0
Oiri
0
2
0
-4
0
1
2 -2 lj|_0 0 lJ[o
15.
P
-1
1
0
1
0
0
0
0
1
0
is the
1
0
1
0
2
0
0
1
0
3
0
1
0
1
and P
b =
1
3
0
> i
0
0
1
2
21’ X,
1
0
1
0
0
1
4
^2
2
3
-5
1
0
0
17
-^3
5
0
0
3'!
0
1
0
^2
2
3
-5
1
y3
5
, the system to solve is
4 nx,]
3
0
0
0
2
0
3
0
1
3
0
0
>’i
-2
0
2
0
>'2
4
3
0
1
>’3
2 , so the
5
0
1
1
0
0
1
-2
-2 =
4
1
•^3
= -2
=4
2^2
IS
+ >>3=1
2
which has the solution >’1 = -—, y2 = 2,
>’3
= 1
Tl
IS
4
0 1_
1
0 oiri
0
1
system P A\ = P b is
ri
0
0
Since Ph =
1
I
3
0
-2
2’
LDIZ-decomposition of 4.
0
0
1
1
0ir3
0
. Since P = P ', we have
= 2
1
-5^2+ ^3 =5
0
1
1
0
0
1
^2
which has the solution y\=\, y2=2. ^3=12.
201
i 04.1
2
3
-2 =
2
-3
3
IS
Chapter 9: Numerical Methods
SSM: Elementary Linear Algebra
2
•^1 ---^2
(b) There is no dominant eigenvalue since
?ii|=|X4|=5.
3
which gives the solution
^2+-^3= 2
3. The characteristic polynomial of A is
^3 =3
X~ -4X-6-0, so the eigenvalues of A are
1
to the original system: X| =
1
X = 2±yf\0. 3.= 2+ Vio = 5.16228 is the
— > -T? — — >
2
2
dominant eigenvalue and the dominant
X3 =3.
1
eigenvector is
-0.16228 ■
3-vro
19. (a) If A has such an Lf/-decomposition, it can
be written as
-1
'a
b
1
0
X
y
X
y
-1
c
d
w
1
0
Z
wx
wy + z
Ax0
=
5
0
5
0.98058
-0.19612
which leads to the equations
x=a
5
y=b
Ax 1
~
1
wx = c
-lir 0.98050
-1
Ax,
wy + z = d
Q
,
5
Ax, =
solution X = a, y = ^7, m = —, and
a
ad - be
a
a
z=d
=
Because the solution is unique, the
Lf/-decomposition is also unique.
b
1
c
d
4^
a
Olfa
1
0
0.98837
b
X4 =
ad-be
1
0.98837
5.09391
1
-0.15206
-0.83631
Ax2
0.98679
Ax2
-0.16201
AX3 =
(b) From part (a) the Lf/-decomposition is
a
5.09902
-0.78446
-0.15206
Since a^O, the system has the unique
be
-0.19612
5
-1
0.98679
5.09596
-1
-1
-0.16201
-0.82478
AX3
0.98715
ll^^3
-0.15977
a
AX4 =
5
-1
0.98715
5.09552
-1
-1
-0.15977
-0.82738
True/False 9.1
0.98709
(a)
The dominant unit eigenvector is =
False; if the matrix cannot be reduced to row
echelon form without interchanging rows, then it
does not have an L{/-decomposition.
-0.16018J’
which X4 approximates to three decimal place
accuracy.
(b)
False; if the row equivalence of A and U requires
interchanging rows of A, then A does not have an
Lf/-decomposition.
(c)
True
(d)
True
(e)
True
X‘''^ =(Axi/x, 5.24297
=(Ax2/x2 =5.16138
=(Ax3/x3 =5.16226
=(Ax4/x4 =5.16223
approximates X to four decimal place
accuracy.
Section 9.2
5. The characteristic equation of A is
'■y
A- -6A-4 = 0, so the eigenvalues of A are
Exercise Set 9.2
A = 3±Vl3.
A = 3 + Vl3 =6.60555 is the dominant
1. (a) A3 =-8 is the dominant eigenvalue.
eigenvalue with corresponding scaled
202
Section 9.2
SSM: Elementary Linear Algebra
2-713
Ax0 -
-0.53518
Xi -X,
1
1
-3
1
-2
-3
5
I
2
Ax0
^(2) AX2-X2 - 2.976
X2-X2
;^(3) _ '^XyXg = 2.997
-1
Xl =
AX| -Xj
5^3 •^3
1
max(AxQ)
(c) The characteristic polynomial of A is
=6
X^-4?^+3 =(A,-3)(?i-l)=0 so the
Xi -X,
Ax I
-
Axi -Xj = 2.8
(b)
3
eigenvector
1
-3
-3
5
-4
1
Ax I
dominant eigenvalue of A is X = 3, with
8
1
dominant eigenvector -I ■
-0.5
^2 =
max(Axi)
3-2.997
Ax2•X2 = 6.6
3
X2-X2
or 0.1%.
-3.5
1 -3]r-0.5'
Ax2
-3
5
6.5
1
Ax2
-0.53846
max(Ax2)
1
^3 =
9. By Formula (10),
^5 =
^(3) ^X3-X3 - 6.60550
0.99180
A^xo
max(A^Xo) -
Thus
X3-X3
1
AX3 =
=
AX5 • X5
= 2.99993.
X5'X5
-3' -0.53846
-3
= 0.001
(cl) The percentage error is
5
-3.53846
Ax3
-0.53488
max(Ax3)
1
X4 =
1
6.61538
1
13. (a) Starting with Xg = 0 , it takes 8 iterations.
0
0.229
AX4 • X4
= 6.60555
= 0.668 ,
X4 ■X4
2
7. (a) Axg =
= 7.632
0.668
-1
1
2
0
'0.507]
2
X2
= 0.320 , X^^'> = 9.968
0.800
Ax 0
*1 =
Ax 1
max(Axo)
' 2
-
'0.380]
-0.5
X3 - 0.197
-f
1
2.5
2
-0.5
2
'0.344]
Ax 1
^
X4 = 0.096 ,
max(AX|)
X
-1
-0.8
1
2
Axt =
10.622
0.904
2
-0.8
- 10.827
_0.934_
2.8
'0.317]
-2.6
X5 = 0.044 ,
-10.886
_0.948_
Ax2
^ max(Ax2) [-0-929
'0.302]
^6 - 0.016 ,
0.953
203
-10.903
Chapter 9: Numerical Methods
SSM:Elementary Linear Algebra
Section 9.3
0.294
= 0.002 ,
= 10.908
Exercise Set 9.3
0.956
■ 0.290]
^8 - -0.006 ,
= 10.909
1. The row sums of A are h0 - 2 . The row sums
0.957
2
2
1
of A
T
are a0 - 0
0
(b) Starting with Xq =
0
3
, it takes 8 iterations.
0
0.577
1
0
x, =
0.577
,
0
5
1
3
5
,
0.65094
ll^aol
0.65094
1.30189
- 8.062
T
0
4'h, =
0.830
1.69246
0.193
0.60971
4^h1
0.041
0.376
1
0
4ao
0
*3 =
0
3
0.39057
h,=
0.249
0.498
2
= 6.333
0.577
X2 “
1
0 0
3. 4a0 - 1
a
,
- 8.382
0
1 -
0.79262
0.905
X4
0.175
2
1
0.073
1
2 0 0
0.305
,
5. 4^4 =
=8.476
0.933
0
0
0
0
0 0
0
0
0
2
0.167
0.091
X5
S5
0.266
2
The column sums of4 are
, X^^'> = 8.503
1
0
0.945
2
0.162
0.101
^^6
0.245
3
2
,X^^^ =8.511
a
0-
3
1
0.951
3
0
0.159
0.107
0.234
0.70225
, X^'^'> =8.513
r
4' 4a0
a
0.953
1-
T
A' 4a0
0.70225
0.11704
0
0.158
0.70656
0.110
xg
=8.513
0.228
4^4a
0.70656
4^4a|
0.03925
a, =
0.954
204
0
, so we start with
Section 9.4
SSM: Elementary Linear Algebra
0.04370
0.70705
aj
_ A^Aa2 _ 0.70705
0.73455
0.01309
T
A' Aa3
s=
a3 =
0.32670
0
0.07075
0.70710
T
A' Aa3
34
0.58889
0.70710
0.02273
0.00436
0.73630
SS
A^ Aaj
0
a4 =
0.7071 1
A^Aa4
A^Aa4
ss
0.59045
T
A Aa3
0.32766
0.03677
0.70711
0.01179
0.00145
0
0.73679
T
A' Aa4
0.59086
^>5 =
0.7071 1
A^Aa4
_ A^Aaj _ 0.70711
^6
0.00048
A^ Aa^
0
0.32790
0.01908
0.00612
From ag, it is apparent that sites 1 and 2 are tied
^6 =
for the most authority, while sites 3 and 4 are
irrelevant.
0.73693
A^Aaj
A^Aag
0.59097
0.32796
0.00990
1
7. A^A =
0 0 0
1
3
2
1
0
authority, followed by sites 3 and 4, while sites 1
0
2
2
1
0
and 5 are irrelevant.
0
1
1
1
0
0 0 0
2
0
From ag, it is apparent that site 2 has the most
Section 9.4
1
Exercise Set 9.4
3
The column sums of A are
1. (a)
2 , so we start with
For n = 1000 = 10^, the flops for both
phases is
2
-(10-^)^ +-(10^)^ --(10^)= 668,165,500,
3
1
0.22942
3
0.68825
2
0.45883
^9
0.22942
2
(b)
0.45883
n
= 10,000=10'':
^(104)3+1(104)2
0.15250
3
y
A' Aa0
a, =
llA^Aa
0
6
0.6681655x10“' =0.067 second.
1
ao =
2
which is 0.6681655 gigaflops, so it will take
0.71166
2
2(10")
6
0.55916
= 666,816,655,000 flops
or 666.816655 gigaflops
0.30500
The time is about 66.68 seconds.
0.25416
(c)
0.08327
a, =
A^Aa,
n
= 100,000 = 10^
0
0.72861
n
-(10-“')^+-(10-^)^—(10-“')
0.58289
3
2
6
T
A' Aa1
0.32267
= 666,682x10^ flops
0.13531
or 666,682 gigaflops
Tbe time is about 66,668 seconds which is
about 18.5 hours.
205
Chapter 9: Numerical Methods
SSM: Elementary Linear Algebra
in AB requires 2n - 1 Hops, for a total of
3. « = 10,000 = 10“*
n^(2n-\)= 2n^-n^ flops.
'y
3
Note that adding two numbers requires 1 flop,
adding three numbers requires 2 flops, and in
general, n - 1 flops are required to add
(a) ^tr = -(10'2)= 666.67x10^
3
666.67 gigaflops are required, which will
n numbers.
666.67
take
= 9.52 seconds.
70
(b)
Section 9.5
= 10* =0.1x10^
Exercise Set 9.5
0.1 gigaflop is required, which will take
1. The characteristic polynomial of
about 0.0014 second.
rn
^
a‘'a= 2 [1 2 0]= 2 4 0 isX^{X-5y,
(c) This is the same as part (a); about
9.52 seconds.
0 0 0
0
thus the eigenvalues of A^ A are X| =5 and
^2 = 0, and cT] = Vs and (72 =0 are singular
(d) 2n* =2x10‘2 =2000x10®
2000 gigaflops are required, which will take
about 28.57 seconds.
values of A.
5
5. (a) n = 100,000 = 10
2 3
— n
3. The eigenvalues of
= 2xI0'5
3
1
2
1
-2
5
0
-2
1
2
1
0
5
A^A =
3
are
=5
I
= 0.667x10'-“*
and X2=5 (i.e., 3. = 5 is an eigenvalue of
= 6.67x10-^x10®
multiplicity 2); thus the singular value of A is
cr, = Vs.
Thus, the forward phase would require
about 6.67x10-“* seconds.
n
5. The only eigenvalue of
2 =|o'0 =10x10®
A^ A =
The backward phase would require about
10 seconds.
2 3
= 2x10*2
3
1
1
2
0
1
1
0
2
is?i = 2
and
(multiplicity 2), and the vectors V| =
(b) n = 10,000 = 10''
—n
1
-1
0
0
V2 =
3
form an orthonormal basis for the
1
12
= 0.667x10'-"
eigenspace (which is all of R^). The singular
= 6.67x102x10®
values of A are cr| = V2 and (J2 = V2. We have
About 667 gigaflops are required, so the
computer would have to execute
2(667)= 1334 gigaflops per second.
1
1
J_ 1 -1 1]- n/2
U| = — Av I -
V2h
cr1
iJLo
7. Multiplying each of the n entries of A by c
i
2
requires n flops.
u
AV2 =
2-
cr-y
9. Let C =[Cjj]= AB. Computing
multiplying each of the n entries
requires first
-1
V^Li
1
0
. This
I
A = UI.V'^ =
must be summed, which
requires n - 1 flops. Thus, each of the
1
1
I
results in the following singular value
decomposition of A:
by the
corresponding entry bj^j, which requires n flops.
Then the n terms
, and
1
entries
206
1
^
I
1
S
^/5
IV2 0 [1 O'
0
[0 1.
Section 9.5
SSM: Elementary Linear Algebra
7. The eigenvalues of
16
4 0]r4 6
6
24
are Xi = 64
6 4j[o 4
24
52
2n/2
u
3-
. This results in the following
3
and ^2 = 4, with corresponding unit
6
2
1
eigenvectors v^ =
2 and V2 =
singular value decomposition:
~T5
A = UI.V'^
1
^5
2
I
respectively. The singular values of A are cT] = 8
3
VI
and CT2 = 2. We have
i
0
3
1
1
_1 4 6
U| = — Av I “o
VI _ VI , and
2
VI
1
u
2-
(To
J_ 4 6
Av2 = 2 0 4
I
VI
VI
A = UI.V'^
i
VI
VI
0
J_
0
VI
0
VI
Vil
1
1
VI
0
I
1
-1
1
“2
1
2
1
3
0
0
2
and X2 = 2, with corresponding unit
0
eigenvectors V| = 0 and V2 =
.1- j
9
-9
-9
9
-1
2
respectively. The singular values of A are
are
-2
(Ji = >/3 and £72 = V2. We have
-2
I
^1=18 and 3-2 = 0, with corresponding unit
eigenvectors Vj =
1
I
VI
VI
and V2 =
1
VI
1
0
1
1
1
1
0
VI
-1
i
VI
0
1
-2
2
-1
1
3V2 2
-2
0
1
I
3
VI
1
i
3
VI
2
0
1
VI . We choose
1
U2 =
2
1
and
1
VI
A is £7| = ^/l8 = 3V2, and we have
U| =
VI
1
u, = — Av 1 -
respectively. The only nonzero singular value of
Av I -
VI
are X, =3
1
1
L VI
1
1
T
J_
VI
6
1
0
9. The eigenvalues of
"
J_
11. The eigenvalues of
i
2
2
VI
0
one possibility.
VI.
This results in the following singular value
decomposition:
2
0
0
(extended) orthonormal basis for R^. This is just
I
VI
1
VI
Note: The singular value decomposition is not
unique. It depends on the choice of the
VI
2
VI
_L
L 3 VI
i
8 0 4
0
3
2.
_2
I
VI
3y/2 o1 2VI
1
i
-1
VI
2
V6
3
1
u
3-
We must choose the vectors U2 and U3 so that
lUl-
VI
i
orthonormal basis for R^. This results in the
VI
A possible choice is U2 =
SO that {u|, U2, U3} is an
i
U2, U3} is an orthonormal basis R^.
0
VI
following singular value decomposition:
and
A = UI.V'^
1
VI
I
0
VI
207
^
VI
1
i
i
VI
VI
VI
i
i
1
VI
VI
"x/3 0 Ip 1
0
x/2
0
0 ^
0
0
1
Chapter 9: Numerical Methods
SSM:Elementary Linear Algebra
True/False 9.5
5. The reduced singular value expansion of A is
(a) False; if A is an w x n matrix, then
2
3
is an
3V2
nxm matrix, and A^ A is an n x n matrix.
i -
1
I
2
3
(b) True
7. The reduced singular value decomposition of A
(c) False; A^A may have eigenvalues that are 0.
I
0
T
3
(d) False; A would have to be symmetric to be
orthogonally diagonalizable.
[1 0]+ V2 ^ [0 1].
i
is
I
1
(e) True; since A^ A is a symmetric nxn matrix,
9. A rank 100 approximation of A requires storage
space for 100(200 + 500 + 1)= 70,100 numbers,
(f) False; the eigenvalues of A^A are the squares of
the singular values of A.
while A has 200(500)= 100,000 has entries.
(g) True
True/False 9.6
Section 9.6
(a) True
Exercise Set 9.6
(b) True
1. From Exercise 9 in Section 9.5, A has the
(c) False; V[ has size nxk so that V\ has size
singular value decomposition
A=
1
2
1
3
42
1
0
6
2V2
3
kxn.
3V2 0 --L
3
2
I
3
0
0
0
0
1
V2
di
1
1
Chapter 9 Supplementary Exercises
1. Reduce A to upper triangular form.
6
-6
2
-3
6
0
6
-3
1
0
2
=U
where the lines show the relevant blocks. Thus
0
the reduced singular value decomposition of A is
2
The multipliers used were — and 2, so
3
A = UylyVl =
3 [3V2][-;^
2 0
L=
2
2 0
-3
1
and A =
-2
-2
1
1
0
2'
3
3. Reduce A to upper triangular form.
3. From Exercise 11 in Section 9.5, A has the
2
singular value decomposition
1
A=
i
1
2
0
x/3 0 Ip 1
i
i
I
i
0
V2
0
0
0
1
s
Thus the reduced singular value decomposition
of A is
0
1
1
1
2
4 7^ 0
2
4
1
3
7
[1 2 3’
1
4
7 ^1
1
3
7
_1 3 7
3
0
where the lines indicate the relevant blocks.
1
4 6
’1 2 3’
1
2
3
1
2
3
0
2
4 ^0
1
2^ 0
1
2
0
1
4
1
4
0
0
2
1
2
0
1
0
0
0
3
2 =U
1
1
A=
=
VI
1
I
VI
VI
The multipliers used were —, -1,
>/3 0l[l o'
0 V2 [o 1.'
208
’ 2’
1,
Chapter 9 Supplementary Exercises
SSM: Elementary Linear Algebra
2 0
0
AX4
1
and — so L =
2
1
2
0
1
1
2
2 0 0
A=
1
1
2
3
2
0
0
1
2
1
2
0 0
1
X5 =
and
0.9918
max(Ax4)
as compared to the exact
^5 = 0.9918
1
eigenvector v =
5. (a) The characteristic equation of A is
7. The Rayleigh quotients will converge to the
?^^-4A,+3=(?i-3)a-l)=0 so the
dominant eigenvalue ^4 =-8.1. However,since
dominant eigenvalue of A is 3.| =3, with
A.4
the ratio
corresponding positive unit eigenvector
8.1
= 1.0125 is very close to 1,
8
I
0.7071
V =
the rate of convergence is likely to be quite slow.
0.7071 ■
I
9. The eigenvalues of
(b) A\q =
2
1
1
1
2
0
A=
2
2
2
2
are X,1 =4
2
and X2 =0 with corresponding unit
1
Ax0
1 [2
xi =
eigenvectors V| =
0.4472
Ax 1
X2 =
0.7809
Ax2
||ax2
0.6805
X4 =
IS CTf
I
I
U| = — Av 1 “T
2
(^1
0
1
1
I
ViJ
^/i
AX3
0.6983
AX4
0.7100
We must choose the vectors U2 and U3 so that
0.7042
(U|, U2, U3) is an orthonormal basis for
0.7042
A
i
0
as compared to
possible choice is U2 = 1
0.7071'
Vi
I
Vi
0.7071 ■
This results in the following singular value
decomposition:
2
A = UI.V^
I
Ax0
max(Axo)
0.5
Ax 1
1
max(Ax2) [0.9286
AX3
^[ 1
0.9756
209
1
Vi
0
0
i
Ax2
0
Vi
Vi
^ max(AX|) _0.8
max(Ax3)
0
and U3 =
0
(c) Axo= j
*3 =
Vi _
0 0
0.7100
V =
1
= yf4 = 2, and we have
1
AX3 __ [0.7158'
X5 =
X5 =
and V2 =
1
respectively. The only nonzero singular value A
||Ax,|| [0-6247
0.7328
=
1
0.8944
2 0
0
0
0 0
0
Vi
Vi
Vi
_L
^
Vi
Vi
Chapter 9: Numerical Methods
SSM: Elementary Linear Algebra
11. A has rank 2, thus 1/| =[U| U2] and
T
r
V,
^ and the reduced singular value
V2
decomposition of A is
T
i
I
0
2
i
i
2
1
24
i
1
0
2
i
01 f
I
2
3
3
2
2
1
3
3
3
12
i
2
210
Chapter 10
Applications of Linear Algebra
Section 10.1
3. Using Equation (10) we obtain
2
Exercise Set 10.1
1. (a) Substituting the coordinates of the points
3' 1
X
=0
into Equation (4) yields
2
2
1
2
xy
0
0
0
1
0-1
1
4
0
0
2
0
1
4
-10
25
2
-5
1
16
-4
1
4-1
1
which, upon cofactor expansion along the
first row, yields -3x + y + 4 = 0; that is,
y = 3x - 4.
y
0
X
0
y
=0 which is the
2
xy
y
0
0
1
0-1
1
=0 yields
y
0
x2
same as 4
X
(b) As in (a). 0
1
X
0
X
y
0
0
2
0=0 by
4
-10
25
2
-5
16
-4
1
4
-1
expansion along the second row (taking
advantage of the zeros there). Add column five
to column three and take advantage of another
2x + y - 1 = 0 or y = -2x + 1.
row of all but one zero to get
2. (a) Equation (9) yields
x^
xy
X
y
1
4
0
0
2
2
6
I
4
-10
20
2
4
2
0
1
16
-4
0
4
34
5
3
I
7
7
X +y
40
= 0 which, upon first-
y^ + y x
= 0.
Now expand along the first row and get
row cofactor expansion, yields
160x^ +320xy +160(y^ + y)-320x = 0; that is,
18(x2 + y2)-72x-108y +72=0 or,
dividing by 18, x^ + y^-4x-6y+4= 0.
7
1
X +2Ay + y -2x + y = 0, which is the equation
of a parabola.
Completing the squares in x and y yields the
4. (a) From Equation (11), the equation of the
standard form (x-2)^ +(y-3)^ =9.
^
2
x" + y
plane is
X
y
2
-2
1
34
3
5
1
52
-4
6
1
(b) As in (a).
=0 yields
X
y
z
1
1
1
-3
1
1
-1
1
1
0-1
2
1
= 0.
Expansion along the first row yields
-2x - 4y - 2z = 0; that is, x + 2y + z = 0.
50(x- + y^)+1OOx-200y -1000 = 0; that
7
7
X
is, x“ + y+2x-4y-20 = 0. In standard
form this is (x+1)^+(y-2)^ = 25.
y
z 1
(b) As in (a), ^
3
1
1
-1
-1
1
1
2
1
1
=0 yields
-2x + 2y - 4z + 2 = 0; that is -x + y - 2z + 1
= 0 for the equation of the plane.
211
Chapter 10: Applications of Linear Algebra
SSM: Elementary Linear Algebra
5. (a) Equation (11)involves the determinant of
the coefficient matrix of the system
+ y^ +
5
0
X
y
z
•^1
>"1
^1
^2
0
X2
yj
^2
^3
0
.^3
^3
Z3
C4
0
1
(b) As in (a),
passing through the origin parallel to the
plane passing through the three points, the
constant term in the final equation will be 0,
which is accomplished by using
y\
•^2
3’2
22
X3
y3
Z3
1
=0
0
1
1
-24(x^ + y^ +z^)+48x +48y +72 = 0; that
is, x^+y^+z^-2x-2y-3=0 or in
standard form, (x-l)“ +(y-l)^ + z^ =5.
c,jr + C2y + C3Z + C4 =0. For the plane
•^1
z 1
-2
yields
i- 1, 2, 3, while row 1 gives the equation
z
2-1
3
through the three points (jc,-, y,-, z,),
y
1
3
5
Rows 2 through 4 show that the plane passes
X
X y
0
7. Substituting each of the points (xj, y|),
(X2, y2), (X3, y3), (X4, y4), and (xj, y5) into
the equation
0
C[X^ +C2xy+ C3y^ +C4X+ C5y+ C5 = 0 yields
vf + ^"2-^1 y\ + cjyf + C4X, + C5y, + cg =0
1
=0
<^1^5 +<^2-^5>'5 +^33'! +<^4-^5 +‘^53'5 +^6 -0-
(b) The parallel planes passing through the
origin are x + 2y + z = 0 and -x + y - 2z = 0,
respectively.
These together with the original equation form a
homogeneous linear system with a non-trivial
solution for Cj, C2,..., c^. Thus the determinant
6. (a) Using Equation (12), the equation of the
x^+y^+z^ X
sphere is
of the coefficient matrix is zero, which is exactly
Equation (10).
y z 1
14
1
2
3
1
6
-1
2
1
1 =0.
2
1
0
1
1
6
1
2
-1
1
8. As in the previous problem, substitute the
coordinates (x,-, y,-, z,) of each of the three
points into the equation CiX + C2y+ C3Z + C4 =0
to obtain a homogeneous system with nontrivial
Expanding by cofactors along the first row
yields
solution for q,..., C4. Thus the determinant of
16(x^ + y^ + z^)-32x-64y -32z+32= 0;
that is, (x^ + y^+z^)-2x-4y-2x +2= 0.
the coefficient matrix is zero, which is exactly
Equation (11).
9. Substituting the coordinates (x,-, y,-, z,)of the
Completing the squares in each variable
yields the standard form
four points into the equation
O
9
T
(x-l)2+(y-2)2+(z-l)2=4.
C[(x +y +z )+ C2X + C3y+ C4Z + C5 =0 of the
Note: When evaluating the cofactors, it is
useful to take advantage of the column of
ones and elementary row operations; for
sphere yields four equations, which together with
the above sphere equation form a homogeneous
linear system for q,..., C5 with a nontrivial
example, the cofactor of x^ + y^ + z^ above
solution. Thus the determinant of this system is
can be evaluated as follows:
zero, which is Equation (12).
1
2
-1 2
1 0
1
3
1
1
2
3
1
1 1^-2
1
2-1
0-2 0
1 “ 0 -2 -2 0 = 16 by
1
0
0-40
cofactor expansion of the latter determinant
along the last column.
212
Section 10.2
SSM: Elementary Linear Algebra
Section 10.2
10. Upon substitution of the coordinates of the three
points (x^,
(xj, >’2) and (X3, ^3), we
Exercise Set 10.2
obtain the equations:
C| y + Cjx^ + C3X + C4 =0
C|>’1 + C2xf + C3X, + C4 = 0
2
Xj 2.V|-a-2=0
1.
A, =2
r,
^1 yi +
S
+ C4 = f>
(2 A
1
C|y3+C2X3+C3X3+C4 =0.
1
KT’'J
/
Kl
This is a homogeneous system with a nontrivial
J
A2=1
solution for q, C2, C3, C4, so the determinant of
2,2)
the coefficient matrix is zero; that is,
2x|+ 3^2 = 6
t
y
X"
X
1
yi
(0,0)
In the figure the feasible region is shown and the
extreme points are labeled. The values of the
objective function are shown in the following
= 0.
^2
^2
y2 4 ^2
table:
11. Expanding the determinant in Equation (9) by
cofactors of the first row makes it apparent that
2
yi
1
X2
y2
1
3’3
1
^3
in the final equation is
(X|, X2)
z = 3x| + 2x2
(0,0)
0
so the coefficient of x^ + y^ =0 and the
1
•^1
0
0
1
X2
0
0
1
^3
0
0
1
13
2
22
3
6
(2, 0)
12. If the three distinct points are collinear then two
of the coordinates can be expressed in terms of
the third. Without loss of generality, we can say
that y and z can be expressed in terms of x, i.e., x
is the parameter. If the line is (x, ax + b, cx + d),
then the determinant in Equation (11) is
-cx - d
2
(^•f)
resulting equation is that of the line through the
three points.
—ax - b
7
(!■ '
columns are linearly dependent (y,- = mx,- +b),
equivalent to
Value of
(f
. If the points are collinear, then the
X
Extreme point
2
the coefficient of x + y
^1
I
5
(2,0)
22
Thus the maximum, —, is attained when
3
2
^1
2.
= 2 and X2 = 3
•^2
A, =3
3'
2,v,-A2=-2y
Expanding along the first row, it is clear that the
determinant is 0 and Equation (11) becomes
4a|-A2=0
0-0.
13. As in Exercise 1 1, the coefficient of
'y
0
1
X + y +z will be 0, so Equation (12) gives
the equation of the plane in which the four points
2
X,
The intersection of the five half-planes defined
by the constraints is empty. Thus this problem
lie.
has no feasible solutions.
213
Chapter 10: Applications of Linear Algebra
SSM: Elementary Linear Algebra
Thus, it has a minimum, which is attained at the
3.
14
5
point where
—A'l + A'-j — 1
25
— • The value of
and -^2
335
z ther^is
3a'| -X2=-5
18
• In the problem’s terms, if we use
1
25
— cup of milk and
2a|+4a2 = 12
ounces of corn flakes, a
18
minimum cost of 18.60 is realized.
+
-»-'i
The feasible region for this problem, shown in
6. (a)
the figure, is unbounded. The value of
Both X| > 0 and X2 ^ 0 are nonbinding. Of
the remaining constraints, only
z = -3X|+ 2^2 cannot be minimized in this
region since it becomes arbitrarily negative as
we travel outward along the line -X|+X2 =1;
2x|+3x2 - 24 is binding.
(b) X| -X2 < V will be binding if the line
i.e., the value of z is
X| -X2 = V intersects the line X2 =6 with
-3xi + 2^2 = -3xi + 2(X| +1)= -X]+2 and X|
0 < X| < 3, thus if -6 < V < -3. If V < -6,
can be arbitrarily large.
then the feasible region is empty.
■'^2
4.
A| = 6000
A'l +a2 =10.000 >
(c)
iSOOO, 5000)
X2 ^ V will be nonbinding for v > 8 and the
feasible region will be empty for v < 0.
7. Letting X| be the number of Company A’s
(6000, 4000)
V X6000. 2000)
(2000, 2000;
containers shipped and X2 the number of
Company S’s, the problem is:
X2 =2000
Maximize z = 2.20x|+3.00x2 subject to
40X|+50x2
2x| +3x2
X,
X2
X|
.V| - a'2 = 0
The feasible region and vertices for this problem
are shown in the figure. The maximum value of
the objective function z = .lX| + .07x2
attained
<37,000
^ 2000
>0
> 0.
A2
when X| = 6000 and X2 = 4000, so that z = 880.
In other words, she should invest $6000 in bond
A and $4000 in bond B for annual yield of $880.
(
\
3
5.
A'2
3a'| - 2
2
'3
40a'|+50a'2 =37,000
2j
0, 666^
3y
=0
9'
(550, 300)
,2’ 4,
2a-, +3a2 =2000
S
!■
'14
_
(0, 0)
A, - 2a2 = 0
'A- -
4a| + 2x2 = 9
1
^1
vertex at which the maximum is attained is
X| = 550 and X2 = 300, where z = 21 10.
A truck should carry 550 containers from
Company A and 300 containers from Company
B for maximum shipping charges of $21 10.
,21 ' 2L
-A,I H
V
(925, 0)
The feasible region is shown in the figure. The
^2’2,
,9 ’ isj"
I-' I
1
1
A-, = —
10 ^
3
The feasible region and its extreme points are
shown in the figure. Though the region is
8. 'We must now maximize z = 2.50x|+3.00x2.
unbounded, X| and X2 are always positive, so
The feasible region is as in Exercise 7, but now
the objective function z = 7.5xi+5.0x2 'S also.
the maximum occurs for x, = 925 and X2 = 0,
where z = 2312.50.
214
Section 10.3
SSM: Elementary Linear Algebra
925 containers from Company A and no
containers from Company B should be shipped
Section 10.3
for maximum shipping charges of $2312.50.
Exercise Set 10.3
1. The number of oxen is 50 per herd, and there are
9. Let X| be the number of pounds of ingredient A
used and X2 the number of pounds of ingredient
7 herds, so there are 350 oxen. Hence the total
number of oxen and sheep is 350 + 350 = 700.
B. Then the problem is:
Minimize z = 8x, +9x2 subject to
2. (a) The equations are fi = 2A, C = 3(A + 5),
D ^ 4{A +B+ Q,300 = A + B+C+D.
2X|+5x2 - 10
2jr| + 3^2 > 8
6x|+4^2 ^ 12
X| >0
Solving this linear system gives A = 5(and
B= 10, C = 45,D = 240).
(b) The equations are B = 2A, C= 3B,D = 4C,
]32 = A + B + C+D.Solving this linear
system gives A = 4(and 5= 8, C = 24,
X-, >0
^2 (0,3)
\
D = 96).
(1 11'
is’ 5,
3. Note that this is, effectively, Gaussian
elimination applied to the augmented matrix
YS^2jci + 3x-y =8
“l 1 10]
\
_> i 7;
2xi +5x2 ='0
4. (a) Let x represent oxen and y represent sheep,
then the equations are 5x + 2y = 10 and
2x + 5y = 8. The corresponding array is
(5,0)
6.V1 +4.V2 = 12
Though the feasible region shown in the figure is
unbounded, the objective function is always
positive there and hence must have a minimum.
This minimum occurs at the vertex where
2
2
5
5
2
12
10
X, = — and X9 =—. The minimum value of z is
5
^
5
124
and the elimination step subtracts twice
column 2 from five times column 1, giving
— or 24.80.
Each sack of feed should contain 0.4 lb of
ingredient A and 2.4 lb of ingredient B for a
5
minimum cost of 24.80.
10. It is sufficient to show that if a linear function
21
2
20
10
has the same value at two points, then it has that
value along the line connecting the two points.
20
Let ax^ +bx2 be the function, and
and so y =
aXj'+ bx2 = ax"+ bx2 = c. Then
{tx\ +{\-t)xl, txo +(l-r)x2) is an arbitrary
point on the line connecting (xj', X2) to (x^, X2)
and a(tx[ +(1 -t)x")+ b(tX2 +(1 -0-^2)
= t{ax[ + bx2)+{\-t){axl+bx2)
21
unit for a sheep, and
34
x =-— units for an ox.
21
(b) Let X, y, and z represent the number of
bundles of each class. Then the equations
are
= tc +{\-t)c
2x+y
3y + z = l
= c.
X
4z = l
and the corresponding array is
215
Chapter 10: Applications of Linear Algebra
2
3
SSM:Elementary Linear Algebra
(b) Exercise 7(b) may be solved using this
1
technique,
1
represents gold, X2 represents
brass, x^ represents tin, and X4 represents
1
1
4
1
much-wrought iron, so n = 4 and
= 60,
1
|(60)= 40,
«2 =
Subtract two times the numbers in the third
column from the second column to get
03 =1(60)= 45,
1
^4= 1(60)= 36.
•^1
3
1
1
-8
4
1
-1
1
(a,-i-a3+a4)-a|
;;
n-2
40-h 45-^36-60
4-2
Now subtract three times the numbers in the
61
second column from the first column to get
2
1
61
19
2
2
61
29
X2 = ^2 - Xj =40
1
25
4
X3 = 03 -x'l =45
4
-8
1
-1
X4 = ^4 - xi = 36-
This is equivalent to the linear system
2
2
61
11
2
2
The crown was made with 30— minae of
2
x + Az = \
y-8z = -l.
1
1
gold, 9— minae of brass, 14— minae of
25z=4
2
2
From this, the solution is that a bundle of the
tin, and 5I2 minae of iron.
9
first class contains — measure, second
25
6. (a) We can write this as
class contains — measure, and third class
25
5x-(->’-(-z-A'=0
x +ly +z-K^O
4
contains — measure.
x-i-yH-8-A:= 0(a3x4 system). Since the
coefficient matrix of equations(5) is
25
invertible (its determinant is 262), there is a
5. (a) From equations 2 through n, Xj = aj-Xj
unique solution x, y, 2 for every K-, hence, K
is an arbitrary parameter.
(j= 2,..., n). Using these equations in
equation 1 gives
2ia:
X, +{a2-x^)+{ai,-x^)+ ---+ia„-x^)= a^
a2+aj-i
Xj =
(b) Gaussian elimination gives x =
1-
^r^2
■
\4K
J =
First find X] in terms of the known
,
131
131 ’
\2K
z =
131
, for any choice of K.
Since 131 is prime, we must choose K to be
an integer multiple of 131 to get integer
quantities n and the a,. Then we can use
Xj = Uj- X| (/■ = 2, ..., n) to find the other
solutions. The obvious choice \sK= 131,
giving x = 21, y = 14, z = 12, and K= 131.
X;.
I
216
Section 10.4
SSM: Elementary Linear Algebra
(c) This solution corresponds to
6(yi -2y2+y^)= -1.1880
K=262 = 2- 131.
/,2
6(j’2 ~2>’3 + >’4)
7. (a) The system is x + y = 1000,
/1 \
-2.3295
1
1
X-
5J
; = 10, with solution x = 577
4r
6(y3-2>’4 + y5) = -3.3750
2
and -
= 422 and
9’>'=
receives 5779
The legitimate son
6(>’4-2y5+y6)= -4.2915
staters, the illegitimate son
and the linear system (21)for the parabolic
runout spline becomes
-1.1880
■5 1 0 o' M2
2
receives 422—.
9
1
2 1
(b) The system is G + B = - 60,
4
1
0
0
1
4
1
0
0
1
5
-2.3295
M3
M4
M5
-3.3750
-4.2915
3J
Vi
G+T =
Solving this system yields M2 =-.15676,
M3
- 60, G + /= - 60,
4
35
Ay3 =-.40421, M4 =-.55592,
M5 =-.74712.
G + B + T + I = 60, with solution G = 30.5,
B = 9.5, T = 14.5, and I = 5.5. The crown
From (19) and (20) we have
was made with 30-^ minae of gold,
The specific interval .4 < x < .6 is the third
= M2 =-.15676, Mg =M5 =-.74712.
interval. Using (14) to solve for a^, b^,
9— minae of brass, 14— minae of tin, and
2
C3, and ^3 gives
(M4-M3) = -.12643
2
1
5— minae of iron.
«3 =
2
6h
M3
1
(c) The system is 4 =
-C,
V3
y
^3 =
B = C+ - A, C= - 5+10, with
33
33;
= -.20211
(^4-^3)
(M^+2M2)h
h
6
= .92158
J3 = 3,3 =.38942.
The interpolating parabolic runout spline for
solution A = 45, 5 = 37.5, and C = 22.5.
.4 <x < .6 is thus
The first person has 45 minae, the second
S{x) = -. 12643(x - .4)^ - .2021 l(x - Af
has 37- , and the third has 22—.
2
2
+ .92158(x-.4) + .38942.
Section 10.4
(b) S(.5) = -. 12643(. 1)-^ - .20211(. 1)^
+ .92158(.1) + .38942
Exercise Set 10.4
= .47943.
Since sin(.5) = S(.5) = .47943 to five
2. (a) Set h = .2 and
decimal places, the percentage error is zero.
=0, y, =.00000
X2 =.2, y2 =.19867
X3 =.4, y3 =.38942
X4 =.6, >’4 =.56464
3. (a) Given that the points lie on a single cubic
curve, the cubic runout spline will agree
exactly with the single cubic curve.
X5 =.8, >’5 =.71736
xg =1.0, 3-6=.84147
Then
217
Chapter 10: Applications of Linear Algebra
SSM: Elementary Linear Algebra
(b) Set h - I and
d4 = y4=19.
x^=0,y^=\
X2 = I, >>2 = 7
X3=2, ^3=27
X4 = 3, )'4 = 79
^5 =4> ^5 = 181.
For 0 < X < 1 we have
5(x)= S,(x)= 3x-^ -2x2 +5x +1.
For 1 < X < 2 we have
S(x)= S2(x)
= 3(x -1)2 + 7(x -1)2 +10(x -1)+7
Then
= 3x2-2x2+5x+1.
6(J| -2^2+>'3) = 84
/r2
6(>’2 ~2>'3 + T4)= 192
/r2
6(>3-2y4 + y5) = 300
/r2
For 2 < X < 3 we have
5(x)= S3(x)
= 3(x -2)2 + 16(x-2)2 + 33(x-2)+ 27
= 3x2-2x2+5x+I.
For 3 < X < 4 we have
and the linear system (24)for the cubic
runout spline becomes
S(x)= S4(x)
"6 0 0]|'M2
= 3x2-2x2+5x + 1.
Thus Si(x)= S2(x)= S3(x)= S4(x), or
1
4
1
= 3(x-3)2 + 25(x-3)2 +74(x-3)+79
84
M3
192
0 0 6j[A/4
300
S(x)= 3x2-2x2+5x + 1 for0<x<4.
Solving this system yields M2 =14,
M3 =32, M4 =50.
4. The linear system (16) for the natural spline
■4 1 0]['M2
From (22) and (23) we have
M, =2M2-M3 =-4
M5 =2M4-M3 =68.
Using (14) to solve for the a/s, /?/’s, c,’s, and
Solving this system yields M2 = -.0000252,
dj’& we have a, =(M2-M|)_^
6/r
M3 =-.0000108, M4 =-.0000132.
«2 =
a
(M3-M2)_,
6h
(M5-M4)
4
r:
6h
^1 =
M1
2
^3 =
(M4-M3)
^
a
1 -
a^ =
2
03 =
04 =
C| =
c, =
=
C4 =
(M2+2M|)/7.
/7
6
(y3-y2)
(/U3 + 2M2)/7
h
6
(3-4-T3)
(M4 + 2M3)/7.
h
6
(}’5-3'4)
(M5+2M4)/7
h
6
-.0000816
_0 1 4j[M4
-.0000636
(M2-M1)
^ _ QQQQQQ42^
6/j
(M3-M2)
(M4-M3)
= .00000024,
= -.00000004,
6/7
= 25,
(y2 -Tl)
M3
6/7
2
^
1
Solving for the a/’s, i)/’s, c/’s, and di’s from
Equations (14) we have
6/1
M.,
= -2, ^ = - = 7,
2
4
From (17) and (18) we have M| = 0, M5 = 0.
= 3,
■ - '■>>
M4
M3
— = 16, /74 =
1
-.0001116
becomes
(M5-M4)_
= -.00000022,
6/!
= 5,
/7,I =
= 10,
M,
^
= 0, /72 = —
= --0000126,
2
2
/7^3 =^
/74 = M4 = - .0000066,
2 = -.0000054,^2
= 33,
M3
= 0.
= 74,
d\ =3’| =1. d2 =y2 =2, ^2^3 =3’3
Cl =
27,
218
(3-2->'i)
/?
= .000214,
6
Section 10.4
SSM: Elementary Linear Algebra
(33-^2) {M^+2Mj)h
C-, =
6
h
(3’5-3'4)
C4 -
+2M^)h
h
iyA-y^) {M^+2Mj,)h = -.000092,
=.000088, C3
h
6
= -.000212,
6
= 3., =.99815, d2 = 3’2 =.99987, d^, = 3-3 =.99973, a'4 = 3'4 =.99823.
The resulting natural spline is
5(3)=
-.00000042(3:+10)^+.000214(3+10)+.99815, -10 < 3 < 0
.00000024(3)^ -.0000126(3)“ +.000088(3)+.99987,0 < 3 < 10
-.00000004(3-10)^ -.0000054(3-10)^ -.000092(3-10)+.99973, 10 < 3 < 20
-.00000022(3-20)^ -.0000066(3-20)^ -.000212(3-20)+.99823, 20 < 3 < 30.
Assuming the maximum is attained in the interval [0, 10] we set S\x) equal to zero in this interval:
S\x)=.000000723^ -.00002523+.000088 = 0.
To three significant digits the root of this quadratic equation in the interval [0, 10] is 3= 3.93, and
5(3.93)= 1.00004.
-.0001116
'6 0 0][M2
5. The linear system (24)for the cubic runout spline becomes
1
4
1
M-i
-.0000816
[0 0 6JLM4
-.0000636
Solving this system yields Af2 =“0000186, M3 =-.0000131, M4 =-.0000106.
From (22) and (23) we have M,= 2^2-M3 =-.0000241, M5 = 2M4-M3 =-.0000081.
(M2-M,)
=.00000009,
Solving for the a/s,(i,’s, c,’s, and (7,’s from Equations(14) we have aj =
6/j
(M3-M2)
(M4-M3)
(M5-M4)=.00000004.
=.00000004, ^4 =
=.00000009, 03 =
fl
,=
6/j
^
= -.0000121,
2
b,1 =
^
M4
=-.0000053,
=—
=-.0000093, b.=^
= -.0000066, b.=^
2
-^
2
^
2
(>’2->'|) (M2+2M|)/7,
C| =
h
/j
(J3-52) (M3+2M2)/r
=.000282, c, =
6
(>’4-}3) (M4+2M3)/7
C3 =
6/t
6/j
6
h
=.000070,
6
(55-J4) (M5+2M4)/7_ -.000207,
= -.000087, C4 =
h
6
=.99815, d2=y2 = -99987, d2=y2,= -99973, J4 = >-4 =.99823.
The resulting cubic runout spline is
5(3)=
.00000009(3+10)^-.0000121(3+ 10)^+.000282(3+10)+.99815, 10<3<0
.00000009(3)-^ -.0000093(3)- +.000070(3)+.99987, 0 < 3 < 10
.00000004(3-10)^ -.0000066(3-10)“-.000087(3-10)+.99973, 10 < 3 < 20
.00000004(3-20)-'^ -.0000053(3-20)^ -.000207(3- 20)+.99823, 20 < 3 < 30.
Assuming the maximum is attained in the interval [0, 10], we set 5'(3) equal to zero in this interval:
5'(3)= ,000000273- -.00001863+.000070 = 0.
To three significant digits the root of this quadratic equation in the interval [0, 10] is 4.00 and 5(4.00)= 1.00001.
219
Chapter 10: Applications of Linear Algebra
SSM: Elementary Linear Algebra
6. (a) Set h = .5 and
x\ =0,
For 1 < X < 1.5, we have
=0
^(x)= ^2(x)= -2(x -1)+0= -2x + 2.
The resulting natural spline is
j:2 =.5, yj = 1
X3 = 1, ^3 =0
-2x +2
S{x)=
For a natural spline with n = 3,
0.5<x<l
-2x +2 l<x<1.5'
M|=3/3 =0 and
6(yi -2^2+ >’3)
(c) The three data points are collinear, so the
spline is just the line the points lie on.
= -48 = 4M2. Thus
7. (b) Equations(15) together with the three
equations in part(a) of the exercise
statement give
M2 =-12.
M2-M,
a^ =
6ti
M1
b, = —
'
2
c\ =
= -4, a2 =
M^-Mo
-= 4,
6h
^
h
2
6
h
6(yi -2^2+y3)
i?
M,+4M2+Af3 =
{M2+2M^)h ^
"
6(^2 -2^3+ y4)
M2+4M^+M^ =
>’3-^2 {M^+2M2)h
^2 =
6(y„ ,-2y|+y2)
4Mi+Af2+Af„_, =
M2
= 0, b2= — = -6,
= 0, ci,=0.
6
d2=\
^«-3+4Af„_2+M n-[ -
For 0 < X < .5, we have
5(x)= 5'|(x)= -4x^+3x.
A/,1 +M n-2 +4M n-\ -
6(}'„-3-2y„-2 + y,;-i)
6(y„2-2y„ i+y,)
For .5 < X < 1, we have
5(x)= ^2(x)= 4(x-.5)^ -6(x-.5)^ +1
The linear system for M\, M2,
= 4x^-12x^+9x-l.
The resulting natural spline is
-4x^+3x
5(x)=
0 < X < 0.5
■4
1
0
0
•••
0
0
0
1
M1
1
4
1
0
0
0
0
0
0
1
4
1
■•■
0
0
0
0
M2
M3
0
0
0
0
■■ ■
0
1
4
1
M n -2
1
0
0
0
■■■
0
0
1
4
M «-l
4x^-12x2+9x-1 0.5<x<r
(b) Again h = .5 and
X, = .5, y, = 1
X2 = 1, y2 = 0
X3=1.5,y3=-1
6
^.^^^-6(3'i-2y2 + y3)
y„_l-2y,+y2
y\ -2y2 + 3'3
}'2-2>'3 + )'4
=0
y„_3-2y„_2 + y„_i
Thus Mj =M2 =M3 =0, hence all a, and
. >'«-2-2y„_i+3'i
bi are also 0.
q =
h
in
matrix form is
c2=^i::^
= -2
^
h
d[ = y’l = 1, ^f2 = ^2 = 0
For .5 < X < 1, we have
S(x) = 5] (x) = -2(x - .5) +1 = -2x + 2.
220
Section 10.5
SSM: Elementary Linear Algebra
8. (b) Equations (15) together with the two
equations in part(a) give
2A/| +A/2 =
5
.6<7i -.5^2
-~^2- Solutions are
6
6(y2-y|-hy{)
5
I?
6(y| -2^2+^3)
M,+4M2+A/3 =
thus of the form q = 5 6 . Set
1
5
6
1
M2+4Mj+M^ =
or
= — to obtain q = 'J
II
^ -6
s=
6(^2-2^3+3^4)
II
.1
M„_2+4M„_,+M
6(y„-2-2y„ i+y„)
2. (a)
n
.23
= .2 ; likewise x^^^ ^ .52
.1
A/„_| + 2M
.25
6(3»-i->V,+^yn)
.273
n
and x^^^ = .396
This linear system fox MM2,
in
.331
matrix form is
“2
1
0 0 ••• 0 0 0 0
1
4
1
0
0
1
4
1
1
4
0 0
0 0
0
0
•••
0 0
0
0
•••
0 0 0 0
0 0 0 0
•••
1
4
1
0
0 0 0 0
•••
0
1
4
1
0 0 0 0
•••
0 0
1
(b) P is regular because all of its entries are
positive. To solve (/- P)q = 0, i.e.
M1
M2
M,
M n-2
-.7
0
.6
-.2
<i2
0 , reduce the
-.2
-.5
.9
3/3
0
■ 8 -1 -1
Mn
-hy'\ -y]+y2
6
-.1
-.6
coefficient matrix to row-echelon form:
M n-\
2
■ .8
-6
6
-2
-5
y^-2y2+y2
3'2-2y3 + 3'4
1
0
0
1
2
5
-9
-2 ^ 0
-21
29
0
0
9
0
21
,2
29
21
y„_2-2y„_i+y
0 0
n
>'«-i
0_
This yields solutions (setting
= s) of the
22
Section 10.5
21
form
Exercise Set 10.5
29 5.
21
1
1. (a) x(')=Px(°) =
.4
x(2)=p^(l)^
.6 ’
.46
To obtain a probability vector, take
.54 ■
22
72
21
Continuing in this manner yields
■.454'
x(3) =
x(4) =
5 =
f + f + I 72,
.546 ■
.4546
.5454
yielding q =
29
72
2i
72
and x^^^ =
.45454
3. (a) Solve (/-P)q = 0, i.e.,
.54546 ■
2
3
-2
(b) P is regular because all of the entries of P
are positive. Its steady-state vector q solves
(/ - P)q - 0; that is.
_3lr
4
3
<?l
q
-|
_ 0
0 ■
The only independent equation is
.6 -.5]rg|1_[0'
■5JU2j^ 0 ■
-.6
This yields one independent equation.
221
Chapter 10: Applications of Linear Algebra
2
^<72- yielding q =
SSM: Elementary Linear Algebra
8 i'. Setting
P"x =
9
8
17
5 =— yields q =
Already true for ;? = 1,2. If true for « - I,
17
17
then P" = P(P"“‘x)
H-l
.19
(b) As in (a), solve
(i)
=P
.26 ^7| _[0
.26j^2j“ 0
-.19
Xi
n-l\
i>-m
i.e., .19r/| =.26q2- Solutions have the form
X|+X2
H-l
26
26
19
q= 19 5. Set s =— to get q =
45
•^1
45
19
45
H
2
_I
3
7
1
(c) Again, solve
0
3
L
0
II
1
1
4
1
1
1
3
2
f) ^1
^2
0
<73
0
'-(I
by
4J
echelon form: 0
n—^oo
0
X|+X2
1
I
if X is a
state vector.
4
_i yielding
1
0
Since lim P"x =
reducing the coefficient matrix to row-
"l 0
X| +X2
0 0
(c) The Theorem says that the entries of the
steady state vector should be positive; they
0
0
i
are not for
4
solutions of the form q =
i
s.
3
i
1
k
3
19
12
Set ^ = — to get q =
19
I
5. Let q =
k
. Then
4
19
1
11
k
19
j1 *
k
(P^)i=llPij^j = ZTPij-7
4. (a) Prove by induction that ^[2^ = 0: Already
7=1
true for « = 1. If true for n - 1, we have
7=1 ^
* 7=1
1
k’
since the row sums of P are 1. Thus (Pq),- = qi
p" = />"-'p, so
for all i.
ill)
Pi2 = dr"P12 + P12 '^P22- But
n - »('!-')
P12 “ P12
=0 SO p\'^^ =0+0= 0. Thus,
6. Since P has zeros entries, consider
1 1 i
2
P2 =
no power of P can have all positive entries,
so P is not regular.
4
4
■^1
,
Px =
•^2
4
, SO P is regular. Note that all
4
2
1
1
^Xi+X2
3
Exercise 5 implies q =
t -^1
1
2
the rows of P sum to 1. Since P is 3 x 3,
t-^l
1
P-X =
4
111
1
(b) If X =
4
111
1
etc. We use
1
3
1
3
_^X|+^X| +X2
induction to show
222
Section 10.6
SSM: Elementary Linear Algebra
Section 10.6
7. Let X =[X| X2 be the state vector, with
x^ = probability that John is happy and
X2 = probability that John is sad. The transition
■4
matrix P will be P =
Exercise Set 10.6
1. Note that the matrix has the same number of
2~
rows and columns as the graph has vertices, and
that ones in the matrix correspond to arrows in
the graph.
5
I ^ since the columns
5
3_
must sum to one. We find the steady state vector
for P by solving
I
2
5
3
I
I3
5
0
0
0
92
(a)
10
2
^9l =-92. so q =
3
, so — is the probability that John will
1
(b)
13
13
be happy on a given day.
8. The state vector X = [X| X2 3:3]^ will
(c)
transition matrix, pjj will represent the
proportion of the people in region j who move to
r.90 .15 .10'
region i, yielding P = .05
.75
.05 .
.05
.10
.85
.10
-.15
-.10
9i
0
-.05
.25
-.05
92
0
-.05
-.10
.15
93
0
1
0
13
0
1
4
0
0
7
7
0
13
7
Set s =
— and get
24
7
(b)
13
24
9 =
4
24
13
, i.e., in the long run
or 54-%
24
6
1
L24J
A (
;
2
of the people reside in region L — or 16—%
3
7
in region 2, and —
24
1
0
1
0
0
0
0
or 29—%
6
y
0
I
0
0
0
0
0
1
1
0
0
1
0
0
0
0
0
0
0
0
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
1
1
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
1
0
1
0
2. See the remark in problem 1; we obtain
(a)
First reduce to row echelon form
4
1
1
0
represent the proportion of the population living
in regions 1, 2, and 3, respectively. In the
yielding q =
1
13
10
13
0
3-. Let s = — and get
10
q =
0
0
, i.e..
in region 3.
223
P,
P2
Chapter 10: Applications of Linear Algebra
SSM:Elementary Linear Algebra
1
0
4. (a) A/^M =
0 0 0
1
0 0
0
0
0 0
1
1
0 0
1
2
1
1
2
0 0 0
0
T
(b) The A:th diagonal entry of M M is
5
i.e., the sum of the squares of the
i=i
entries in column k of M. These entries are 1
if family member i influences member k and
0 otherwise.
(c) The y entry of
is the number of
family members who influence both
member i and memberj.
(b) m^i =1, so there is one 1-step connection
from Pj to Pi1
2
1
0
1
1
1
1
1
1
0
1
1
0
1
5. (a) Note that to be contained in a clique, a
vertex must have “two-way” connections
with at least two other vertices. Thus,
1
could not be in a clique, so {fj. Pi, P^] is
and
=
the only possible clique. Inspection shows
that this is indeed a clique,
2 3 2 2'
(b) Not only must a clique vertex have two-way
1
2
1
1
1
2
1
2
but the vertices to which it is connected
1
2
2
1
must share a two-way connection. This
connections to at least two other vertices,
=
consideration eliminates P] and Pi, leaving
So 17^2 = 2 and 111^2 -3 meaning there
{P3, P4, P5} as the only possible clique.
are two 2-step and three 3-step connections
Inspection shows that it is indeed a clique.
from P] to Pi by Theorem 10.6.1. These
are:
(c) The above considerations eliminate /], P^
1-step: P\^ Pi
and P7 from being in a clique. Inspection
2-step: Pj —> P4 —> Pi and
> P3
P2
shows that each of the sets {Pi, P4, P5},
{P4,P6,P8}, {Pi,Pe,P^], {P2,P4^Ps}
and {P4, P5, Pg} satisfy conditions (i) and
3-step: /j —^ Pi —^ /j —^ Pi,
Py —> P^ —> P^ —> Pi, and
P, ^ P4 ^ P3 ^ Pi.
(ii) in the definition of a clique. But note
that P^ can be added to the first set and we
(c) Since rrii^ = 1,
= 1 and
= 2, there
still satisfy the conditions. P^ may not be
are one 1 -step, one 2-step and two 3-step
added, so [Pi, P4, P^, Pg} is a clique,
connections from P^ to P4. These are:
containing all the other possibilities except
1-step: P| -> P4
{P4, P5, fg), which is also a clique.
2-step: /[ —> P^ —> P^
3-step: Pj ^ P2 ^ Pi ^ P4 and
P|
P4 —> P3 -4 P4.
224
Section 10.7
SSM: Elementary Linear Algebra
8. Associating vertex Pj with team A, P2 with
6. (a) With the given M we get
0 1 0 1 o'
1
0
5= 0
1
1
1
0 0
1
1
1
0
0
3
1
3
3
0
3
1
1
1
3
0
1
3
3
5^ =
P5 with E, the game results yield the
following dominance-directed graph:
1 ,
0 0 0
0 0
B,
0 0
1
3
1
1
0
1
1
3
3 0
which has vertex matrix
Since s'-p =0 for all i, there are no cliques
0
in the graph represented by M.
0 0
M= 0
0
1
0
1
1
0
1
0
1
0
0
1
0
1
0
1
1
0
1
0
1
1
(b) Here S =
0
1
0
0 0
1
1
1
7
0
0 0
1
6
0
7
1
6
3
1
7
2
8
1
4
7
1
8
2 7
5 ’
0
6
1
7
2
1
1
0
1
1
1
1
1
M+M^ =
2 3 4 5 2 2_
0
1
1
1
0 0
2
1
1
1
0
1
0
0
0
1
0
1
0
2
1
1
0
2
2
2
Then m'^ =
2
0
0 0 0
0 0
0
0 0
6
0
1
1
0 0
0 0
0
5^ =
0
1
2
0
2
1
0
1
2
2
1
1
0
2
1 .
0
1
1
0
1
1
2
1
2 0
The elements along the main diagonal tell us
that only P3, P4, and
Summing the rows, we get that the power of A is
are members of a
8, of B is 6, of C is 5, of D is 3, and of E is 6.
clique. Since a clique contains at least three
Thus ranking in decreasing order we get A in
first place, B and E tie for second place, C in
fourth place, and D last.
vertices, we must have {P3, P4, P5} as the
only clique.
0 0
1
1.
1
Section 10.7
1
0 0 0
M=
Then
0
1
0
0
1
0 0
Exercise Set 10.7
1
0
2 0
1
0
0
1
1
1. (a) From Equation (2), the expected payoff of
the game is
1
and
=
1
1
1
0 0 0
4
0 0
0
2
1
2
1
0
1
1
1
2
0
1
1
1
0
0
-4
pAq =[l 0 1
6-4
5
-7
3
-8
0
6
1
i
4
I
-2
4
I
m+m'^ =
4
5
2
By summing the rows oi M + M , we get that
the power of fj is 2 -1- 1 -1- 2= 5, the power of
P2 is 3, of P3 is 4, and of P4 is 2.
225
Chapter 10: Applications of Linear Algebra
(b) If player/? uses strategy [Pi
SSM: Elementary Linear Algebra
P2 P3]
(c) Here, ^32 is a saddle point, so optimal
l
0
4
1
4
against player C’s strategy
and
strategies are p* =[0 0 I], q* =
0
his payoff
I
y = ^32 = 2.
4
I
4
(d) Here, 021 is a saddle point, so
9 1
will be pAq = — Pi + - P2 -P3.Since
4j
Uy'
0
p* =[0 1 0 0], q* =
p^, P2, and P3 are nonnegative and add
and
0
up to 1, this is a weighted average of the
V
9
-021 -“2.
numbers —
—, and -1. Clearly this is the
4’ 4
4. (a) Calling the matrix A, the formulas of
largest if p| = P3 =0 and P2 = 1; that is,
p =[0
1
3
Theorem 10.7.2 yield
|
P* = 8j’
0],
i
R
(c) As in (b), if player C uses
[?! <72
27
q* = ® , V = — (A has no saddle points).
^4f against ^ 0 ^ ,
1
we get pAq = -6q^ + 3q2 +^3
(b) Asin(a),p* =[f f]=[f i_.
2‘?4-
Clearly this is minimized over all strategies
q* =
by setting <71 = 1 and ^2 = <73 = <?4 = 0.
That is q =[1 0 0 0]^.
10
60
i
1400
70
= — (Again, A
50
60
60
L6.
has no saddle points).
2. As per the hint, we will construct a 3 x 3 matrix
(c) For this matrix,
is a saddle point, so
with two saddle points, say O]
|
= 033 = 1. Such a
1
matrix is A = 0
1
2
7
2
p* =[l
1
0], q*-
0
, and y = O]1 =3.
0 . Note that
1
(d) This matrix has no saddle points, so, as in
zik
(a), p* = =1
-5
-5
<3|3 = <331 =1 are also saddle points.
-3
3. (a) Calling the matrix A, we see 022
^ saddle
q* =
point, so the optimal strategies are pure,
1
1
5
5j’
3
-5
5
_2
0
L5j
■5
-19
19
-5
5
, and y =
0
namely: p* =[0 1], q* =
, the value of
(e) Again, A has no saddle points, so as in (a).
I
the game is y = <322 = 3.
3.
p* = 13
iO
I3_’ ^
(b) As in (a), <221 is a saddle point, so optimal
strategies are p* = [0
1
0], q* =
0
„*
—
13
11
-29
, and V=
13
13
, th e
5.
Let
a| I = payoff to 7? if the black ace and black two
value of the game is v = a2\ =2.
are played = 3.
fl|2 = payoff to R if the black ace and red three
are played = -4.
(321 = payoff to R if the red four and black two
are played = -6.
022 ~ payoff to R if the red four and red three
226
Section 10.8
SSM: Elementary Linear Algebra
are played = 7.
So, the payoff matrix for the game is
3 -4'
4=
1
which reduces to
0
54
1
0 0
13
^.1*
P2
0
0
78
79
ii
20
0
0
A has no saddle points, so from Theorem 10.7.2,
p* =
P\
79
1_ ■
-6
78
79
0
Solutions are of the form p =
; that is, player R
54
. Let
79
1
20
should play the black ace 65 percent of the time,
and player C should play the black two
55 percent of the time. The value of the game is
78
5 = 79 to obtain p =
54
79
3
20’
that is, player C can expect to collect on
2. (a) By Corollary 10.8.4, this matrix is
productive, since each of its row sums is .9.
the average 15 cents per game.
Section 10.8
(b) By Corollary 10.8.5, this matrix is
productive, since each of its column sums is
Exercise Set 10.8
less than one.
1. (a) Calling the given matrix E, we need to solve
I
(I-E)p= J
2
(c) Try x =
P\ _ 0
3
0 ■
yLP2j
1.9
2
I
1 . Then Cx = .9
, i.e., X > Cx,
.9
1
so this matrix is productive by Theorem
1
1
10.8.3.
This yields —pi = — p2, that is, p = j- 3
2
^
L2
3. Theorem 10.8.2 says there will be one linearly
independent price vector for the matrix E if some
positive power of E is positive. Since E is not
f2~
Sets-2 and get p = ^ .
positive, try E^.
(b) As in (a), solve
1
r.2 .34 .r
I
0
2
2
U-E)p =
I
P\
0
P2
0
P2
0
I
1
2
3
I
e'^= .2 .54 .6 >0
.6
.12
.3
4. The exchange matrix for this arrangement (using
111'
In row-echelon form, this reduces to
“1 0
0
1
0 0
-1
-f P2 =
0
p-i
2
0
6
3
2.
For equilibrium, we must solve (/- E)p = 0.
6
p = 5 I . Set ^ = 6 and get p = 5 .
That is
6
I
i
2
3
4
i
2
_i
3
3
i
L 6
I
P\
0
P2
0
4
I
i
ft
0
3
2J
Row reduction yields solutions of the form
(c) As in (a), solve
.65
-.50
-.30
P\
0
-.25
.80
-.30
P2
0
LftJ
0
-.40
4
1 i i
0
Solutions of this system have the form
(/-£)p =
3
A, B, and C in that order) is 111
3
3
4'
0
P\
-.30
.60
16
1600
P = 11 5. Set 5 =
16
227
and obtain
15
Chapter 10: Applications of Linear Algebra
SSM: Elementary Linear Algebra
(b) The increase in value is the second column
120
P = 100
542
; i.e., the price of tomatoes was
of (/-C)“',
106.67
1
690 . Thus the value of
503
$120, corn was $100, and lettuce was $106.67.
170
the coal-mining operation must increase by
5. Taking the CE, EE, and ME in that order, we
542
form the consumption matrix C, where Cjj = the
503
amount(per consulting dollar) of the i-th
engineer’s services purchased by they'-th
n
7. The i-th column sum of £■ is
To .2 .3]
engineer. Thus, C = .1
7=1
0 .4 .
.3
.4
elements of the i-th column oil-E are the
0
negatives of the elements of E, except for the
We want to solve (/ - C)x = d, where d is the
demand vector, i.e.
" 1
-.2 -.3' ^1
500
-.1
1
-.4
•1^2
700
-.3
-.4
1
L-^3j
600
ii-th, which is 1
1
-.2
-.3
1
-.43877
0
0
1
/-£is l-^cj, =1-1 = 0. Now, (I-Ef has
7=1
zero row sums, so the vector x = [1
^1
500
^2
765.31
-IL'^3 J
1556.19
1
•••
1]
T
solves (/-£)^x = 0. This implies
det(/-£)^ =0. But det(/-£)^ =det(/-£),
so (/- £)p = 0 must have nontrivial (i.e.,
1256.48
nonzero) solutions.
The CE received $1256, the EE received $1448,
1448.12
and the ME received $1556.
Back-substitution yields the solution
X =
So, the /-th column sum of
n
In row-echelon form this reduces to
0
and the
1556.19
8. Let C be a consumption matrix whose column
The CE received $1256, the EE received $1448,
sums are less than one; then the row sums of
and the ME received $1556.
are less than one. By Corollary 10.8.4,
6. (a) The solution of the system (/ - Qx = d is
productive so (/-C^)“’ >0. But
(/-C)-'=(((/-C)^))-')^
= ((/-C^)-‘)^
X = (/ - C)“'d. The effect of increasing the
demand
for the tth industry by one unit
0
>0.
Thus, C is productive.
0
is the same as adding
1
is
to d where the 1
Section 10.9
0
Exercise Set 10.9
0
1. Using Equation (18), we calculate
is in the ith row. The new solution is
0
0
(/-C)
-1
d+
1
= x + (I-C)
0
-1
305
0
Yld2= —
0
Yld^ =
^
1
= 155
2
505 _1005
2 + 1° 7 ■
So all the trees in the second class should be
0
harvested for an optimal yield (since 5 = 1000)
0
J/
of $15,000.
0
which has the effect of adding the ith
column of (/ -C)
-1
to the original so lution.
228
Section 10.10
SSM: Elementary Linear Algebra
2. From the solution to Example 1, we see that for the fifth class to be harvested in the optimal case we must have
P55
(.28“'+.31“'+.25“'+.23“')
>14.7^, yielding P5 > $222.63.
-—=.285. Thus, for all the yields to be the same we must have
-1
3. Assume P2 = 1, then Yld2 =
(.28)
P2S
=.285
I
I
(.28“' +.31“')
P4S
=.285
I
(.28“'+.31“'+.25“')
P55
=.285
1
1
(.28“'+.31“'+.25“'+.23“')
=.285
I
(.28“' +.31“' +.25“' +.23“' +.37“')
Solving these successively yields /J3=1.90, /74=3.02, p^=4.24 and p(,=5.00. Thus the ratio
Pi ■ Pi ■ P4 ■ P5 ■ P(,
n
is the number of trees removed from the forest. Then Equation (7) and
5. Since y is the harvest vector, N =
/=!
the first of Equations(8) yield N = g^x^, and from Equation (17) we obtain
5
8\^
N=
L +...+
1 +1L +...+
S’2
6. Set
S’A-1
S'l
= • • • = g„_| = g, and P2 ='■ T"hen from Equation (18), Yld2 =
P2S
1
= gs. Since we want all of the Yldi^ ’s
S’l
to be the same, we need to solve Yldj^ =
Pk^
= gs for pi^ for 3 < ^ < n. Thus
=k-\. So the ratio
(k-Dj;
{>
Pi ■ P?, -P4- --- -Pn =l:2:3:-..:(n-l).
Section 10.10
Exercise Set 10.10
0
1
1. (a) Using the coordinates of the points as the columns of a matrix we obtain 0 0
0 0 0
229
0
I
0
Chapter 10: Applications of Linear Algebra
SSM:Elementary Linear Algebra
f 0 0
(b) The scaling is accomplished by multiplication of the coordinate matrix on the left by 0 i 0 > resulting in
0
the matrix
0 I- I2 0
/^
0 0 0 0
^
0
f3 1
I
,
0 0--’ which represents the vertices (0, 0, 0), —,0, 0 , —, —,0 and 0, —,0 as
2 2
J
{ 2
shown below.
(c) Adding the matrix
-2
-2
1
-1
3
3
-2
-2
3
3
-2
-1
-1
-2
1
-1
0
0
3
3
3
3
to the original matrix yields
, which represents
the vertices (-2,-1, 3),(-1,-1, 3),(-1, 0, 3), and (—2,0, 3) as shown below.
cos(-30°) -sin(-30°) 0
(d) Multiplying by the matrix
sin(-30°)
cos(-30°) 0 , we obtain
0
0
1
0 cos(-30°) co,s(-30°)-sin(-30°) -sin(-30°)| fO
0 sin(-30°) cos(-30°)+ sin(-30°)
0
0
cos(-30°) = 0
0
0
.866 1.366 .500"
.366
-.500
0 0
0
.866 .
0
_
The vertices are then (0, 0, 0),(.866, -.500, 0),(1.366,.366, 0), and (.500,.866, 0) as shown:
1
2. (a) Simply perform the matrix multiplication
i 0
0
1
0
0
0
1
230
I
AV;
■^1
V
-r,+^y/
3'/
■7 .
Section 10.10
SSM: Elementary Linear Algebra
4. (a) The formulas for scaling, translation, and
rotation yield the matrices
(b) We multiply
i 0
0 1 I i
0
1
1
0
0
1
1
0 0
1
1
0 0
0
0 0
0
0
0
1
0
0
0
0
1
0
'^0 0
M1 - 0
3
■I
-, 1, 0 , and -, 1, 0 .
U
tW2 =
(c) Obtain the vertices via
0 OirO
.6
1
0
0
0
1
1
0 0
1
0
1
1
0
0
I
I
2
2
2
2
0
0
0
0
0
0
0
0
0
0 .6
0 0 0 0
0
0 0 i
yielding the vertices (0,0, 0),(1,0,0),
'l
2
0
i
1
0
1.6
1
0
cos 20°
-sin 20°
0
0
0
sin 20°
cos 20°
yielding (0, 0, 0),(1, .6, 0),(1, 1.6, 0), and
cos(-45°)
(0, 1,0), as shown:
0 sin(-45)°
0
1
0
M4 =
, and
_-sin(-45°) 0 cos(-45°)
'cos 90° -sin 90° O'
M5 = sin 90°
cos90°
0.
0
0
1
(b) Clearly P'= M
3. (a) This transformation looks like scaling by the
factors 1,-1, 1, respectively and indeed its
'1 0 o'
matrix is
0
-1
0
0
+ M2).3
5. (a) As in 4(a),
0
0
= 0 .5 0
0
0
1
0 .
0
0
1
M2= 0 cos45° -sin 45°
0
sin 45°
1
I
cos 45°
(b) For this reflection we want to transform
(x,-, y,-, Zi) to (-X,-, y,-, z,) with the matrix
■-1
0
0
1
0
0
M3
o'
0 .
1
M4 =
Negating the x-coordinates of the 12 points
in view 1 yields the 12 points
(-1.000, -.800, .000), (-.500, -.800, -.866),
1
0
0
0
0
0
0
0
0
cos35°
0
sin 35°
0
1
0
[-sin 35° 0 cos35°
cos(-45°)
etc., as shown:
•••
M5 = sin(-45°)
-sin(-A5°)
0
cos(-45°)
0
0
1
0
0
(c) Here we want to negate the z-coordinates,
'1 0
O'
with the matrix
0
1
0
0
,
0
0
0
Mg = 0 0
0
0 , and
1
1
1
1
2
0
0
M7= 0
1
0
0
0
1
(b) As in 4(b),
0 .
P' = M-j{M^+M^M^{M2+M2M^P)).
-1
This does not change View 1.
231
Chapter 10: Applications of Linear Algebra
SSM: Elementary Linear Algebra
Section 10.11
6. Using the hing given, we have
cosy? 0 siny?
R, =
0
1
0
Exercise Set 10.11
,
-siny? 0 cos/?
coscr
-sin or
??2 = sin or
cos or
0
0,
0
0
sin^
0
1
0
-sin^ 0
^(^2+^3)
t1
1
cosd
/?3 =
1. (a) The discrete mean value property yields the
four equations
0
■'■^4 + * + l)
h~
,
cos^
h =~(^i +^4)
cos(-or) -sin(-or) 0
??4 = sin(-or)
0
cos(-y?)
0
1
Translated into matrix notation, this
sin(-y?)
0
^5 =
^4=^(f2+f3 + l + l).
cos(-or) 0 , and
0
0
0 74 74 0
h
_-sin(-y?) 0 cos(-y?)_
becomes
h
74 0
0 74
h
h
74 0
0 74
h
7. (a) We rewrite the formula for v- as
I-Xi+Xq I
1 • y,- + yo • 1
l-z,+zo-l
V,- =
0
/
V;/ =
0
0
1
0
U
0
2
0
I
2
(/ - M)t = b for t:
1
0
1
+
(b) To solve the system in part (a), we solve
. So
1 1
1
74 74
0
h
X;I
■^0
yo
y-i
I
1
0
1
Zo
2/
4
0
0
0
1
1
0
I
4
1
4
0
I
4
0
t1
1
0
0
i
4
4
i
4
h
I
h
0
/4
2
4
1
0
2
I
In row-echelon form, this is
(b) We want to translate x,- by -5, y,- by -t-9,
Zi by -3, so xq = -5,
1
The matrix is
0
0-5'
0
1
0
9
0
0
1
-3 ■
0
0
0
1
1
1
4
4
0
1
0
0
1
=9, Zq =
0
0
t
-15
4
h
0
1
2
1
h
0
0
1
1
0
1
1
4
I
4
3
8. This can be done most easily performing the
Back substitution yields the result t =
multiplication RR^ and showing that this is /.
3
4
axis we obtain
1
0
0
0
0
0
COS0
-sind
0
COS0
sin0
0
sin^
COS0
0
-sin^
COS0
1
0
o'
0
1
0 .
0
0
1
1
4
For example, for the rotation matrix about the x-
rr'^ =
4
232
Section 10.12
SSM: Elementaty Linear Algebra
-.0371
(c)
^3 was
'o i i 0
0
_i 0 0 i
i 0 0 i
0 i i 0
X100% = -12.9%, and for
.2871
0
0371
1
0
+
0
t2 and
2
X100% = 5.2%.
was -
.7129
0
1
0
2. The average value of the temperature on the
7
circle is
0
l/rr
^ f{6)rd0, where r is the radius
I
of the circle and f{0) is the temperature at the
point of the circumference where the radius to
2
0
that point makes the angle ^with the horizontal.
1
2
n
-K
t(2)=M<'>+b =
<0<— and is zero
2
i
Clearly/(6^ = 1 for ^
.s
otherwise. Consequently, the value of the
integral above (which equals the temperature at
1
the center of the circle) is -.
5
2
■j.‘
3. As in 1(c), but using M and b as in the problem
16
II
statement, we obtain
16
t<'>=A/t(°>+b
.S
16
if
_ri 1 1 1 I 1 1
ii
_4
Ll6j
4
2
4
2
4
4
t^2)=Mt('^+b
7
.32
isf
2.3
t(">=M<'>+b =
16
16
32
16
16
16
2
32
Section 10.12
23
.32.
Exercise Set 10.12
\5
64
1. (c) The linear system
47
64
t'5)=M<">+b =
■^31 =^[28 + X3| -X32]
iS,
64
47
L64j
t
(-“i)
-t =
I.S
1
1
64
4
64
47
3
1
64
4
64
15
I
I
64
4
64
47
1
i
64
4
64
■^32 =^[24 + 3x3, -3x32]
can be rewritten as
19x3, +X32 =28
-.3x31 +23x32 = 24,
which has the solution
*
31
-*^31 -77:
22
27
•*32 -~
22
(d) Using
percentage error
computed value-actual value xl00%
actual value
we have that the percentage error for
2. (a) Setting
and
using part (b) of Exercise 1, we have
233
8
Chapter 10: Applications of Linear Algebra
SSM: Elementary Linear Algebra
I
= —[28]= 1.40000
=^[24]= 1-20000
= —[28-+-1.4-1.2]= 1.41000
A-® =^[24+3(1.4)-3(1.2)]= 1.23000
(3) _
X
3!
=-^[28+1.41-1.23]
= 1.40900
20
X® =^[24+3(1.41)-3(1.23)]= 1.22700
■14) =J-[28+
1.409-1.227] = 1.40910
20
X
31
=^[24+ 3(1.409)-3(1.227)]
= 1.22730
a4-?
=
31
20
[28+ 1.4091-1.2273] = 1.40909
= -!-[24
+ 3(1.4091) - 3(1.2273)]
20
= 1.22727
a(9
=
31
[28 + 1.40909-1.22727]
20
= 1 .40909
= ^[24 + 3(1.40909) - 3(1.22727)]
= 1 .22727.
(b) x(‘>=(l, l) = (40),40))
4l =^[28 + l-l] = l .4
4? =^[24+ 3(1)-3(1)] = 1.2
in this part is the same as x^’^ in part (a), we will get x (2)
3
Since
will also be the same as in part (a),
(c) x^') =(148,-15) = (x(°>,Ag4
=3-[28+
148-(-15)] = 9.55000
4'
20
Ji) 32
20
[24 + 3(148) - 3(-l 5)] = 25.65000
=^[28 + 9.55-25.65] = 0.59500
234
as in part (a) and therefore
..., \
Section 10.12
SSM: Elementary Linear Algebra
=^[24+3(9.55)-3(25.65)]
=-1.21500
3I
[28+0.595+ 1.215]
20
= 1.49050
x® =^[24+3(0.595)+ 3(1.215)]
= 1.47150
= —[28+1.4905 -1.4715]= 1.40095
20
=^[24+ 3(1.4905)-3(1.4715)]
= 1.20285
= —[28+1.40095 -1.20285]
= 1.40991
.20285)]
= 1.22972
ji-(6) = J_
r 28+1.40991 -1.22972]= 1.40901
20
= —[24+ 3(1.40991)-3(1.22972)]
= 1.22703
4. Referring to the figure below and starting with Xq^ =(0, 0):
(1)
(1)
is projected to x on L|,
is projected to x^j^ on L2< *9^ is projected to x^*^ on L3, and so on.
X
I
0
L.
X,-.V2=0
As seen from the graph the points of the limit cycle are x]' = A, X2 = B, X3 = A.
Since xj" is the point of intersection of L^ and L3 it follows on solving the system
X|2 =1
xn-x['2=0
that x*=(l, l). Since X2=(X2|,X22) is on Lj, it follows that X2i-X22=2. Now XjX^ is perpendicular to Lo,
235
Chapter 10: Applications of Linear Algebra
therefore
SSM: Elementary Linear Algebra
(x* -0
— (I)= -l so we have -V22 - I =
or X21 +X22 = 2.
.•^2!
Solving the system
*
*
-
X21 -X22 =2
^2l +A22 -2
gives X21 = 2 and ^22 = 0. Thus the points on the limit cycle are x]" =(I, I), X2 =(2, 0), X3 =(I, I).
7. Let us choose units so that each pixel is one unit wide. Then
= length of the center line of the /-th beam that
lies in they-th pixel.
If the (-th beam crosses the y'-th pixel squarely, it follows that Cjj = I. From Fig. 10.12.11 in the text, it is then
clear that
%7 =^18 =^19 = 1
^^24 = ^25 - <326 = 1
«31 =^32 =«33 =1
“73 = ^76 = ^^79 = 1
^82 = «85 = ^88 = 1
<791 = 094 =((97 = 1
since beams 1, 2, 3, 7, 8, and 9 cross the pixels squarely. Next, the centerlines of beams 5 and 11 lie along the
diagonals of pixels 3, 5, 7 and 1, 5, 9, respectively. Since these diagonals have length V2, we have
053 =055 =057 =x/2 =1.41421
^11. I = “11, 5 = ^11, 9 = V2 = 1.41421.
In the following diagram, the hypotenuse of triangle A is the portion of the centerline of the 10th beam that lies in
the 2nd pixel. The length of this hypotenuse is twice the height of triangle A, which in turn is -Jl-l. Thus,
a
10, 2
= 2(V2-l)=.82843.
By symmetry we also have
^10, 2 = <^10, 6 = ^12, 4 =‘^n,8 = ^62 = <^64 = <^46 = ^48 = -82843.
Also from the diagram, we see that the hypotenuse of triangle B is the portion of the centerline of the 10th beam
that lies in the 3rd pixel. Thus, 0,0 3 = 2-V2 =.58579.
236
Section 10.12
SSM: Elementary Linear Algebra
i
By symmetry we have a|o_ 3 = a^2, l
The remaining
^49 “-58579.
’s are all zero, and so the 12 beam equations (4) are
X7 + Jtg + X9 = 13.00
+JC5 + Xg =15.00
X| + X2 + X3 = 8.00
.82843(x6+X8)+.58579x9 =14.79
I.41421(x3+X5+X7)= 14.31
.82843(x2+x4)+.58579x, =3.81
X3 H-x^ + X9 = 18.00
X2 +X3 +Xg = 12.00
X| + X4 + Xy = 6.00
.82843(x2+X6)-4-.58579x3 = 10.51
1.41421(x, +X5+X9)= 16.13
.82843(x4+X8)+.58579x7 =7-04
8. Let us choose units so that each pixel is one unit wide. Then
= area of the /-th beam that lies in they-th pixel.
Since the width of each beam is also one unit it follows that a^j = 1 if the i-th beam crosses they-th pixel squarely.
From Fig. 10.12.11 in the text, it is then clear that
a^J =a|8 =ai9 =1
£224 - ^25 = «26 ='
a3| =£232 =££33 =1
£273 = £276 = <379 = 1
<^82 =
= ^88 “*
£291 =094 =£297 =1
since beams 1,2, 3, 7, 8, and 9 cross the pixels squarely.
For the remaining £2,y’s, first observe from the figure that an isosceles right triangle of height /?, as indicated, has
'j
area h . From the diagram of the nine pixels, we then have
237
Chapter 10: Applications of Linear Algebra
SSM: Elementary Linear Algebra
-,2
Area of trianale
B=
C?
4(3-2V2)
= 0,4289
I
=.25000
Area of triangle C = —
4
1
-.9
Area of triangle F =
=.38604
We also have
Area of polygon A = l-2x(Ai-ea of triangle B)
2
=.91421
238
Section 10.12
SSM: Elementary Linear Algebra
Area of polygon D = I -(Area of triangle C)
4
3
4
=.7500
Area of polygon E = \-(Area of triangle F)
=^(l8V2-23)
=.61396
Referring back to Fig. 10.12.1 1, we see that
a
I I, I = Area of polygon A =.91421
^10. 1 =
triangle B =.04289
(2| I 2 = Area of triangle C =.25000
flio 2 = Area of polygon D =.75000
fl|0_ 3 = Area of polygon E =.61396
By symmetry we then have
1, I =
1, 5 =
I, 9 = ^53 = “35 = ^51
=.91421
«10, 1 = '^10, 5 - ‘^'10, 9 ='^12, 1 = ^\2, 5 = <^12, 9
= 063 = <353 = £267 ='^43 = <^45
= <347 =.04289
«1 1, 2 =«ll. 4 =«l l, 6 =‘^1 1, 8
= a^2 =
- <^'56 = ‘^58 ~ -25000
‘^10, 2 =<^10,6 ='^12, 4 =<^12,8
=(362 = “64 = “46 = “48 = -75000
“10.3 = “12, 7 = “61 = “49 = -61396.
The remaining a^j’s are all zero, an(J so the 12 beam equations(4) are
X7 +xg + A9 = 13.00
a-4 +X5 + A6 = 15.00
A'l -f A2 + A3 = 8.00
0.04289(a3 + A5 -f A7)+0.75(a6 + Xg)+0.61396x9 =14-79
0.91421(X3 -f X5 + X7)+0.25(x2 + X4 + X6 + Xg) = 14.31
0.04289(x3 + X5 + X7)+0.75(x2 + X4)+0.61396xi =3.81
X3 + X6 + A9 =18.00
X2 + X3 + Xg = 12.00
X| + X4 + X7 = 6.00
0.04289(x, + A5 + Xg)+0.75(x2 + Xg)+0.61396x3 ='0-5'
0.91421(X| + X5 + Xg)-t- 0.25(x2 + X4 + X6 + Xg) = 16.13
0.04289(X| +X5+X9)+0.75(x4+Xg)+0.61396x7 =2.04
239
SSM: Elementary Linear Algebra
Chapter 10: Applications of Linear Algebra
Section 10.13
Exercise Set 10.13
12
1. Each of the subsets S'], S2, S^, ‘S'4 in the figure is congruent to the entire set scaled by a factor of —. Also, the
rotation angles for the four subsets are all 0°. The displacement distances can be determined from the figure to
find the four similitudes that map the entire set onto the four subsets 5|, S2, S^, ^4. These are, respectively.
y
12
I
25
25
1
53
X
/r
“iV
n
I
0
X
t;- : 25L0 iJL)'.
12
Because s =
25
1^// (‘5) -
+
e,-
fi
0
€■
, i[■
= 1, 2, 3, 4, where the four values of
'
fi
are
13
25
0 ’
0
0
13
, and
25
13
25
ii
.25.
and 1: = 4 in the definition of a self-similar set, the Hausdorff dimension of the set is
\n(k} _ ln(4) = 1.888.... The set is a fractal because its Hausdorff
dimension is not an integer.
2. The rough measurements indicated in the figure give an approximate scale factor of s
m
—- = .47
to two
ln(A:)
ln(4)
decimal places . Since lc = 4, the Hausdorff dimension of the set is approximately dj-i (S) =
'"(i)
'
to two significant digits. Examination of the figure reveals rotation angles of 180°, 180°, 0°, and -90° for the sets
5,, 82,82, and S'4, respectively.
240
Section 10.13
SSM: Elementary Linear Algebra
3. By inspection, reading left to right and top to bottom, the triplets are:
(0, 0,0)
(1,0,0)
(2, 0,0)
(3, 0,0)
(0, 0, 1)
(0, 0, 2)
(1,2,0)
(2, 1,3)
(2, 0, 1)
(2, 0, 2)
(2, 2, 0)
(0, 3, 3)
none are rotated
the upper right iteration is rotated 90°
the upper right iteration is rotated 180°
the upper right iteration is rotated 270°
the lower right iteration is rotated 90°
the lower right iteration is rotated 180°
the upper right iteration is rotated 90° and the lower left is rotated 180°
the upper right iteration is rotated 180°, the lower left is rotated 90°, and the lower right is rotated 270°
the upper right iteration is rotated 180° and the lower right is rotated 90°
the upper right and lower right iterations are both rotated 180°
the upper right and lower left iterations are both rotated 180°
the lower left and lower right iterations are both rotated 270°
4. (a) The figure shows the original self-similar set and a decomposition of the set into seven nonoverlapping
congruent subsets, each of which is congruent to the original set scaled by a factor 5 =-. By inspection, the
3
rotations angles are 0° for all seven subsets. The Hausdorff dimension of the set is
\n(k)
ln(7)
'"(i)
l n(3)
d„(S)=
= 1.771.... Because its Hausdorff dimension is not an integer, the set is a fractal.
241
Chapter 10: Applications of Linear Algebra
SSM: Elementary Linear Algebra
I
1
I!"
li
li if!
(b) The figure shows the original self-similar set and a decomposition of the set into three nonoverlapping
congruent subsets, each of which is congruent to the original set scaled by a factor
By inspection, the
rotation angles are 180° for all three subsets. The Hausdorff dimension of the set is
dj-j(S)-
In(L') _ ln(3) = 1 .584..,. Because its Hausdorff dimension is not
an integer, the set is a fractal.
ln(2)
Hi Vi
Hi Vi
'ft
(c) The figure shows the original self-similar set and a decomposition of the set into three nonoverlapping
congruent subsets, each of which is congruent to the original set scaled by a factor
By inspection, the
rotation angles are 180°, 180°, and-90° for ^i, S2, and ^3, respectively. The Hausdorff dimension of the
set is di-iiS)=
\n{k) _ ln(3) = 1.584.... Because its Hausdorff dimension is not
an integer, the set is a fractal.
ln(|)~ln(2)
S3
5,
242
Section 10.13
SSM: Elementary Linear Algebra
(d) The figure shows the original self-similar set and a decomposition of the set into three nonoverlapping
congruent subsets, each of which is congruent to the original set scaled by a factor
By inspection, the
rotation angles are 180°, 180°, and -90° for 5,, iS'2, and ^3, respectively. The Hausdorff dimension of the
set is dfi(S)=
ln(/:) _ ln(3) = 1 .584.... Because its Hausdorff dimension is not an integer, the set is a fractal.
'"(i)
ln(2)
^3
S2
*^1
V
s, ^
4F
1
4^1
5. The matrix of the affine transformation in question is
cos^
-sin (9
sin 0
CQsd
form 5
.85
.04
-.04
.85
. The matrix portion of a similitude is of the
. Consequently, we must have 5 cos $= .85 and
s sin 6=
-.04. Solving this pair of equations gives 5 =
0= tan
if-.04
-I-(-.04)^ =.8509... and
= -2.69...°.
.85
n\
X
be the vector to the tip of the fern and using the hint, we have
6. Letting
.85
= 72
or
3'
3',
X
ffx
X
.04
.x
-.04 .85j[y
.075
4-
.180
. Solving this matrix equation gives
=
3'
.15
-.04
.075
.766
.04
.15
.180
.996
rounded to three decimal places.
7. As the figure indicates, the unit square can be expressed as the union of 16 nonoverlapping congruent squares.
I
each of side length —. Consequently, the Hausdorff dimension of the unit square as given by Equation (2) of the
text is dH(S)=
Injlc) _ ln(16)
Wi)
ln(4)
243
Chapter 10: Applications of Linear Algebra
SSM: Elementary Linear Algebra
I
4
8. The similitude T^ maps the unit square (whose vertices are (0, 0),(1, 0),(1, 1), and (0, 1)) onto the square whose
3 3
vertices are (0, 0), -.0 ,
— , and
4 j U 4
1
A
f
-.0 , (1,0), L- , and
vertices are
V
N
3
0, — . The similitude T2 maps the unit square onto the square whose
1
3
— . The similitude T-^ maps the unit square onto the square whose
4’ 4
/
1
3 \]
vertices are 0, — ,
I 4
4’ 4/
3
—, 1 , and (0, 1). Finally the similitude
maps the unit square onto the square
4
\
c
1
1
I \
T,
- , 1, - , (1, 1), and
1 . Each of these four smaller squares has side length of -,
4’ 4/ \ 4J
,4 J
4
1
whose vertices are
3
so that the common scale factor of the similitudes is s= —. The right-hand side of Equation (2)of the text gives
4
In(^) _ ln(4) = 4.818.... This is not the correct Hausdorff dimension
of the square (which is 2) because the four
WfW)
smaller squares overlap.
1
9. Because ^ = —
ln(/:) _ ln(8) = 3
and ^ = 8, Equation (2) of the text gives dfj(S)=
for the Hausdorff dimension
h^“i^
of a unit cube. Because the Hausdorff dimension of the cube is the same as its topological dimension, the cube is
not a fractal.
10. A careful examination of Figure Ex-10 in the text shows that the Menger sponge can be expressed as the union of
20 smaller nonoverlapping congruent Menger sponges each of side length Consequently, k = 20 and •5' =^and so the Hausdorff dimension of the Menger sponge is d[^{S)=
In(L')
ln(20)
Mi)
ln(3)
= 2.726.... Because its
Hausdorff dimension is not an integer, the Menger sponge is a fractal.
11. The figure shows the first four iterates as determined by Algorithm 1 and starting with the unit square as the
initial set. Because k = 2 and ^ = —, the Hausdorff dimension of the Cantor set is
3
dfj{S)=
\n{k) _ ln(2) = 0.6309.... Notice that the Cantor set is a subset
of the unit interval along the x-axis and
ln(i)“ln(3)
that its topological dimension must be 0(since the topological dimension of any set is a nonnegative integer less
than or equal to its Hausdorff dimension).
244
Section 10.14
SSM: Elementary Linear Algebra
Initial set
First Iterate
Second Iterate
Third Iterate
Fourth Iterate
12. The area of the unit square Sq is, of course, 1. Each of the eight similitudes Tj, 72,..., 7g given in Equation (8)of
the text has scale factor
^nd so each maps the unit square onto a smaller square of area ^
. Because these
g
eight smaller squares are nonoverlapping, their total area is -, which is then the area of the set
. By a similar
g
argument, the area of the set S2 is --th the area of the set S']. Continuing the argument further, we find that the
8 f
areas of Sq, 5,, S2, S^, S^,
form the geometric sequence 1, -, -
9 V 9>
/r.\3
/r.\4
8
9J
9J
, .... (Notice that this
fsr
implies that the area of the Sierpinski carpet is 0, since the limit of — as n tends to infinity is 0.)
v9y
Section 10.14
Exercise Set 10.14
1. Because 250= 2• 5‘^ it follows from (i) that n(250)= 3-250 = 750.
Because 25 = 5^ it follows from (ii) that 11(25) = 2• 25 = 50.
Because 125 = 5^ it follows from (li) that 0(125) = 2-125 = 250.
Because 30 = 6 - 5 it follows from (ii) that 11(30)= 2 - 30 = 60.
Because 10 = 2 - 5 it follows from (i) that 11(10)= 3-10 = 30.
245
Chapter 10: Elementary Linear Algebra
SSM: Elementary Linear Algebra
Because 50 = 2-5^ it follows from (i) that n(50) = 3-50 = 150.
Because 3750= 6-5'^ it follows from (ii) that 0(3750)= 2•3750 = 7500.
Because 6 = 6-5° it follows from (ii) that 0(6)= 2 •6 = 12
Because 5 = 5' it follows from (ii) that 0(5)= 2-5 = 10.
2. The point (0, 0) is obviously a 1-cycle. We now choose another of the 36 points of the form — — , saw
6 6J
0, — . Its iterates produce the 12-cycle
6)
I
0
i ^ 9
6
6
3
6
0
1
i
6
, 6
6
1
I
4
5
6
6
6
6
5
0
3
4
6
—>
3
6
1
6
1
4
i
1
2
I
6
6
6
6
6
6
2
So far we have accounted for 13 of the 36 points. Taking one of the remaining points, say 0, —
V
foi ri] roi rf
the 4-cycle 62 ^ 4
6
in
4
°
6
6
, we arrive at
6,
continue in this way, each time starting with some point of the form
n
6’ 6
that has not yet appeared in a cycle, until we exhaust all such points. This yields a 3-cycle;
3
4
4
2
6 ^ 6 ^ 3 ; another 4-cycle: 6
6
6
3
0
oj i
i
I
6 ^ 6
0
—>
2
1
6
6
3
2
3
1
6
6
6
.3
4
5
s-
6
6
2 ; and another 12-cycle:
6
6
6
6
0
4
6
1
5
4
6
6
6
0
0
6
")
5
I
6
-4
6
4
3
6
6
9
6
5
6
in
The possible periods of points for the form
n
U 6
are thus 1,3,4 and 12. The least common multiple of these
four numbers is 12, and so 0(6)= 12.
246
Section 10.14
SSM: Elementary Linear Algebra
3. (a) We are given that Xg = 3 and
X2 = ^1 + ^0
modl5 = 7+3
= 7. With p = 15 we have
mod 15 = lOmod 15 = 10,
X3 =X2+Xi
modl5 = 10+7 mod 15 = 17mod 15 = 2,
M = x^ + X2
mod 15 = 2+10 mod 15 = 12mod 15 = 12
X^ = ^4 + X3
modl5 = 12 + 2 mod 15 = 14mod 15 = 14,
X(, =X5+X4
mod 15 = 14 +12 mod 15 = 26 mod 15 = 1 1,
Xj =X6+X5
mod 15 = 11 +14 modl5 = 25modl5 = 10
•^8 = ■^'7 + ^6
mod 15 = 10 + 1 1 mod 15 = 21 mod 15 = 6,
Xg =X8+X7
mod 15 = 6 +10 mod 15 = 16mod 15 = 1,
X|o =X9+X8
mod 15 = 1 + 6
mod 15 = 7 mod 15 = 7,
■^'l l -■^'10+-^
mod 15 = 7 + 1
mod 15 = 8 mod 15 = 8
X|2
X|3
X|4
X15
X|5
= Xj I +X|Q mod 15 = 8 + 7
= X|2 +X| I mod 15 = 0 + 8
= X|3 +X12 mod 15 = 8 + 0
= X|4 +X|3 mod 15 = 8 + 8
= X|5 +X|4 mod 15 = 1 + 8
X|7
X|8
X|9
X20
X21
X22
X23
X24
X25
X25
X27
X28
X29
X30
= X|g +X|5 mod 15 = 9 + 1
= X|7 +Xi5 mod 15 = 10 + 9
= X|g +Xi7 mod 15 = 4+10
= ^19 +^18 mod 15 = 14 + 4
= X20 +X19 mod 15 = 3+14
=^21+ X20 mod 15 = 2 + 3
= X22 + X21 mod 15 = 5 + 2
= X23 + X22 mod 15 = 7 + 5
= X24 + X23 mod 15 = 12 + 7
~ ''-2“) ^24 mod 15 = 4 +12
= X26 + X25 mod 15 = 1 + 4
= X27 + X26 mod 15 = 5 + 1
= X28 + X27 mod 15 = 6 + 5
= X29 + X28 mod 15 = 11 + 6
mod 15 = 15 mod 15 = 0,
mod 15 = 8 mod 15 = 8,
mod 15 = 8 mod 15 = 8,
mod 15 = 16mod 15 = 1,
mod 15 = 9 mod 15 = 9,
mod 15 = 10 mod 15 = 10,
mod 15 = 19 mod 15 = 4,
mod 15 = 14 mod 15 = 14,
modl5 = 18modl5 = 3,
mod 15 = 17 mod 15 = 2,
mod 15 = 5 mod 15 = 5,
mod 15 = 7 mod 15 = 7,
mod 15 = 12 mod 15 = 12,
mod 15 = 19 mod 15 = 4,
modl5 = 16modl5 = l,
mod 15 = 5 mod 15 = 5,
mod 15 = 6 mod 15 = 6,
mod 15 = 11 mod 15 = 1 1,
modl5 = 17 modl5 = 2.
= X30 + X29 mod 15 = 2 + 11 modl5 = 13modl5 = 13,
= -^31 + X30 mod 15 = 13 + 2 mod 15 = 15 mod 15 = 0,
= X32 +X31 mod 15 = 0+13 mod 15 = 13 mod 15 = 13,
= X33 + X32 mod 15 = 13 + 0 modl5 = 13modl5 = 13.
= X34 +X33 mod 15 = 13 + 13mod 15 = 26mod 15 = 1 1,
= X35 + X34 mod 15 = 11 +13mod 15 = 24mod 15 = 9,
= X3g + X35 mod 15 = 9 + 11 mod 15 = 20 mod 15 = 5,
= X37 +X3g mod 15 = 5 + 9 mod 15 = 14 mod 15 = 14,
= X38 + X37 mod 15 = 14 + 5 mod 15 = 19 mod 15 = 4,
= X39 + X38 mod 15 = 4 + 14 mod 15 = 18 mod 15 = 3,
= X4Q+X39 mod 15 = 3 + 4 mod 15 = 7 mod 15 = 7,
and finally; X4Q = Xq and X41 = X|. Thus this sequence is periodic with period 40.
X31
X32
X33
•^34
X35
X36
X37
X38
X39
X40
•^41
247
Chapter 10: Elementary Linear Algebra
SSM:Elementary Linear Algebra
(b) Step (ii) of the algorithm is
^n+\ =Xn+Xn-l ^od p.
Replacing n in this formula by n + 1 gives
Xn+2 =x„+^+x„modp
•^12
_ 1
■^13
1
1
4
2
6
mod 21
10
mod 21
16
=(x„+x„_l)+ x,^mod p
=
+ a:„_i mod p.
These equations can be written as
=x„_^+x„modp
^n+2 = ^n-l +
mod p
10
16
Xi4
10
_
1
,-*15
2
26
which in matrix form are
mod 21
42
1
1
_^n+2
Xn-\
2
X
mod p.
5
n
0
Xq _ 5
(c) Beginning with
5
, we obtain
'^16
_ 1
1
5
.•^17
1
2
0
5
^2
_ 1
1
5
^3
1
2
5
mod 21
16
mod 21
mod 21
5
mod 21
5
10
5
mod 21
15
10
and we see that
15
X4
1
1
10
^5
1
2
15
25
101
C
\L
4
C2
19
0
1
4
Xj
2
19
23
1
2
0
i
\L
mod 21
/r
0
1
0
\L
/r
^8
_ 1
1
2
X9
1
2
0
C5
mod 21
0
1
2
J/
101
2
1
-iN
1
101
1
2
3
101
J/
-IN
1
1
1
2
5
1
1
1
2
J/
13
5.
101
8.
2
■*11
1
2
2
4
21
LioiJ
34
11
101
mod 1 =
21
101
101
55
101
1
, 0
101
be greater than 5.
2
101
mod 1 =
period of the periodic point
2
1
8
101
LioiJ
Jy
101
modi =
Because all five iterates are different, the
mod 21
1
3
101
2
1
2
•^10
101
modi =
101
i -|V
0
i
0
i
101
V _
101
modi =
101
101
2
I
i
101
J/
101
42
2
1
101
/r
mod 21
1
rr_L
\L
1
X,
-IN
fr i
mod 21
_ 1
_ -^0
4. (c) We have that
mod 21
40
^6
•^16
Xi7
mod 21
5. If 0 < X < 1 and 0 < y < 1, then
7(x, y)= x-\
mod 21
6
-r2,
4
. f
T (x, y) = xH
6
248
5
12
^
,y modi, andso
10
12
y
)
, y modi.
must
Section 10.14
SSM: Elementary Linear Algebra
/
1C
and so by part (ii) of the definition, this is
\
r\x, >■)= JC+ —,
y modi, ...,
12
not the matrix of an Anosov automorphism.
T'^(x,y)= x + —, >’ modi
\
12
(c) The eigenvalues of the matrix
>
= (x+5, y) mod 1 = (x, y).
0
1
-1
0
are
±/; both of which have magnitude 1. By part
(iii) of the definition, this cannot be the
matrix of an Anosov automorphism.
Thus every point in S returns to its original
position after 12 iterations and so every point in
5 is a periodic point with period at most 12.
X
Starting with an arbitrary point
Because every point is a periodic point, no point
in the
can have a dense set of iterates, and so the
interior of S, (that is, with 0 < x < 1 and
mapping cannot be chaotic.
0 < y < 1) we obtain
0
6. (a) The matrix of Arnold’s cat map
’ [1 2
, IS
-1
one in which (i) the entries are all integers,
(ii) the determinant is 1, and (iii) the
mod 1 = ^ mod 1 = . ^
l -x ’
0. y
0
1
y
-1
0
l-x
modi =
l-JC
mod 1
-y
Ti + yfS
l-x
eigenvalues, —-— = 2.6180... and
1-y
3-a/5
0
^—-— = 0.3819..., do not have magnitude
1
l-x
0j[l-y
1. The three conditions of an Anosov
1-y
X
0
0
1
are
1
-I
0
3
2
1
1
center point ^ of S. Consequently, each
Consequently, this is the matrix of an
Anosov automorphism.
9
point in the interior of S has period 4 with
0
are
0
1
the exception of the center point ^ which
both equal to 1, and so both have magnitude
1. By part (iii) of the definition, this is not
the matrix of an Anosov automorphism.
5
1
2
3
■7
is a fixed point.
For points not in the interior of S, we first
0
are
observe that the origin
integers; its determinant is 1; and neither of
2
5
2
is a fixed poin t.
X
point of the form
Consequently, this is the matrix of an
Anosov automorphism.
6
0
which can easily be verified. Starting with a
its eigenvalues, 4±Vf5, has magnitude 1.
The determinant of the matrix
mod I
i
its eigenvalues, 2±V3, has magnitude 1.
The entries of the matrix
X
-1 + y
X
Thus every point in the interior of 5 is a
periodic point with period at most 4. The
geometric effect of this transformation, as
seen by the iterates, is to rotate each point in
the interior of S clockwise by 90° about the
integers, its determinant is 1; and neither of
1
1-y mod 1 =
y
are
The eigenvalues of the matrix
0
X
±1, both of which have magnitude 1. By
part (iii) of the definition of an Anosov
automorphism, this matrix is not the matrix
of an Anosov automorphism.
The entries of the matrix
mod 1
-1 + x
automorphism are thus satisfied.
(b) The eigenvalues of the matrix
1-y
modi =
0
with 0 < x < 1, we
obtain a 4-cycle
is 2,
X
0
l-x
0
l-x
0
X
—>
0
249
^ if
0
Chapter 10: Elementary Linear Algebra
A'
SSM: Elementary Linear Algebra
, otherwise we obtain the 2-cycle
2JL>’2
i
0
b
0
I
0
1
1
0
a
-t-
2
2
A,
0
2JL3'2
i
a
1
+
Similarly, starting with a point of the form
1
0
b
•^3
+
b
2JL3’3
with 0 < y < 1, we obtain a 4-cycle
1
a
y
0
—>
y
0
1-y
0
1-v.
0
T.
-4
0
2
Q
1
i
0
2
.
2
-1
-1
a
1
b
this inverse and set
. Notice
d
0
2
2
We multiply the above three matrix equations by
y ^ , otherwise we obtain the 2-cycle
i
1
1
IS
if
1
0
1
The inverse of the matrix
that c and d must be integers. This leads to
. Thus every point not in
0
c
c
0
d
d ’
2
>’i
the interior of S is a periodic point with 1,2,
or 4. Finally because no point in S can have
i
2
At
0
C
2
a dense set of iterates, it follows that the
I
d
T
c
d ’
mapping cannot be chaotic.
c
9. As per the hint, we wish to find the regions in S
that map onto the four indicated regions in the
c
0
d
1
d ■
The only integer values of c and d that will give
figure below.
values of a,- and y,- in the interval [0, 1] are
I
(i,i)
(0, 1)
-1
3’3
c = d = b. This then gives a = b = 0 and the
(1, 1)
1
A
mapping
1
X
1 2
maps the three
y
points (0, 0), 0, — , and (1, 0) to the three
V
2)
[\
points (0,0), —, 1 , and (1, 1), respectively.
U .
1
The three points (0, 0), 0,
I
2
, and (1,0) define
the triangular region labeled I in the diagram
(0,0)
(1,0)
(i,0)
below, which then maps onto the region T.
(0, 1)
We first consider region T with vertices (0, 0),
fl
2’
(1, 1)
1 , and(l, 1). We .seek points (A|, y|).
(a2, y2), and (A3, y3), with entries that lie in
1
[0, 1], that map onto these three points under the
A
1
3’
1
mapping
A
2
3’
(o,i)
(i,i)
fb for certain
integer values of a and b to be determined. This
leads to the three equations
(0, 0)
(1,0)
For region 11', the calculations are as follows:
250
Section 10.14
SSM: Elementary Linear Algebra
I
1
a
b
2JL>’lJ
2
0
a
X2
2
i
-
c
c
>’2
d ’
c
1
c
d
0
d ’
c
2
d
-1
2
0
>’3
i
2
2
iirx
1 2j[y.
vertices
0
-1
2
-1
0
c
c
-1
1
0
d
d ’
X2
2
-1
1
c
yi
-1
1
1
d
1
c
I
d
■J 0
X
1
llfx
3’ ,
1
2
d ■
2
-1
maps region IV with
-2
y
1,—
onto region
22
IV'.
0,— , (0, 1), and (1,0) onto region
22
12. As per the hint, we want to fi nd all solutions of
r
the matrix equation
X|
a
0
b
0 ’
+
>’i
X-,
nonnegative integers. This equation can be
2
+
rewritten as
b
2J 23
a
0
b
1
1 3
Xg ^
r
, which has the
4Jl_yg_
s
3r-5
-4r + 35
solution xg =
’
2-1
0
c
c
1
0
d
d '
>’i
,3'oJ [2 5j[yoJ
where 0 < xq < 1, 0 < yg < L and r and s are
I
a
2JL>’2_
_
c
d'
0
vertices (0, 1), (1, 1), and
maps region II with
+
0
+
follows;
X]
2
Only c = 0 and d = -\ will work. This leads to
a = -[, b = -2 and the mapping
C
d ■
ir. For region III', the calculations are as
2
I
a
+
.>’3]
1
v
1 ’
b
a-3 ^ 2 -1 ^ _ c
Only c = 1 and d = -\ will work. This leads to
a = 0,b = -\ and the mapping
1
X3
y\
1
d
J 0
y\
X
1
•^'1" _
1
0’
2
•^3
I
+
b
b
2JL>’3j
X,
>'2
a
X2
b
>2
_
1
2
+
a
^1
I
1
2
+
and yg =
5
. First
5
trying r = 0 and x = 0, 1,2, ..., then r = 1 and
^ = 0, 1,2, ..., etc., we find that the only values
of r and x that yield values of Xg and yg lying
I
2
At
c
2
d
>’2
X3 _
2
-1
-1
>'3
0
c
1
d
0
C
i
L2j
in [0, 1) are;
d ’
2
r = 1 and s = 2, which give xg = — and yg = -;
1
I
I
/• = 2 and 5 = 3, which give Xg = — and yg = —;
2
4
r=2 and 5 = 4, which give
~^’
maps region III with
+
5’
5
X
1 2jLy
3
1
d ■
Only c = -1 and d = 0 will work. This leads to
a = -\,b = -\ and the mapping
X
5’
5
c
3
r
4
= 3 and 5 = 5, which give xg =— and yg = —.
5
vertices (1, 0),
1,— , and (0, 1) onto region
’ 2
W e can then check that
(2 1
III'.
For region IV', the calculations are as follows;
1
and
5’5
1
X|
2
3’1
a
0
b
0 ’
form one 2-cycle and
another 2-cycle.
251
1
3
b’ 5
and
(2 4^
5’ 5
4
2
5’ 5
form
Chapter 10: Elementary Linear Algebra
SSM: Elementary Linear Algebra
14. Begin with a 101 x 101 array of white pixels and
add the letter ‘A’ in black pixels to it. Apply the
2. (a) For A ==
mapping T to this image, which will scatter the
black pixels throughout the image. Then
superimpose the letter ‘B’ in black pixels onto
this image. Apply the mapping T again and then
superimpose the letter ‘C’ in black pixels onto
the resulting image. Repeat this procedure with
the letter ‘D’ and ‘E’. The next application of the
mapping will return you to the letter ‘A’ with the
pixels for the letters ‘B’ through ‘E’ scattered in
the background.
9
1
7
2’
det(A)= 18-7= 11,
which is not divisible by 2 or 13. Therefore
by Corollary 10.15.3, A is invertible. From
Eqation (2):
2
-1
-7
9
-I
A-' =(11)
38
-19
-133
171
Section 10.15
12
23
Exercise Set 10.15
3 (mod 26).
15
Checking:
1. First group the plaintext into pairs, add the
dummy letter T, and get the numerial equivalents
9
-1
AA
4 1
RK
18
NI
11
14 9
GH
130
20 20
1
3
(a) For the enciphering matrix A = 2 1 ’
2 1
1
3
G
1 ~ 9
18
51
1
11
47
21
U
1
3
14
41
15
O
2
1
9
37
11
K
7
31
2
3
1
8
22
5
22
_2 lj[20j [60,
(b) For A =
E
9
1
23 15j[7 2
53
1
0
0
1
(mod 26).
3
1
5
3
, det(A)= 9-5=4, which
10.15.3, A is not invertible.
H
4
(b) For the enciphering matrix A = 1
8
"4 3]r4]^ri9
3
18
105
1
A
1
2
11
40
14
N
4
3
14
83
5
E
1
2
9
32
6
F
4
3
7
52
0
Z
1
2
8
23
23
VT
4 3 20 ^ 140 ^ 10
9’
is not divisible by 2 or 13. Therefore by
Corollary 10.15.3, A is invertible. From (2):
-I
A-‘ =(9)
E
4
11
det(A)= 72 - 11 = 61 = 9(mod 26), which
2’
5
6
1
3
reducing everything mod 26, we have
1
7
is divisible by 2. Therefore by Corollary
(c) For A =
2
(mod 26)
V
The Hill cipher is GIYUOKEVBH.
1
1
312
B
1 3 20 ^ 80 _ 2
0
Y
2
1
1
0
157 26"
I
25
79
12
-1
A"‘A =
reducing everything mod 26, we have
“1 3ir4]_r7'
7
131 78'
TT
7 8
12
7 2j[23 15
from Table 1.
DA
1
=3r-19
27
-3
1
23
J
Checking:
1 2j[20j |_ 60j [ sj H
The Hill cipher is SFANEFZWJH.
252
9 -if
8
-11
-33
24
19"
24
(mod 26).
Section 10.15
SSM: Elementary Linear Algebra
AA
8
-I
11
1
19
1
-I
15
12
1 9Jl23 24
261
416'
183 52'
208
235
1
0
0
1
1
-1
78
(mod 26)
19
8
11
208
1
0
0
1
2
(d) For A =
0
1
(mod 26)
12
1
182'
27 156'
469
26
(mod 26).
det(A)= 14 - 1 = 13,
7 ’
19
which is divisible by 13. Therefore by
Corollary 10.15.4, A is not invertible.
183
1
0
0
1
(mod 26).
1
11
14
15 24
1
15
10 24
4
Now we have to find the inverse of A =
3
r
3 2 ■
Since det(A) = 8 - 3 = 5, we have by (2);
det(A) = 6-6 = 0, so that
6 2j’
8
3. From Table 1 the numerical equivalent of this
ciphertext is
1
1
1
0
21 5j[l 3
23 24j[l 9
27
27
15
-1
A~'A =
A“'A =
(e) For A =
8
1 3J1_2I 5
2
-I
-(5)
-3
A is not invertible by Corollary 10.15.4.
4
2
= 21
(f) For A =
1
3’
-3
det(A)= 3-8=-5 = 21
42 -21'
(mod 26), which is not divisible by 2 or 13.
Therefore by Corollary 10.15.4, A is
" -63
84
^'16 5'
invertible. From (2):
-1
4
15
6
(mod 26).
3 -8'
-1
A~‘ =(21)
To obtain the plaintext, multiply each ciphertext
vector by A“' and reduce the results modulo 26.
3
=5
'16 5jl9l [309] [2.31
15
-40
-5
5
15
12“
21
5
5
E
246
12
L
249
15
O
'16 5iri5'
360
22
F
15 6j[24
369
5
E
6
16
5
11
15 6jLl4_
(mod 26).
Checking:
W
291
15
91
13
M
105
1
A
"16 5]r r
15 6jl_15_
'16 5]ri0'
_15 6j[24
280
294
20
T
H
The plaintext is thus WE LOVE MATH.
4. From Table 1 the numerical equivalent of the
known plaintext is
AR
MY
13 25
and the numerical equivalent of the
corresponding ciphertext is
253
Chapter 10: Elementary Linear Algebra
SL
SSM: Elementary Linear Algebra
HK
l
Since det(A“‘)= 35-90 = -55 = 23(mod 26),
19 12
A=(A-')-'
so the corresponding plaintext and ciphertext
vectors are
5
-15
-6
7
-1
Pi =
1
19
18
OC|= 12
= 23
= 17 ^
-15
-6
7
13
^ Co =
P2 =
25
1 1
T
19
We want to reduce C = *'!
-255
119
12
II
C7
85
-102
to / by
elementary row operations and simultaneously
7
5
2
15
(mod 26)
is the enciphering matrix.
T
apply these operations to P =
Pi
T
P2
13
5. From Table 1 the numerical equivalent of the
known plaintext is
25 ■
AT
The calculations are as follows:
'19 12
1
1
18
Form the matrix
8
11
13
11
II
JY
8
1 1
11
Replace 132 and 198 by
25
10
1
Pi =
16
13
15
vectors are:
by 19“' =1 1 (mod 26).
2
17
The corresponding plaintext and ciphertext
Multiply the first row
25
13
QO
10 25
198
13
15
and the numerical equivalent of the
corresponding ciphertext is
25
[C I P].
132
OM
20
^c
20
I -
15
P2 =
their residues modulo
13
25
17
^C2
15
26.
1
2
-8 times the first row to
0
-5
-75
10
25
17
15
We want to reduce C =
16
1 1
-103
elementary row operations and simultaneously
the second.
0
2
11
16
21
3
1
Replace the entries in
0
16
1
15
5
-19
6
1
15
5
13 ■
20
Form the matrix
17
15
15
13
Multiply the second row
[C I P].
by 21“' =5(mod 26).
0
20'
'10 25
residues modulo 26.
II
1
apply these operations to P = 15
The calculations are as follows:
the second row by their
2
27
40
16
33
17
15
15
13
Add the second row to
-I
Add -2 times the
0
the first (since 10
does not exist mod 26).
second row to the first.
0
0
1
7
I
14
16
7
17
15
15
13
6
15
Replace -19 by its
5
ur
matrix is A
-1
7
15
^ so the deciphering
7
15
6
Replace the entries in
the first row by their
residue modulo 26.
Thus (/!“')
to / by
residues modulo 26.
0
14
16
7
-223
-257
-106
Add -17 times the first
^ (mod 26).
row to the second.
254
Section 10.15
SSM: Elementary Linear Algebra
14
0
16
3
11
which yields the message THEY SPLIT THE
7
Replace the entries in
24
ATOM.
the second row by their
6. Since we want a Hill 3-cipher, we will group the
letters in triples. From Table 1 the numerical
equivalents of the known plaintext are
residues modulo 26.
1
14
16
7
0
1
57
456
Multiply the second row
I H A
9 8
by 11 '= 19 (mod 26).
1
14
16
7
0
1
5
14
V
1
E C
22 5
3
O
M
E
15
13
5
and the numerical equivalent of the
corresponding ciphertext are
Replace the entries in
HPAFQGGDU
8
the second row by their
16
1
6 17
7
7
4
21
The corresponding plaintext and ciphertext
residues modulo 26.
vectors are
1
0
-54
-189
Add -14 times the
0
I
5
second row to the first.
0
0
24
19
5
14
1
9
8
p, = 8
C| = 16
14
1
Replace -54 and -189
5 OC2 = 17
P2 =
by their residues
3
modulo 26.
24
19
5
14
Thus (A“‘)
, and so the
24
-I
deciphering matrix is A
GI
HG
7
15
7
P3 = 13
C3 = 4
5
21
5
16
19 14_’
We want to reduce C = 6 17
From Table 1 the numerical equivalent of the
given ciphertext is
LN
6
22
YB
VR
EN
12 14 7 9 8 7 25 2 22 18 5 14
_7
JY
7 to / by
4 21_
elementary row operations and simultaneously
10 25
9
apply these operations to P =
QO
22
5
3
15
13
5
17 15
The calculations are as follows:
To obtain the plaintext pairs, we multiply each
ciphertext vector by A~^:
“24 5]ri2
358
20
19 I4j[l4
424
'24 5 [7'
213
5
259
8
16
1
9
8
1'
6
17
7
22
5
3
7
4
21
15
13
5
Form the matrix
T
H
[C I P],
E
19 14 [9
"24
5 [8
19 14 7
25
Y
15
20
22
24
21
221
19
S
6
17
1
22
5
3
250
16
P
7
4
21
15
13
5
24
5
25
610
12
L
Add the third row
19
14
2
503
9
j (mod 26)
to the first since 8
24
5
22
618
20
T
does not exist modulo
19
14
670
20
T
24
19
5
14
5
14
291
24
5
10
365
19
14
25
540
24
5
17
483
19
14
15
533
-1
26.
H
190
5
6
E
1
140
154
168
147
6
\1
1
22
5
42
3
7
4
21
15
13
5
A
T
Multiply the first row
15
O
by 15“' =7 (mod 26).
13
M
20
255
Chapter 10: Elementary Linear Algebra
1
10
24
12
17
16
6
17
7
22
5
3
7
4
21
15
13
5
SSM:Elementary Linear Algebra
Add -5 times the third
row to the second.
1
0 0
4
15
1
0
9
17
0
0 0
1
15
6
17
0
Replace the entries in
the first row by their
24
residues modulo 26.
1
10
24
12
17
0
-43
-137
-50
-97
0 -66
-147
-69
-106
Replace the entries in
the second row by their
16'
residues modulo 26.
-93
4
15
9
17
0
15
6
17
-107
Thus,
Add -6 times the first
row to the second and
24
and so the
-7 times the first row to
-I
the third.
1
10
24
12
17
16
0
9
19
2
7
11
0
12
9
9
24
23
deciphering matrix is A
4
9
15
17
24
15
6 .
0 17_
From Table 1 the numerical equivalent of the
given ciphertext is
Replace the entries in
H
the second and third
8
P A
16
1
F
Q
G
G D
6
17
7
7
U
4
G G D
21
7
4 4
rows by their residues
H
P
G
O
D
Y
16
8
16
7
15
4
25
modulo 26.
1
10
0
0
24
12
17
15
18
57
6
21
33
To obtain the plaintext triples, we multiply each
9
9
24
23
ciphertext vector by A”’ :
12
■ 4 9 I5]r 8l r191] [9] /
Multiply the second row
15
by 9“' = 3 (modulo
17
6
16 = 398 = 8
■ 4 9 15]r 6] [282] [22] V
1
10
24
12
17
16
0
1
5
6
21
7
15
17
6
0
12
9
9
24
23
24
0
17
7
4
9
15
7
5
£
263
3
C
379
15
O
15
17
6
4
299
13
M
residues modulo 26.
24
0
17
21
525
5
E
0
-26
-48
-193
-54
0
1
5
6
21
7
-63
17 = 421 =
Replace the entries in
the second row by their
1
-51
H
_24 0 nJL ij [209J [ij A
26).
0 0
NOR
14
-228
4
9
15
7
124
20
T
15
17
6
4
197
15
O
24
0
17
4
236
2
B
4
9
15
8
281
21
U
15
17
6
16
434
18
R
0 njL 7_
311
25
Y
471
3
C
443
1
A
785
5
E
461
19
5
-61
Add -10 times the
second row to the first
and -12 times the
_24
second row to the third.
1
0
0
4
15
24
0
1
5
6
21
7
0
0
1
15
6
■ 4 9 15][ 15'
15
Replace the entries in
by their residues
4
15
4
24 0 17j[l8
modulo 26.
0 0
6
_24 0 I7j[25
'4 9 15][14'
15 17
6 15
17
the first and second row
1
17
1
0
-69
-9
-78
0 0
1
15
6
17
642
18
R
Finally, the message is / HAVE COME TO
24
BURY CAESAR.
0
A
573
256
Section 10.16
SSM: Elementary Linear Algebra
7. (a) Multiply each of the triples of the message
by A =
1
1
0
0
1
1
1
I
1
Thus A
1
0
1
1
1
1
1
1
0
1
and reduce the results
0
1
0
1
0
0
0
1
1
1
1
1
1
2
1
0
0
1
0
1
0
1
1
1
0
1
1
0
1
1
modulo 2.
“1
0
-1
1
1
o'
0
2
0
1
I
1
1
0
0
1
I
1
1
I
1
1
0
0
1
1
2
0
0
1
0
1
1
2
0
2
1
0
1 = 2 = 0
0
The decoded message is 1 10101111, which
is the original message.
_1 1 lj[lj [3J [l_
The encoded message is 010110001.
8. Since 29 is a prime number, by Corollary
(b) Reduce [A|/] to /
'l
1 0
0
1
0
1
0
1
0 0
1
10.15.2 a matrix A with entries in Z29 is
modulo 2.
invertible if and only if det(A)^0(mod 29).
1 0 0
Form the matrix
Section 10.16
Exercise Set 10.16
[A|/].
1
I
0
0
2
2
1
1
0 0
0
1
0
1
0
1
1. Use induction on n, the case n = 1 being already
given. If the result is true for n- \, then
Add the first row to
M" =M
the third row.
1
0
I
1
I
0
1
0
I
I
0
1
I
0
I
0
0
I
2
1
0 0
1
I
0
1
0
1
0 0
1
0
1
Replace 2 by its
= PD'’~^DP-1
= PD"P~\
residue modulo 2.
0
proving the result.
Add the third row
2. Using Table 1 and notations of Example 1, we
derive the following equations:
to the second row.
0
0 0
1
1
M
=(PD'‘~'P~'){PDP~')
-1
= PD"-\p-'p)DP
0 0
0 0
0
«-i
I
Replace 2 by its
0
2
4
I
I
b.,
= —an-l +-b„_] +
II
2
2
2
Cn-\
residue modulo 2.
I
2 0
0
1
0
0 0
2
1
1
1
1
0
U, =tVi+tU!-i4
2
Add the second row
f 1 0
The transition matrix is thus M = 1 1 i
to the first row.
2
1
0
0 0 0
1
0 0
0
1
1
2
2
1
0 i i
Replace 2 by its
The characteristic polynomial of M is
0
residue modulo 2.
257
Chapter 10: Elementary Linear Algebra
SSM: Elementary Linear Algebra
1
det(^/-M)= -f-2J +
x(2'')
A X
2J
and
x(2"+i)
We have
= X(X-1) X—L ,
1
2
M2M\ =
so the eigenvalues of 4/ are X = 1, X2 = —, and
111
2
2
2
1 11
1 0
0 i
2
1
0 ^ i [o 0 0
X3 = 0. Corresponding eigenvectors (found by
4
111
2
2
2
0
i
I
4
The characteristic polynomial of this matrix is
solving (X/-l/)x = 0)are e, =[l 2 l]"^ ,
63=[1 0 -lf,ande3=[l -2 l]^. Thus
X'* -—X^ + —
4
so the eigenvalues are Xi = 1,
4
'
1
M
n
X2 - 4’ X3 = 0. Corresponding eigenvectors
-I
= PD"P
=[5 6 if, 63 =[-l 0 if, and
e3=[l -2 if. Thus,
are e
0
1
0
n
2
0
-2
1
0
1
0
(i)
0
0
0
I
I
I
4
4
4
i
0
I
2
2
1-1
4
I
II
1
4
-I
4
= PD'‘P
This yields
a
0
5
-1
1
6
0
-2
n
x('0 =
0
n
b11
0
1
0
c
n
n+\
;i+l
1 i_m
4 4 \2)
I
i
i
2
0
2
Qq + ^0
^0
4 i+li)
U-(i)
0
X
_L
12
12
12
1
1
1
3
6
3
1
_1
1
4
4
4
Using the notation of Example 1 (recall that
%
a+1
a+l
0
1-
~
obtain
5
^2n = — +
Cq
12
(2ao-^0-4co)
6-4n
1
Remembering that OQ+bQ+CQ = 1, we obtain
f\
N/J+l
^2a = 2
\a+l
1
a
%
n
(2^0 “^0 “4co)
in
12
6-4''
\n+l
1
I
and
=-+
4
(Qq-Cq)
V2
2
{2aQ-bQ-4cQ)
'^2/j+l -T+
3 6-4n
1
1
1
1,
1
(\
c„=J«o+2l>o + 2C„- -
^2a+l -T
\/i+l
\/i+l
3 6-4"
1
"^0+ -
Co
(2^0 ~^0 “4Co)
C2a+1 -0-
VZ
N/l+1
4 V2j
(«o-co)-
4. The characteristic polynomial of M is
(1-1) X— , so the eigenvalues are X, =
1
Since
V
approaches zero as 11 —> 00, we
2J
2j
and 'L2= — - Corresponding eigenvectors are
obtain a..
n —
n
, b,, — , and c„ ^ —
4 "
2
"
4
as
easily found to be e| =[1 of and
00.
e-,=[] -1]^. From this point, the verification
3. Call A4| the matrix of Example l,and M2 the
of Equation (7) is in the text.
matrix of Exercise 2. Then
258
Section 10.16
SSM: Elementary Linear Algebra
9
i
1
2
1
= —, and, in general,
then b2=-^
=-,
=-;t
5. From Equation (9), if bQ=.25 =-, we get b^ =-^ = —,
Q
iO
-s
li 11
1
9
2
bn
10
2
—. We will reach — =.10 in 12 generations. According to Equation (8), under the controlled program
20
the percentage would be
=.00006 =.006%.
in 12 generations, or
2'4
16,384
1
2
i
2
0
6.
=
0
i
{\-S)
12
1
2
1
2
0
£)'ip-lx(0) =
0
-in+\
-\n+\
n+\
i-^f^L(-3-V^)(i+75r"'+(-3+V5)(i-VI)
i-^[(i+vir"'+(i-vi)
i-^[(i+vir+(i-virj
PD’‘P~'x^°^ =
n+\
i _!_
n
(i+7ir+(i-vi)
3 4"+'.
«+l
i-^L(i+V5r'+(i-V5)
n+]
3 4''+2
(-i-VsKi+Vsr'+l-j+TSKi-Vs)
1
2
0
[As n tends to infinity.
and
4/1+2
4//+1
approach 0, so x^"^ approaches
0
0
0
1
2
7. From (13) we have that the probability that the limiting sibling-pairs will be type (A, AA)is
2, 1
2, 1
‘^o+-^o+-‘^o+-^o+“^oThe proportion of A genes in the population at the outset is as follows: all the type
259
Chapter 10: Elementary Linear Algebra
SSM: Elementary Linear Algebra
2
1
857
{A, AA)genes, — of the type (A, Aa)genes, -
(c)
=
285
the type (A, aa) genes, etc. ...yielding
855
=
2.
1
2^
1
287
%+“^0 +^<^0 ■'■^^0 3 ^05.
8. From an (A, AA) pair we get only (A, AA) pairs
and similarly for (a, aa). From either (A, aa) or
in the first age period. 02^1
0
0
0
0
0
0
0 '
0
0
0
1
the number of
offspring produced in the second period times
the probability that the female will live into the
second period, i.e., it is the expected number of
offspring per female during the second period,
and so on for all the periods. Thus, the sum of
these, which is the net reproduction rate, is the
expected number of offspring produced by a
given female during her expected lifetime.
(a, AA) pairs we must get an Aa female, who will
not mature. Thus no offspring will come from
such pairs. The transition matrix is then
'1 0 0 O'
0
fli is the average number of offspring produced
1^
9. For the fi rst column of M we realize that parents
of type (A, AA) can produce offspring only of
that type, and similarly for the last column. The
19
7. R = 0 + 4 1 +3 2J
fi fth column is like the second column, and
8.
follows the analysis in the text. For the middle
two columns, say the third, note that male
offspring from (A, aa) must be of type a, and
females are of type Aa, because of the way the
genes are inherited.
— = 2.375
2A4J
/? = 0 + (.00024)(.99651) + - + (.00240)(.99651)---(.987)
= 1.4961 1.
Section 10.18
Exercise Set 10.18
Section 10.17
1. (a) The characteristic polynomial of L is
Exercise Set 10.17
f
1. (a) The characteristic polynomial of L is
2
8
2
1
2J
2
so
4 ’
Thus h, the fraction harvested of
A,= —.
2
4
3
K
- . -2 = -1 so the yield i s
each age group, is 1 - —
thus A) =—
and
2’
2
2
f 'i\
A^+ — Ah—
3
A -A —, so the eigenvalues of Lare
A = - and A = --
^\r
yj’—2X — = A.—
3
3
1
i
xi =
1
2
= I
3
3
33^% of the population. The eigenvector
is the corresponding
1
2
3 .
eigenvector.
(b) x(')=Lx(°) =
x®=Lx(2) =
x‘")=Lx® =
corresponding to A = ^
100
50 ’
x(2)=Lx“) =
3
; this is the
1
18
age distribution vector after each harvest,
50 ’
(b) From Equation (10), the age distribution
88 ’
125 ’
IS
175
250
382
i
r
x-‘*=Lx(4) =
1
vector X] is 1 ^
570
ii^
. Equation (9) tells
11
191
—, so we harvest —
us that h\ =
19
T
19
or 57.9% of the youngest age class. Since
Lx I 260
I£
1
I
8
2
8
iT
, the youngest class
Section 10.19
SSM: Elementary Linear Algebra
contains 79.2% of the population. Thus the yield is 57.9% of 79.2%, or 45.8% of the population.
=.975, b^ =.965, etc. This, together with the harvesting data
2. The Leslie matrix of Example I has b^ =.845,
from Equations(13) and the formula of Equation (5) yields
=[l .845 .824 .795 .755 .699 .626 .5323 0 0 0 of
Lx| =[2.090 .845 .824 .795 .755 .699 .626 .532 .418 0 0 of.
The total of the entries of LX| is 7.584. The proportion of sheep harvested is h^(Lx^)2 +hg{Lx^)g =1.51, or
19.9% of the population.
= • • • = /?„ =0. Equation (4) then takes the form
4. In this situation we have ly ^0, and h^ = /t^ = • • ■ = h|_^ =
cj) +<32^1 + <33^1^2
+
•••/3/_](l-/t/)+ a/+|^,i'2
-/'/)+ •••+ «,
(1 - h,)[a,b^b2 ■ • • fe/_i + a/+i^i/’2 ■■■b,+- - - + a,fyb2 ■ ■ ■ b„_^ ] = 1 - a, - fii2^i
So h l
a^ +^2^1 "I
-
^^/-1^1^2 "'^1-2
= ’•
«/-i^i^2 ' '' ^/-2 •
+1
a,b^b2---b,_^ +- -- + a,^b\b2---b,^_y
a^ +(32^1 "I
^'2/-1^I^2 "■^/-2 - I + «/^|^2
aib\b2---b,_^ + ■■■ + a,Jb^b2■■■b,^_^
R-\
a,b^b2- -bi_^ + --- + a,M2-- -b„_^
5. Here hj =\, hj ^ 0, and all the other /t^’s are zero. Then Equation (4) becomes
<3] +a2bi H
We solve for
\-ai_\b^b2 ■■ ■bj_2 +(l-/t/ )[(3/i^|^’2
'■'^y-1^1^2 ■"^7-2] = *•
fit]
+
^2^]
1"
rt/
|^|^2
'
"^1—2
~^ +1
h,/ to obtain h / aib^b2 ■■■bi_^ H
^^j-\b\t>2'"bj-2
a^ +02^1 ■!— + ‘3y-i^i^2■■■^y-2
aib^b2 ■■ ■bl_^ H
\-aj_\b\b2 ■■■bj_2
Section 10.19
Exercise Set 10.19
1. From Theorem 10. 19. 1, we compute <3o"="“K ^
4
cosktdt =
—, and
sin ktdt =0. So the least-squares trigonometric polynomial of order 3 is
K
n
= —n:^,
cik~^n ^f ^(t
3
2
4
-I-4 cos r-I-cos 2f-h—cos 3r.
3
9
2
7'
2
2. From Theorem 10.19.2, we compute
K-\i
sin
2km
2 rT 2
'
T
2kKt
i'
r/f =
and
T
dt = -
T
kTT
So the least-squares trigonometric polynomial of order 4 is
j2
j2 /
+—
3
2m
cos—
T
1
4;rt
-I- —cos
4
1
6m
h —cos
T
9
1
1
T
%m\
T
2 /
cos —
\6
T
261
.
2m
sin
;r
I
.
4m
H —sm
T
2
1
.
6m
-r-sm
7-
3
I
.
^m
h—sm
T
4
T
Chapter 10: Elementary Linear Algebra
I
SSM: Elementary Linear Algebra
An:
2
3. From Theorem 10.19.2, aQ= —
K •’0
n
sin f Jr = —. (Note the upper limit on the second integral),
n
'f
sin r cos fa Jr
a^=n
K
[/: si n fa sin r + cos fa cos r
0
[0+(-l)^-'-l]
^U“-I
if k is odd
0
2
if
is even.
7r(k--\)
N • , ■
j
- if'^' = 1
h=- I Sinktsintdt = 2
n
0 \1k>\
So the least-squares trigonometric polynomial of order 4 is —+ —sinrK
\
An
4. From Theorem 10.19.2, Aq =—
\
2
2
3k
cos 2r - — cos4r.
\5k
4
sin —rJr= —.
K
I
r2;r
=-l
K
sin—r cosfa Jr
2
4
K{4k--\)
4
K(2k -lX2k +1)’
K
1 An . 1
sin—rsinfaJr = 0.
=-
K
"" 2 ^
0
So the least-square trigonometric polynomial of order/; is
2
4rcosr
n ;rl, l -3
2 At
2 A
5. From Theorem 10.19.2, aQ= —
T JO
2
r
O/t =T
rcos
fb)dt=-
T JO
2kKt j 2 fT
Jr + — , (7-r)cos
T
Jr
T
4T
4k^K~
if/: is even
0
87
if k is odd ’
(2kfK^
So the least-squares trigonometric polynomial of order n is:
T
cos
4 ;rH2-
1- —cos
T
lOrrr
bnt
2Kt
87
6-
h
T
io2
2Knr
1
+...+
tCOS
cos3r
cosrrr
3-5
5-7
(2«-l)(2« + l)
tdt+^l^iT-t)dt =^.
2kKt
T ^T
cos2r
7
cos
(2/1)-
if n is even; the last term involves n - I if n. is odd.
262
7
Section 10.20
SSM: Elementary Linear Algebra
(c) As in part(a) the system for C], C2, and C3
6. (a) | i| =^|j^r =V^
■2^
3
Cl
0
-2 0
C2
0
3 -2
IS
3
I
cos^ kt dt
(b) icosfall
which has the
C3
3
l + cos It
unique solution q =-^, C2 =-j, and
dt
2
C3 = 0. Because these coefficients are all
■Itt
(c) llsinL'fll-
nonnegative, it follows that v is a convex
combination of the vectors V|, V2, and
sin^ ktdt
''3-
•2;rn-cos2r"i dt
(d) As in part (a) the system for q, C2, and C3
2
IS
3-2
3
3
0
-2
I
I
Section 10.20
Exercise Set 10.20
Cl
I
C2
0
C3
1
unique solution q = —
15’
3
4
1. (a) Equation (2) is q
3
which has the
6
^
C2 = —, and
15
C3 = —. Because these coefficients are all
3
15
and Equation (3) is q +C9 +C3 = 1. These
nonnegative, it follows that v is a convex
equations can be written in combined matrix
combination of the vectors Vi, V2, and
form as
"1
3
4
C|
3
1
5
2
C2
3
1
''3-
2. For both triangulations the number of triangles,
m, is equal to 7; the number of vertex points, n,
is equal to 7; and the number of boundary vertex
points, k, is equal to 5. Equation (7),
C3
1
This system has the unique solution C| = —,
2
C2 =
m = 2n-2- k, becomes 7 = 2(7) - 2 - 5, or
2
—, and C3 = —. Because these
5
7 = 7.
5
coefficients are all nonnegative, it follows
3. Combining everything that is given in the
statement of the problem, we obtain:
that V is a convex combination of the vectors
V2, and V3.
w = Mv + b
= yW(qV| +C2V2 +C3V3) + (q +C2 +C3)b
= q(A7V| +b) + C2(A/v2 +b) + C3(A7v3 +b)
(b) As in part (a) the system for C|, 03, and C3
'1 3 4][q
is
1
5
1
2
2
C')
4
C3
1
unique solution C| =-^,
= qW| +C2W2 +C3W3.
which has the
4
—, and
^^ = 5
1
Cq = —. Because one of these coefficients
5
is negative, it follows that v is not a convex
combination of the vectors V|, V2, and
''3-
263
Chapter 10: Elementary Linear Algebra
(b)
''l
SSM: Elementary Linear Algebra
Solving these two linear systems leads to
V2
3
-I
I
(c) As in part (a), we are led to the following
two linear systems:
''4
Vs
V6
5. (a) Let M =
mil
'«I2
m21
m22
and b =
. Then
bi
the three matrix equations A/v,- + b = w,-,
■-2
I
1 1
0
3
5
m|2
5
I
0
bI
3
-2
I
m21
-2
3
5
m22
2
I
0
h
m
and
-3
Solving these two linear systems leads to
i = I, 2, 3, can be written as the six scalar
M =
equations
+ mi2 +i>i = 4
/n2i +ni22 +^2 “^
m
0
and b =
M =
>^3
I
0
0
I
and b = ^ .
It
(d) As in part (a), we are led to the following
two linear systems:
2mii +3mi2 +bi =9
2m2i +3m22 +^2
5
2mi 1 +mi2 +6| = 5
2m2i +nv22+b2 =3
written in matrix form as
1
2
1
3
_2
1
1
1
2
I
mi l
2
2
1
m
-4
-2
1
1
1
1
3
1
2
1
1
7
12
by
mil
4
"'12
9
b\
5
0
2
1
m2i
2
2
1
m22
2
and
1
-4 -2 IJL/72
-1
3
-9
Solving these two linear systems leads to
and the second.
M =
i 1
2
fourth, and sixth equations as
2
9
2
The first, third, and fi fth equations can be
'1
0
0
i
and b =
2
-1
m2i
3
m22
5
7. (a) The vertices Vi, V2, and V3 of a triangle
3
can be written as the convex combinations
bl _
Vi =(l)vi +(0)V2+(0)V3,
The first system has the solution mi 1 = 1,
V2 = (0)V|+(l)v2+(0)v3, and
mi 2 =2, b^=\ and the second system has
V3 = (0)V| +(0)v2 +(l)v3. In each of these
the solution m21 =0, m22 = 1, i>2=2.
Thus we obtain M =
1
2
0
1
and b =
cases, precisely two of the coefficients are
1
zero and one coefficient is one.
2 ■
(b) If, for example, v lies on the side of the
triangle determined by the vectors V| and
(b) As in part (a), we are led to the following
two linear systems:
■-2 2 1 mil
0
0
1
V2 then from Exercise 6(a) we must have
-8
m|2 =
0
_ 2 1 lj[ b^
Cl +C2 = 1. Thus at least one of the
coefficients, in this example C3, must equal
5
-2
2
1
^21
0
0
1
m22 = 1 •
_ 2 1 lj|_fo2
that V =C|Vi+C2V2+(0)v3 where
and
1
zero.
4
264
Section 10.20
SSM:Elementary Linear Algebra
(c) From part (b), if at least one of the
coefficients in the convex combination is
zero, then the vector must lie on one of the
sides of the triangle. Consequently, none of
the coefficients can be zero if the vector lies
in the interior of the triangle.
8. (a) Consider the vertex v, of the triangle and
its opposite side determined by the vectors
V2 and V3. The midpoint v,„ of this
opposite side is
(V2+V3)
and the point on
2
the line segment from V| to
thirds of the distance to
that is twois given by
2/ V2 +
. V3 \
2
— Vi + —Vin = -v, -I--
3
3 ' 3l
3
2
1
1
= -V,
1 -H -Vt -I--V3.
3 ^ 3'
3
(b)
1[2
3 3
1 5
+-
3 2
1
1 8
3 1
3 6
1
-I--
8
3
2
265
/-
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