Inconsistent Systems
By
Dr. Julia Arnold
An inconsistent system is one which has either no solutions,
or has infinitely many solutions.
Example of No solutions:
-x + y = -22
3x + 4y = 4
4x - 8y = 32
In this example we have only two unknowns but three
equations. To have a solution the third equation must
intersect the first two in the exact same location.
y = x-22
The graph of each equation would
be a straight line. As you can
see, the three do not share a
common intersection point.
-x + y = -22
3x + 4y = 4
4x - 8y = 32
Let’s multiply equation 1 by -1 to get a +1 in the first
position and then write the augmented matrix.
x - y = 22
3x + 4y = 4
4x - 8y = 32
1
3
4
1 22
4 4
8 32
-3R1 +R2 = -3 +3 -66
3 +4 +4
0 7 -62
-4R1 + R3 = -4 +4 -88
4 -8 32
0 -4 -56
1
0
4
1
0
0
1 22
7 62
8 32
1 22 1
7 62 0
4 56 0
1
22
1 8 . 857
1
14
1
0
0
1 22 1
7 62 0
4 56 0
1
22
1 8 . 857
1
14
This final matrix says y = 14 and it says y = -8.85
which is not possible. Thus no solutions.
1
0
0
1
22
1
1 8 . 857 0
1 14 0
1
22
1 8 . 857
0 22 . 857
By multiplying the last row by -1 and adding to row 2, our
last row says 0 = -22.857 which is a false statement and
hence no solutions.
Having too many equations can be a problem but
having the wrong equations can also be a
problem.
Example 2 of No solutions:
x+y+z=3
2x - y - z = 5
2x + 2y + 2z = 7
1
2
2
1
1
2
1 3
1 5
2 7
Set up the augmented matrix and multiply row 1 by -2
-2 -2 -2 -6 then add row 2
2 -1 -1 5
0 -3 -3 -1
Next multiply row 1 by -2 again and add row 3
-2 -2 -2 -6 add row 3 1 1
1 3
2 2 2 7
0
3
3
1
0 0 0 1
0
0
0
1
1
0
0
1
3
0
1
3
3 1
0 1
Since the coefficients of x, y and z are
all 0 in the last row, this makes the
equation 0 = 1 which is a false
statement. Hence No Solutions.
Graphically in 3D,
this would imply
parallel planes
cut by the
transversal plane.
i.e. no common
intersection and
hence no
solutions.
Infinitely many solutions:
Too many equations can cause no solutions, but too
few equations can cause infinitely many solutions
as shown in the next example.
x + y - 5z =3
x
- 2z = 1
Form the augmented matrix and multiply R1 by -1 and
add row 2.
-1 -1 +5 -3
1 1 5 3
1 0 -2 1
1
0
2
1
0 -1 3 -2
1
0
1
1
5 3
3 2
1
0
1
1
5 3
3 2
Since we have done all we can, we must
conclude that the remaining equations
are -y +3z = -2 and x + y - 5z = 3
Graphically we know that the intersection of two planes is
a straight line.
-y +3z = -2 and x + y - 5z = 3
How can we turn this into an answer?
We decide that one variable (in this case, either y or z)
will be a free variable in that it can take on all values in
the Reals.
If we decide it should be z then we write the other
variables in terms of z as follows
-y + 3z = -2
3z = y - 2
3z + 2 = y or y = 3z + 2
Now we write x in terms of z:
x + (3z + 2) - 5z = 3
x - 2z +2 = 3
x = 2z + 1
The solution is z = any real number
y = 3z + 2 and x = 2z +1 and hence infinitely many solutions.
Remember this example from no solutions?
x+y+z=3
2x - y - z = 5
2x + 2y + 2z = 7
By changing one number we can go to infinitely many solutions.
x+y+z=3
2x - y - z = 5
2x + 2y + 2z = 6
By changing the 7 to a 6 I have created a duplicate
equation to equation 1. (Multiply equation 1 by 2)
As we work on the matrix
we would get the final row to
be all zeroes which means
0x + 0y + 0z = 0 which is true.
Thus infinitely many solutions.
1
0
0
1
3
0
1
3
3 1
0 0
1
0
0
1
3
0
1
3
3 1
0 0
The two equations remaining:
-3y - 3z = -1 and x + y + z = 3
should be treated as in the last example.
Let z be any real number
(1 - 3z)/3 = y or y = (1 - 3z)/3
x + (1-3z)/3 + z = 3
3x + 1 - 3z + 3z = 9
3x = 8
x =8/3
(8/3, (1-3z)/3,z) form the points of this solution.
Practice Problems will be located on a separate power
point presentation.
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