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.