Consistent and Inconsistent Systems
When you solve a system of linear equations, what does your solution set (all of your solutions) describe geometrically? In each problem involving 2 by 2 matrices that we solved, our solution set was the point of intersection of the two lines represented by the equations in our system. In each problem with 3 by 3 matrices, the solution set was the point of intersection of 3 planes. However, a system of linear equations does not always have a point as the solution set.
If you solve the system
|
x1 |
+ |
2x2 |
|
= |
|
4 |
|
2x1 |
+ |
4x2 |
|
= |
|
8 |
algebraically, what do you get? You have an infinite number of solutions along the line x1 = 4 - 2x2 because any solution to the first equation, also solves the second equation. Therefore, you can choose any value for one of the variables, and you will be able to find a value for the other variable so that both equations are satisfied. This is called a consistent system because there is a solution. It is further categorized as an underdetermined consistent system because there is not enough information to determine a unique solution.
Definition A system is consistent if there is at least one solution.Definition A system is underdetermined when there are an infinite number of solutions.
For a linear system, if there are two or more solutions, then there are an
infinite number of solutions. These solutions all lie on the same line.
Geometrically, this system represents a line because both equations are
representations of the same line. Try to solve this system with Gaussian
elimination. What do you get? We get
, because the second equation is a multiple of the first. The second equation
requires that 0x1 + 0x2 = 0 which is always
true. Therefore, our second equation made no additional requirements beyond what
the first equation requires. When Gaussian elimination on a system with a square
coefficient matrix leaves you with zeros across an entire row of the augmented
matrix and there are no rows with zeros to the left of the bar and a non-zero
number to the right of the bar, you know that you have a consistent system that
is underdetermined. We move the zero row or rows to the bottom of the matrix.
When we try to get zeros above the main diagonal in Gauss-Jordan elimination, we
do not try to get zeros in a column if the diagonal element of that column is
zero. Try this next system before reading further!
| 2x1 | + | 4x2 | + | 5x3 | = | 47 | ||
| 3x1 | + | 10x2 | + | 11x3 | = | 104 | ||
| 3x1 | + | 2x2 | + | 4x3 | = | 37 |
What is the solution to the system? Our work and solution are
below.

This tells us that x1 = 6.75 - .75x3 and x2 = 8.375 - .875x3. That means that we can choose whatever number we want for one element of x and get corresponding valid solutions for the other two. For instance, if we choose x3 to be 1, then x1 = 6 and x2 = 7.5. Instead, we may choose x1 = 6.75 then x2 = 8.375 and x3 = 0. There are an infinite number of solutions that we can find in this manner. Notice that this system is also consistent and underdetermined.
What do you get if you try to solve
| x1 | + | 2x2 | = | 4 | ||
| 2x1 | + | 4x2 | = | 9 |
algebraically? The result is a requirement that you know cannot be satisfied. We get 0 = 1 (you may arrive at a different contradictory requirement). This system is called inconsistent.
Definition A system is inconsistent if it has no solutions.
If you graph these lines, you will see that they are parallel. Try to solve this system using Gaussian elimination. Our work follows:

This requires that 0x1 + 0x2 = 1. We know that this cannot be true. When using Gaussian elimination on a system, if you have zeros in an entire row to the left of the bar and do not have a zero to the right of the bar on that row, you know that you have an inconsistent system. Therefore, there is no point where the lines (or planes if you are in higher dimensions) intersect, so the system does not have a solution.
With underdetermined and inconsistent systems, you will never be able to get an identity matrix to the left of the bar of the augmented matrix. Therefore, we will not be able to find an inverse for the coefficient matrix. The only coefficient matrices that have inverses are those that have a unique point as the solution to the system, and the only coefficient matrices that have a unique point as the solution to the system are those that have inverses. These systems are consistent because they have a solution and uniquely determined because there is exactly one solution.
Definition A system is uniquely determined if there is exactly one solution to the system.Remark Many pre-calculus texts refer to underdetermined systems as dependent systems and to uniquely determined systems as independent systems.
For the examples above, find the determinant of A in each case. Can you draw a conclusion from this data that relates the determinant of A to whether or not the system has a unique solution? If not, solve some additional systems and try again to draw a conclusion.
Using determinants, we can check to see if a system is uniquely determined.
If the determinant of the coefficient matrix is not equal to zero, the system is
uniquely determined. Thinking of the systems that involve 2 by 2 matrices, can
you tell why systems with determinants equal to zero are not uniquely
determined? Remember that the formula for the inverse of a 2 by 2 matrix is
Does
have a meaning if
det(A) = 0? No, it does not. This is another illustration of the fact
that the square matrix represented by A in Ax = b for an
inconsistent or underdetermined systems does not have an inverse. You will see
that this holds true for all the problems that you do. If det(A) ≠
0 the system is uniquely determined and A is invertible. If
det(A) = 0, the system is either underdetermined or inconsistent;
therefore, A is not invertible. In practice, the determinant is
not used to test systems in this manner because other methods require fewer
steps. However, these relationships are very important to the theory of
matrices.
Remark In this section, occasionally, we refer to systems that have square coefficient matrices. This is only for simplicity. All of these classifications also apply to matrices with non-square coefficient matrices. However, we cannot take the determinant of a non-square matrix. Also, we cannot rely entirely upon the appearance of the last row of the system after Gaussian elimination even though that information is still valuable.
Here is a visual breakdown of the information that you have been given. This breakdown assumes a square coefficient matrix.

Note: A system is inconsistent if ANY row of the matrix has zeros left of the
bar and a non-zero number right of the bar. Therefore, the matrix
| |