The following discussion suggests the relation between the existence of solutions and the shape of the row echelon form of the augmented matrix.
Example Suppose the following is the row echelon form of the augmented matrix [A b] of a system Ax = b.
[ | * | # | # | # | # | ] |
0 | 0 | * | # | # | ||
0 | 0 | 0 | 0 | * | ||
0 | 0 | 0 | 0 | 0 |
The 3rd row is of the form [0 0 0 0 ≠0], which means that the 3rd equation is 0 = *, where * is nonzero. The contradiction shows that the system Ax = b has no solution.
On the other hand, if the row echelon form is
[ | * | # | # | # | # | ] |
0 | 0 | * | # | # | ||
0 | 0 | 0 | * | # | ||
0 | 0 | 0 | 0 | 0 |
which has no rows like [0 0 0 0 ≠0], then the contradiction as above does not appear. In fact, further row operations will give us the reduced row echelon form.
[ | 1 | a | 0 | 0 | b_{1} | ] |
0 | 0 | 1 | 0 | b_{2} | ||
0 | 0 | 0 | 1 | b_{3} | ||
0 | 0 | 0 | 0 | 0 |
The corresponding system has an obvious solution x_{1} = b_{1}, x_{2} = 0, x_{3} = b_{2}, x_{4} = b_{3}. The general solution can also be found by the discussion here. Thus the system has solutions.
You may also check out this example (no solution, row echelon form of the augmented matrix contains a row of the form [0 0 ... 0 ≠0]) and this example (has solutions, row echelon form of the augmented matrix contains no row of the form [0 0 ... 0 ≠0]). In general, we have the criterion for the existence of solutions.
A system of linear equations has solutions ⇔ There is no row like [0 0 ... 0 ≠0] in the row echelon form of the augmented matrix
When a system has solutions, we also say the system is consistent. Otherwise we say the system is inconsistent.
Example Consider the following system of linear equations.
x_{1} | + 3x_{2} | + 2x_{3} | + x_{5} | = | 0 | |
- x_{1} | - x_{2} | - x_{3} | + x_{4} | = | 1 | |
4x_{2} | + 2x_{3} | + 4x_{4} | + 3x_{5} | = | 3 | |
x_{1} | + 3x_{2} | + 2x_{3} | - 2x_{4} | = | 0 |
The augmented matrix is
[ | 1 | 3 | 2 | 0 | 1 | 0 | ] |
-1 | -1 | -1 | 1 | 0 | 1 | ||
0 | 4 | 2 | 4 | 3 | 3 | ||
1 | 3 | 2 | -2 | 0 | 0 |
In this example, we computed its row echelon form.
[ | 1 | 3 | 2 | 0 | 1 | 0 | ] |
0 | 2 | 1 | -1 | 0 | 1 | ||
0 | 0 | 0 | 2 | 1 | 0 | ||
0 | 0 | 0 | 0 | 0 | 1 |
Since the last row is of the form [0 0 0 0 0 ≠0], the system has no solution. On the other hand, if we change the system a little bit,
x_{1} | + 3x_{2} | + 2x_{3} | + x_{5} | = | 0 | |
- x_{1} | - x_{2} | - x_{3} | + x_{4} | = | 1 | |
4x_{2} | + 2x_{3} | + 4x_{4} | + 3x_{5} | = | 2 | |
x_{1} | + 3x_{2} | + 2x_{3} | - 2x_{4} | = | 0 |
then the same row operations give us
[ | 1 | 3 | 2 | 0 | 1 | 0 | ] |
0 | 2 | 1 | -1 | 0 | 1 | ||
0 | 0 | 0 | 2 | 1 | 0 | ||
0 | 0 | 0 | 0 | 0 | 0 |
Since there is no row of the form [0 0 0 0 0 ≠0], the system has solutions.
Next we modify the system again by taking away x_{3} and x_{5}.
x_{1} | + 3x_{2} | = | 0 | |
- x_{1} | - x_{2} | + x_{4} | = | 1 |
4x_{2} | + 4x_{4} | = | 2 | |
x_{1} | + 3x_{2} | - 2x_{4} | = | 0 |
The row echelon form by the same row operations is obtained by dropping [col 3] and [col 5].
[ | 1 | 3 | 0 | 0 | ] |
0 | 2 | -1 | 1 | ||
0 | 0 | 2 | 0 | ||
0 | 0 | 0 | 0 |
Again there is no row of the form [0 0 0 ≠0], and the system has solutions.
Example We change the right side of the system in the previous example to arbitrary numbers,
x_{1} | + 3x_{2} | + 2x_{3} | + x_{5} | = | a | |
- x_{1} | - x_{2} | - x_{3} | + x_{4} | = | b | |
4x_{2} | + 2x_{3} | + 4x_{4} | + 3x_{5} | = | c | |
x_{1} | + 3x_{2} | + 2x_{3} | - 2x_{4} | = | d |
and try to find the exact condition on a, b, c, d that makes the system consistent. By the same row operations as before, we get
[ | 1 | 3 | 2 | 0 | 1 | a | ] |
0 | 2 | 1 | -1 | 0 | b + d | ||
0 | 0 | 0 | 2 | 1 | a - d | ||
0 | 0 | 0 | 0 | 0 | - 3a - 2b + c + d |
For the system to be consistent, we need the last row to be [0 0 0 0 0 0]. Therefore the condition for the the system to have solutions is - 3a - 2b + c + d = 0.