# Solving AX=B when A is not invertible

1. Oct 13, 2012

### drawar

1. The problem statement, all variables and given/known data
This is what I've just been tested and I didn't know what to do :( Sorry I can't remember what the question is since the question paper was collected after the test.

2. Relevant equations

3. The attempt at a solution
When A is invertible just set up the augmented matrix (A|B) and perform eros until we get the identity matrix on the LHS and X is what's on the RHS. What if A is not invertible?

2. Oct 13, 2012

### voko

What do you think will happen if you set up the augmented matrix and try the same algorithm?

3. Oct 13, 2012

### drawar

The LHS is not the identity matrix and X is not unique?

4. Oct 13, 2012

### voko

Non-identity does not mean much. What will it really look like? Try this system:

1 2 3 | 1
1 2 3 | 1
4 5 6 | 0

5. Oct 13, 2012

### drawar

Subtracting R1 from R2 and 4R1 from R3:

$\left( {\left. {\begin{array}{*{20}{c}} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & { - 3} & { - 6} \\ \end{array}} \right|\begin{array}{*{20}{c}} 1 \\ 0 \\ { - 4} \\ \end{array}} \right)$

You mean it has a zero row?

6. Oct 13, 2012

### voko

Yes, the matrix will have one or more zero rows. Their number will depend on the rank of the matrix. How would you solve that?

Now there is one important observation. If B was not (1, 1, 0), but, for example, (1, 0, 1), would you be able to solve that?

7. Oct 13, 2012

### drawar

Err, what is the rank of a matrix?

Nope, since there is a row (0 0 0|1).

8. Oct 13, 2012

### voko

The number of linearly independent rows or columns.

9. Oct 13, 2012

### drawar

Thanks but can this problem be solved without knowledge of linear independence and rank of matrix?

10. Oct 13, 2012

### voko

You don't need the rank to solve. Just continue solving the example.

11. Oct 13, 2012

### drawar

Oh my goodness! If only I had realized it earlier...
The problem actually asked us to solve the system of linear equations. Since there exists zero rows, there are free variables in the solution and hence the solution is not unique.

12. Oct 13, 2012

### voko

Only if B is suitable. Otherwise, no solutions.

13. Oct 13, 2012

### HallsofIvy

Staff Emeritus
If A is not invertible, then there are two possibilities for the equation Ax= B:
1) There are an infinite number of solutions.
2) There is no solution.

The equation Ax= 0 always has x= 0 as solution. If A is not invertible, then there exist an infinite number of solutions. In fact, the set of solutions is a sub-space, called the "null space" of A.

If Ax= B has a solution, and A is invertible, then every solution can be written as that solution plus some vector in the null space of A.