# Having difficulty with an Ax=b problem and understanding it

1. Jul 7, 2008

### blhhblah

I'm having trouble with Ax=b matrices. I get really confused when dealing with matrices that have a different number of rows from columns. An example is this problem:
(A is a 3x2 matrix)
A =
3 1
4 2
-5 -1

(b is a 3x1 matrix)
b =
1
0
-3
I need to find the solution set for x. Aren't there only 2 variables?
Thanks.

2. Jul 7, 2008

### Defennder

Yes there are only two variables but 3 equations. Construct the augmented matrix A|b and perform row reduction until you get the reduced-row echelon form of the matrix. Now you can easily extract the solutions.

3. Jul 7, 2008

### blhhblah

is this the reduced row echelon form?

1 0 1
0 1 -2
0 0 0

4. Jul 7, 2008

### blhhblah

regarding another question...

how do you solve for x given A is a (3x4) matrix and b is a (4x1) matrix?

5. Jul 7, 2008

### Defennder

Yeah I got that as well. Now you can read the solutions for the variables directly from the matrix.

6. Jul 7, 2008

### Defennder

It's by using the same technique as well. But note that this time you have more unknowns than equations. What does that tell you about the nature of the solutions you will have?

7. Jul 7, 2008

### blhhblah

i guess all the values won't be exact, but i'm not sure how to set this up

8. Jul 7, 2008

### Defennder

What do you mean by "exact"? Just write out the augmented matrix and again use row-reduction as before. Note that it is entirely possible that the matrix might be inconsistent, in that case there are no solutions.

9. Jul 7, 2008

### blhhblah

matrix A was

2 8 4 16
1 1 2 2
1 -1 2 -2

I was able to reduce it to
1 0 2 0
0 1 0 2
0 0 0 0

but i'm not sure how to match it up to b, which is

-1
9
5
17

10. Jul 7, 2008

### HallsofIvy

Your first matrix was equaivalent to the set of equations
2x+ 8y+ 3z= 16, x+ y+ 2z= 2, and x- y+ 2z= -2 with b= <x, y, z>.

You have reduced it to a matrix equivalent to
x+ 2z= 0 and y= 2. What possible values of x, y, and z satisfy both of those?

11. Jul 7, 2008

### sennyk

Whenever I have more or fewer equations than unknowns, I use the following to obtain the least squares answer. $A^T A x = A^T b$, where $A^T$ is the transpose of A.