# Matrices problem

1. Sep 24, 2009

### DanielJackins

1. The problem statement, all variables and given/known data

If ATCT+(CA)T=B where:

A=a c b a B = −34 44 −12 18 C = −3 −1 2 1

then what are the values of ab and c?

(These are all two by two matrices, I don't know how to format them on here. The T's are transposes.

3. The attempt at a solution

So I've tried a few times now with no success. As far as I'm aware ATCT+(CA)T=B is equivalent to 2ATCT, but I've tried doing them separately as well with no success. So I get it down to a bunch of a's and c's of b's to be equal to the B matrix. Then (I don't know if i'm actually allowed to do this) I match them with the corresponding values in the B matrix on the other side of the equation. This gives me a system of equations, which I change to an augmented matrix. I can then get it to RRE form, with the exception of a 4th row (only 3 variables?) which always ends up being something = 0, which throws me off.

Any help/suggestions?

2. Sep 25, 2009

### Dick

You can also write the equation as 2(CA)^T=B. If you transpose both sides, that's 2CA=B^T. C is invertible, so just solve for A. I get a matrix for A that doesn't have the two diagonal elements equal. That's probably why you are having problems solving the linear equations. Try it setting A=[[a,b],[c,d]] with 4 variables.

3. Sep 26, 2009

### DanielJackins

Sorry I'm not sure if I follow. So you're getting the same thing as me, where A is 4x3? How do I 'set' A to be four variables as opposed to 3?

4. Sep 26, 2009

### Dick

All the matrices are 2x2. If I read your notation right A=a c b a mean A is a matrix with first row 'a c' and second row 'b a', right? There is no solution to your problem of that form because of the two 'a's in it. Change one of the 'a's to 'd' for example. Now you have four unknowns instead of three. Now there is a solution.

5. Sep 26, 2009

### DanielJackins

Won't that change the context of the question though?

6. Sep 26, 2009

### Dick

Yes, it will. If you don't change the context then the answer is that there is no solution. Show that with your linear equations.

7. Sep 26, 2009

### DanielJackins

Problem is I can't, it's a web-based assignment and it expects me to input values for a, b, and c

8. Sep 26, 2009

### DanielJackins

Nevermind, I tried what you said and it worked out perfectly. Thanks a bunch!