# Linear algebra question

1. Mar 15, 2013

### kwal0203

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

Assuming that all matrices are $n\times n$ and invertible, solve for $D$.

$C^{T}B^{-1}A^{2}BAC^{-1}DA^{-2}B^{T}C^{-2}=C^{T}$

3. The attempt at a solution

I tried to group all like terms and simplify. I'm pretty sure this is not allowed but I'm not really sure how to approach this question. Thanks a lot any help appreciated!

$C^{T}C^{-1}C^{-1}C^{-1}B^{-1}BB^{T}AAAA^{-1}A^{-1}D=C^{T}$

$C^{T}C^{-1}C^{-1}C^{-1}IB^{T}IIAD=C^{T}$

$((C^T)^{-1}C^{T})C^{-1}C^{-1}C^{-1}B^{T}AD=C^{T}(C^T)^{-1}$

$I(C^{-1}C^{-1}C^{-1}CCC)(B^{T}(B^T)^{-1})(AA^{-1})D=ICCC(B^T)^{-1}A^{-1}$

$D=C^{3}(B^T)^{-1}A^{-1}$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 15, 2013

### haruspex

No, you can't change the order like that. What you can do is multiply both sides by the same thing at the same end. I.e. You can go from Y = Z to XY = XZ or to YX = ZX. Start with multiplying both sides on the left by C-T. Use the fact that you can change the order when the two matrices being multiplied are inverses of each other: X-1X = I = XX-1

3. Mar 15, 2013

### Dick

Don't assume the matrices commute. You can't interchange the order like you did. Aside from the fact this problem is needlessly complex, just use patience and cancel each matrix M by M^(-1) on the appropriate side.

4. Mar 15, 2013

### kwal0203

Thanks guys, but what happens when I get down to D after I multiply each term preceding it by its inverse?

5. Mar 15, 2013

### kwal0203

Ahhhh, multiply on the right!

6. Mar 15, 2013

### Dick

Right! No pun intended.

7. Mar 15, 2013

### kwal0203

Lol thanks

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