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

Lol thanks