# Linear algebra question

## Homework Statement

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

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

## 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!

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

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

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

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

$D=C^{3}(B^T)^{-1}A^{-1}$

## The Attempt at a Solution

haruspex
Homework Helper
Gold Member
2020 Award
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

Dick
Homework Helper

## Homework Statement

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

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

## 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!

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

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

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

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

$D=C^{3}(B^T)^{-1}A^{-1}$

## The Attempt at a Solution

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.

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

Ahhhh, multiply on the right!

Dick
Homework Helper
Ahhhh, multiply on the right!

Right! No pun intended.

Right! No pun intended.

Lol thanks