Linear Algerbra. Inverses and Algerbraic Properties of Matrices

[stumpoman]

Homework Statement

Assuming that all matrices, A, B, C, and D, are n x n and invertible, solve for D.

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

Homework Equations

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

The Attempt at a Solution

I must have missed something in the reading of this section. All I can think of is

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

but I don't know where to go from there or if it is even the right way to start.

 

[voko]

Left-multiply the third equation with B.

 

[HallsofIvy]

Homework Statement

Assuming that all matrices, A, B, C, and D, are n x n and invertible, solve for D.

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

Just "undo" every thing that is done to D, one step at a time. For example if you multiply both sides, on the left, by $(C^T)^{-1}$ you get $B^{-1}A^{-2}BAC^{-1}DA^{-2}B^TC^{-2}= (C^T)^{-1}C^T= I$
Then multiply on the left by $B$, then on the right by $C^2$, or do both together to get $A^{=2}BAC^{-1}DA^{-2}B^T= BC^2$.
Continue to "unpeel" D.

Homework Equations

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

The Attempt at a Solution

I must have missed something in the reading of this section. All I can think of is

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

but I don't know where to go from there or if it is even the right way to start.

Yes, that was the first step as I showed. Continue in the same way.