# Thread: Help with a problem

1. ## Help with a problem

Assuming that all matrices are $n\,\,x\,\,n$ and invertible, solve for $D$:

$ABC^{T}DBA^{T}C=AB^{T}$

I realized that I would someone have to reduce everything to the right of $D$ to $B^{T}I_{n}$ and everything to the left would be $AI_{n}$

So for the "right side", I started with $C^{-1}A^{-T}B^{-1} =(BA^{T}C)^{-1}$ since $C^{-1}A^{-T}B^{-1}BA^{T}C=I_{n}$ and therefore:

$B^{T}(BA^{T}C)^{-1}$ would result in $B^{T}I_{n}=B^{T}$

For the "left size", I started with $C^{-T}B^{-1}=(BC^{T})^{-1}$ since $ABC^{T}C^{-T}B^{-1}=AI_{n}=A$

Multiplying the "left" with the "right" to satisfy the whole equation, I ended up with:

$D=(BC^{T})^{-1}B^{T}(BA^{T}C)^{-1}$

Is this correct, or at the very least is my though process correct and if so, what possible errors are there? Thank you.

2. Originally Posted by Pinkk
Assuming that all matrices are $n\,\,x\,\,n$ and invertible, solve for $D$:

$ABC^{T}DBA^{T}C=AB^{T}$

I realized that I would someone have to reduce everything to the right of $D$ to $B^{T}I_{n}$ and everything to the left would be $AI_{n}$

So for the "right side", I started with $C^{-1}A^{-T}B^{-1} =(BA^{T}C)^{-1}$ since $C^{-1}A^{-T}B^{-1}BA^{T}C=I_{n}$ and therefore:

$B^{T}(BA^{T}C)^{-1}$ would result in $B^{T}I_{n}=B^{T}$

For the "left size", I started with $C^{-T}B^{-1}=(BC^{T})^{-1}$ since $ABC^{T}C^{-T}B^{-1}=AI_{n}=A$

Multiplying the "left" with the "right" to satisfy the whole equation, I ended up with:

$D=(BC^{T})^{-1}B^{T}(BA^{T}C)^{-1}$

Is this correct, or at the very least is my though process correct and if so, what possible errors are there? Thank you.
Yes it is

You can check it yourself, By substituting your D in the first equation and seeing if it satisfies

3. Thank you. I checked it and it did work, but I wanted to make sure I was conveying the theorems of matrix invertibility, etc correctly. I'm having a bit harder time grasping linear algebra than I did grasping Calc 1,2, and 3.