Solve for Matrix $X$ given an equation with 3 Matrices $A,B,X$

Question

Given are 3 square matrices $A,B,X$. Furthermore, we know that $A,X$ and $A-AX$ are invertible.

Given the equation $(A-AX)^{-1} = X^{-1}B$, solve for $X$.

So here is what I did:

$(A-AX)^{-1} = X^{-1}B$

$I = (A-AX)X^{-1}B$

$I = AX^{-1}B-AXX^{-1}B$

$I = AX^{-1}B-AB$

$I-AX^{-1}B = -AB$

$-AX^{-1}B = -AB - I$

$AX^{-1}B = AB + I$

But here I'm stuck. I was told I can take the inverse of both sides now, but I don't see how? How am I supposed to know if $B$ is invertible, this information was never given..

Thanks in advance for any help!

Answer 1

From $I=AX^{-1}B-AB$ you get also $I=A(X^{-1}B-B)=A(X^{-1}-I)B$.

Then, $B$ is also invertible and $A^{-1}B^{-1}=X^{-1}-I$.

Can you finish?

Answer 2

Hint: from the original equation: $$(A-AX)^{-1}=X^{-1}B \implies B = X(A-AX)^{-1}$$

Now, take $\det$ on both sides. What can you conclude about $\det(B)$?