How to solve $(A-AX)^{-1}=X^{-1}B$ and prove $B$ is invertible? So this is a homework problem. So I'll give what I got.
Suppose $A,$ $\ B$ & $X$ are $n$ by $n$ matrices with $A,$ $\ X$ & $A-AX$ invertible. Now suppose that $$(1) \quad (A-AX)^{-1}=X^{-1}B$$
$A)$ Explain why $B$ is invertible.
$B)$ Solve for $X$.
Part A:
Multiply both sides of $(1)$ by $B^{-1}$ to get,
$$(2) \quad (A-AX)^{-1}B^{-1}=X^{-1}$$
Now Multiply $(2)$ by $(A-AX)^{-1}$ to get,
$$(3) \quad B^{-1}=(A-AX)X^{-1}$$
$(A-AX)$ exists, and the existence of $X^{-1}$ is a given. The matrices are of the same size so matrix multiplication is defined. Therefore $B^{-1}$ exists.
Part B
From $(3)$ use the fact that,
$C^{-1}D^{-1}=(DC)^{-1}$
To manipulate the RHS of $(3)$ into,
$$(4) \quad B^{-1}=(X(A-AX)^{-1})^{-1}$$
$$\Rightarrow B^{-1}=(X(A(I-X))^{-1})^{-1}$$
$$\Rightarrow B^{-1}=(X(I-X)^{-1}(A)^{-1})^{-1}$$
$$\Rightarrow B^{-1}=(((I-X)X^{-1})^{-1}A^{-1})^{-1}$$
$$\Rightarrow B^{-1}=A(I-X)X^{-1}$$
$$\Rightarrow B^{-1}=A(X^{-1}-I)$$
$$\Rightarrow A^{-1}B^{-1}=(X^{-1}-I)$$
$$\Rightarrow X=(BA)^{-1}+I$$
Seems correct to me. But I'm a bit bias. Is this correct? If not, where's the error?
 A: Your part A is problematic. You can't start with $B^{-1}$ until you have proved that $B$ is invertible.
Approach 1: propose a candidate and show that it works. Let $M=(A-AX)X^{-1}$. Then
$$
MB=(A-AX)X^{-1}B=(A-AX)(X^{-1}B)=(A-AX)(A-AX)^{-1}=I,
$$
and then use this result to infer that $BM=I$ also.
Approach 2: Take determinant on both sides of $(A-AX)^{-1}=X^{-1}B$ to infer that $\det(B)=\det X/\det(A-AX)\neq 0$.
A: Your answer to Part A is not right, because by multiplying by $B^{-1}$, you assumed it exists. Instead, starting from
$$ (A-AX)^{-1}=X^{-1}B, \tag{1}$$
multiply by $X$ on the left, which gives
$$ B= X(A-AX)^{-1}. $$
Then $B$ is a product of invertible matrices (according to the question) and so $B$ is invertible.
A shorter way to do Part B is to multiply the above formula on the right by $A-AX$. Then
$$ BA(I-X)=X, $$
or
$$ BA=(I+BA)X, $$
and then
$$ X=(I+BA)^{-1}BA  $$
(there are probably other ways of writing this, though).
(And by the way, when multiplying by a non-commutative element, you should state on which side you are placing it, since otherwise it can be ambiguous.)
A: In part a) you cannot multiply by $B^{-1}$ because you don't know whether it is invertible or not. You need to write the solution for part a) as:
$$(A-AX)^{-1}=X^{-1}B$$ implies that $$X(A-AX)^{-1}=B=(X^{-1})^{-1}(A-AX)^{-1}=((A-AX)X^{-1})^{-1}$$ so the matrix $B$ is the inverse of matrix $(A-AX)X^{-1}$.
$$$$ Hence B is invertible.
$$$$ The part B seems to be right!
