Given square matrices $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{X}$... ...such that $\mathbf{A}$, $\mathbf{X}$ and $\mathbf{A} - \mathbf{A}\mathbf{X}$ are invertible. Now, suppose
$$(\textbf{A} - \textbf{A}\textbf{X})^{-1} = \textbf{X}^{-1}\textbf{B}$$
I've proven that B is invertible as follows:


*

*$\because \mathbf{A} - \mathbf{A}\mathbf{X}$ is invertible, and given
that $(\textbf{A} - \textbf{A}\textbf{X})^{-1} =
\textbf{X}^{-1}\textbf{B}$,

*$\implies \textbf{X}^{-1}\textbf{B}$ is also invertible, and

*$((\textbf{A} - \textbf{A}\textbf{X})^{-1})^{-1} = (\textbf{X}^{-1}\textbf{B})^{-1}$

*$\mathbf{A} - \mathbf{A}\mathbf{X} = (\textbf{X}^{-1})^{-1}\textbf{B}^{-1}$

*$\mathbf{A} - \mathbf{A}\mathbf{X} = \textbf{X}\textbf{B}^{-1}$.

*Therefore there must exist a matrix $\textbf{B}^{-1}$, and
$\textbf{B}$ is invertible.


Is this 'proof' of sorts sufficient, and logically sound? If not, how could it be improved?
Furthermore, how might I go about solving for $\textbf{X}$ in terms of $\textbf{A}$ and $\textbf{B}$? From the last line in the proof above, I've gotten to:
$$\textbf{A}\textbf{X} + \textbf{X}\textbf{B}^{-1} = \textbf{A}$$
$$\textbf{A}^{-1}\textbf{A}\textbf{X} + \textbf{A}^{-1}\textbf{X}\textbf{B}^{-1} = \textbf{A}^{-1}\textbf{A}$$
$$\textbf{I}\textbf{X} + \textbf{A}^{-1}\textbf{X}\textbf{B}^{-1} = \textbf{I}$$
$$\textbf{X} + \textbf{A}^{-1}\textbf{X}\textbf{B}^{-1} = \textbf{I}$$
 but I'm stuck. I believe I have to somehow 'pull out' the $\textbf{X}$, but I was told by my lecturer that the order of matrix multiplication is important, and is not commutative, unlike normal algebraic/arithmetic multiplication, so I can't just 'factorise out' the $\textbf{X}$ in the middle term.
Thanks very much.
 A: Your proof is flawed because you are assuming $\mathbf{B}^{-1}$ exists, which is what you must prove.
The proof that $\mathbf{B}$ is invertible is easier:
$$
\mathbf{B}=\mathbf{X}(\mathbf{A}-\mathbf{A}\mathbf{X})^{-1}
$$
is the product of two invertible matrices, so it is invertible. Indeed, if $\mathbf{C}$ and $\mathbf{D}$ are invertible, it is readily seen that $\mathbf{D}^{-1}\mathbf{C}^{-1}$ is the inverse of $\mathbf{C}\mathbf{D}$.
Now, how to express $\mathbf{X}$ in terms of $\mathbf{A}$ and $\mathbf{B}$ (and inverses thereof)?
You make another error: from $(\mathbf{A}-\mathbf{A}\mathbf{X})^{-1}=\mathbf{X}^{-1}\mathbf{B}$ you can deduce that
$$
\mathbf{A}-\mathbf{A}\mathbf{X}=(\mathbf{X}^{-1}\mathbf{B})^{-1}=
\mathbf{B}^{-1}\mathbf{X}
$$
(not the other way around). Therefore
$$
(\mathbf{A}+\mathbf{B}^{-1})\mathbf{X}=\mathbf{A}
$$
Since the right hand side is invertible and $\mathbf{X}$ is invertible, also $\mathbf{A}+\mathbf{B}^{-1}$ is and so
$$
\mathbf{X}=(\mathbf{A}+\mathbf{B}^{-1})^{-1}\mathbf{A}
$$
