Show that
for nxn matrices, if $\mathbf{AB}$ is invertible, then $\mathbf A$ and $\mathbf B$ are also invertible.
Theorem.
A square matrix $\mathbf{A}$ is invertible $\iff$ $\mathbf x=\mathbf 0$ is the only solution of the matrix equation $\mathbf{Ax}=\mathbf 0$
Corollary.
For any square matrix $\mathbf A$ and $\mathbf B$,
\begin{align*} \mathbf{AB}=\mathbf I \iff \mathbf{BA}=\mathbf I \end{align*}
Assume $\mathbf{AB=I}$.
Let $\mathbf{Ax=0}$ be a matrix equation.
Then
\begin{align*} \mathbf{Ax=0} &\Rightarrow \mathbf{BAx=B0} \\ &\Rightarrow \mathbf{Ix=0} \\ &\Rightarrow \mathbf{x=0}. \end{align*}Thus, by the above theorem, $\mathbf A$ is invertible.
Hence
\begin{align*} \mathbf{AB=BA=I} \end{align*}
Solution to the problem
Since $\mathbf{AB}$ is invertible, $\exists \mathbf{C} \text{ such that } \mathbf{ABC=CAB=I}$.
Then
\begin{align*} \mathbf{ABC=A(BC)=I} \Rightarrow \mathbf{A(BC)=(BC)A=I} \Rightarrow \exists\mathbf{A}^{-1} \text{ and } \mathbf{A}^{-1}=\mathbf{BC}\\ \mathbf{CAB=(CA)B=I} \Rightarrow \mathbf{(CA)B=B(CA)=I} \Rightarrow \exists\mathbf{B}^{-1} \text{ and } \mathbf{B}^{-1}=\mathbf{CA}\end{align*}
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