Answer
Let $\mathrm{A}$ and $\mathrm{B}$ be square matrices such that $\mathrm{A}\mathrm{B}=\mathrm{I}.$
Then both are invertible and $\mathrm{B}=\mathrm{A}^{-1}$ , $\mathrm{A}=\mathrm{B}^{-1}$.
Work Step by Step
Let $\mathrm{A}$ and $\mathrm{B}$ be square matrices such that $\mathrm{A}\mathrm{B}=\mathrm{I}.$
By Th.8,
$\mathrm{k}.\quad$ There is an $n\times n$ matrix $\mathrm{B}$ such that $A\mathrm{B}=I$.
and
$\mathrm{a}. \quad A$ is an invertible matrix.
are both true.
Now that we know that $\mathrm{A}^{-1 }$exists, multiply
$\mathrm{A}\mathrm{B}=\mathrm{I}\quad$from the left with $\mathrm{A}^{-1}$:
$\mathrm{A}^{-1}\mathrm{A}\mathrm{B}=\mathrm{A}^{-1}\mathrm{I}$
$\mathrm{I}\mathrm{B}=\mathrm{A}^{-1}$
$\mathrm{B}=\mathrm{A}^{-1}$
Applying Th.8 for the square matrix B,
$\mathrm{j}.\quad$ There is an $n\times n$ matrix $\mathrm{A}$ such that $\mathrm{A}\mathrm{B}=I$.
and
$\mathrm{a}. \quad\mathrm{B}$ is an invertible matrix.
are both true, so we know that $\mathrm{B}^{-1}$exists.
Multiply
$\mathrm{A}\mathrm{B}=\mathrm{I}\quad$from the right with $\mathrm{B}^{-1}$:
$\mathrm{A}\mathrm{B}\mathrm{B}^{-1}=\mathrm{I}\mathrm{B}^{-1}$
$\mathrm{A}\mathrm{I}=\mathrm{B}^{-1}$
$\mathrm{A}=\mathrm{B}^{-1}$
