Answer
Applying Th.8(j. and a.) and the associativity of matrix multiplication,
$\mathrm{B}$ is invertible.
Work Step by Step
Let $\mathrm{M}$ be the inverse of $AB$.
Then, by definition of inverse matrix,
$\mathrm{M}(\mathrm{A}\mathrm{B})=\mathrm{I}$
Since matrix multiplication is associative
$\mathrm{M}(\mathrm{A}\mathrm{B})=(\mathrm{M}\mathrm{A})\mathrm{B}=\mathrm{I}$.
What we have is
$\mathrm{B}$ is a square $n\times n$ matrix, and
$\mathrm{j}.\quad$ There is an $n\times n$ matrix $C=\mathrm{M}\mathrm{A}$ such that $C\mathrm{B}=I$.
is true,
which means that, by Th.8,
$\mathrm{a}. \quad\mathrm{B}$ is an invertible matrix.
is also true.