Answer
$B$ is not the multiplicative inverse of $A$.
Work Step by Step
(See p.629)
Let A be an $n\times n$ square matrix. If there is a square matrix $A^{-1}$ such that
$AA^{-1}=I_{n}$ and $A^{-1}A=I_{n},$
then $A^{-1}$ is the multiplicative inverse of $A$.
--------------
The products AB and BA should both be $I_{2}=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right].$
$AB=\left[\begin{array}{ll}
-4(-2)+0(0) & -4(4)+0(1)\\
1(-2)+3(0) & 1(4)+3(1)
\end{array}\right]==\left[\begin{array}{ll}
8 & ...\\
... & ...
\end{array}\right]\neq I_{2}$
Since at least one of the products is NOT $I_{2}$,
$B$ is not the multiplicative inverse of $A$.
