Answer
The product of $ AB={{I}_{2}}$, product of $ BA={{I}_{2}}$ and $ B $ is the multiplicative inverse of $ A $ i.e. $ B={{A}^{-1}}$.
Work Step by Step
The given expression is
$ A=\left[ \begin{matrix}
4 & -3 \\
-5 & 4 \\
\end{matrix} \right],B=\left[ \begin{matrix}
4 & 3 \\
5 & 4 \\
\end{matrix} \right]$
Now, compute the matrix as $\left[ AB \right]$
$\begin{align}
& AB=\left[ \begin{matrix}
4 & -3 \\
-5 & 4 \\
\end{matrix} \right]\left[ \begin{matrix}
4 & 3 \\
5 & 4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
4\times 4+\left( -3 \right)\times 5 & 4\times 3+\left( -3 \right)\times 4 \\
\left( -5 \right)\times 4+4\times 5 & \left( -5 \right)\times 3+4\times 4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{2}}
\end{align}$
Now we will compute the matrix as $\left[ BA \right]$
$\begin{align}
& BA=\left[ \begin{matrix}
4 & 3 \\
5 & 4 \\
\end{matrix} \right]\left[ \begin{matrix}
4 & -3 \\
-5 & 4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
4\times 4+3\times \left( -5 \right) & 4\times \left( -3 \right)+3\times 4 \\
5\times 4+4\times \left( -5 \right) & 5\times \left( -3 \right)+4\times 4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{2}}
\end{align}$
Since both matrix $ AB $ and $ BA $ are equal to the identity matrix.
Now, we will compute the matrix $ B={{A}^{-1}}$.
$ B=\left[ \begin{matrix}
4 & 3 \\
5 & 4 \\
\end{matrix} \right]$
And
$ A=\left[ \begin{matrix}
4 & -3 \\
-5 & 4 \\
\end{matrix} \right]$
Determine the inverse of matrix A.
Now, using inverse formula
${{A}^{-1}}=\frac{1}{\left| ad-bc \right|}\left[ \begin{matrix}
d & -b \\
-c & a \\
\end{matrix} \right]$
Substitute the value to get, $ a=4,b=-3,c=-5,d=4$
So, $\begin{align}
& {{A}^{-1}}=\frac{1}{\left| 16-15 \right|}\left[ \begin{matrix}
4 & 3 \\
5 & 4 \\
\end{matrix} \right] \\
& =\frac{1}{1}\left[ \begin{matrix}
4 & 3 \\
5 & 4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
4 & 3 \\
5 & 4 \\
\end{matrix} \right]
\end{align}$
Yes, $ B $ is the multiplicative inverse of $ A $.