Answer
The product of $ AB={{I}_{3}}$, product of $ BA={{I}_{3}}$ and $ B $ is the multiplicative inverse of $ A $ $ B={{A}^{-1}}$.
Work Step by Step
The provided expression is
$ A=\left[ \begin{matrix}
-2 & 1 & -1 \\
-5 & 2 & -1 \\
3 & -1 & 1 \\
\end{matrix} \right],B=\left[ \begin{matrix}
1 & 0 & 1 \\
2 & 1 & 3 \\
-1 & 1 & 1 \\
\end{matrix} \right]$
Now, we will compute the matrix as $\left[ AB \right]$
$\begin{align}
& AB=\left[ \begin{matrix}
-2 & 1 & -1 \\
-5 & 2 & -1 \\
3 & -1 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 0 & 1 \\
2 & 1 & 3 \\
-1 & 1 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\left( -2 \right)\times 1+1\times 2+\left( -1 \right)\times \left( -1 \right) & \left( -2 \right)\times 0+1\times 1+\left( -1 \right)\times 1 & \left( -2 \right)\times 1+1\times 3+\left( -1 \right)\times 1 \\
\left( -5 \right)\times 1+2\times 2+\left( -1 \right)\times \left( -1 \right) & \left( -5 \right)\times 0+2\times 1+\left( -1 \right)\times 1 & \left( -5 \right)\times 1+2\times 3+\left( -1 \right)\times 1 \\
3\times 1+\left( -1 \right)\times 2+1\times \left( -1 \right) & 3\times 0+\left( -1 \right)\times 1+1\times 1 & 3\times 1+\left( -1 \right)\times 3+1\times 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{3}}
\end{align}$
Now, we will compute the matrix as $\left[ BA \right]$
$\begin{align}
& BA=\left[ \begin{matrix}
1 & 0 & 1 \\
2 & 1 & 3 \\
-1 & 1 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
-2 & 1 & -1 \\
-5 & 2 & -1 \\
3 & -1 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1\times \left( -2 \right)+0\times \left( -5 \right)+1\times 3 & 1\times 1+0\times 2+1\times \left( -1 \right) & 1\times \left( -1 \right)+0\times \left( -1 \right)+1\times 1 \\
2\times \left( -2 \right)+1\times \left( -5 \right)+3\times 3 & 2\times 1+1\times 2+3\times \left( -1 \right) & 2\times \left( -1 \right)+1\times \left( -1 \right)+3\times 1 \\
\left( -1 \right)\times \left( -2 \right)+1\times \left( -5 \right)+1\times 3 & \left( -1 \right)\times 1+1\times 2+1\times \left( -1 \right) & \left( -1 \right)\times \left( -1 \right)+1\times \left( -1 \right)+1\times 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{3}}
\end{align}$
Both matrix $ AB $ and matrix $ BA $ are equal to the identity matrix
Now, we will compute the matrix $ B={{A}^{-1}}$.
$ B=\left[ \begin{matrix}
1 & 0 & 1 \\
2 & 1 & 3 \\
-1 & 1 & 1 \\
\end{matrix} \right]$
And
$ A=\left[ \begin{matrix}
-2 & 1 & -1 \\
-5 & 2 & -1 \\
3 & -1 & 1 \\
\end{matrix} \right]$
Determine the inverse of matrix A.
Now, reduce matrix to row form:
$ A={{\left[ \begin{matrix}
-2 & 1 & -1 \\
-5 & 2 & -1 \\
3 & -1 & 1 \\
\end{matrix} \right]}^{-1}}$
Augment with an identity matrix
$ A=\left[ \begin{matrix}
-2 & 1 & -1 & 1 & 0 & 0 \\
-5 & 2 & -1 & 0 & 1 & 0 \\
3 & -1 & 1 & 0 & 0 & 1 \\
\end{matrix} \right]$
Swap matrix rows: ${{R}_{1}}\leftrightarrow {{R}_{2}}$
$ A=\left[ \begin{matrix}
-5 & 2 & -1 & 0 & 1 & 0 \\
-2 & 1 & -1 & 1 & 0 & 0 \\
3 & -1 & 1 & 0 & 0 & 1 \\
\end{matrix} \right]$
Again swap matrix row
$\begin{align}
& {{R}_{2}}\leftrightarrow {{R}_{3}} \\
& {{R}_{_{2}}}\to {{R}_{2}}-\frac{2}{5}\times {{R}_{1}} \\
\end{align}$
And
$\begin{align}
& {{R}_{3}}\to {{R}_{3}}+\frac{3}{5}\times {{R}_{1}} \\
& {{R}_{3}}\to {{R}_{3}}-1\times {{R}_{2}} \\
\end{align}$
$\left[ \begin{matrix}
-5 & 2 & -1 & 0 & 1 & 0 \\
0 & \frac{1}{5} & -\frac{3}{5} & 1 & -\frac{2}{5} & 0 \\
0 & 0 & 1 & -1 & 1 & 1 \\
\end{matrix} \right]$
$\begin{align}
& {{R}_{2}}\to {{R}_{2}}+\frac{3}{5}\times {{R}_{3}} \\
& {{R}_{1}}\to {{R}_{1}}+1\times {{R}_{3}} \\
\end{align}$
$\left[ \begin{matrix}
-5 & 2 & 0 & -1 & 2 & 1 \\
0 & \frac{1}{5} & 0 & \frac{2}{5} & \frac{1}{5} & \frac{3}{5} \\
0 & 0 & 1 & -1 & 1 & 1 \\
\end{matrix} \right]$
$\begin{align}
& {{R}_{1}}\to 5\times {{R}_{2}} \\
& {{R}_{1}}\to {{R}_{1}}-2\times {{R}_{2}} \\
& {{R}_{1}}\to -\frac{1}{5}\times {{R}_{1}}
\end{align}$
$\left[ \begin{matrix}
1 & 0 & 0 & 1 & 0 & 1 \\
0 & 1 & 0 & 2 & 1 & 3 \\
0 & 0 & 1 & -1 & 1 & 1 \\
\end{matrix} \right]$
${{A}^{-1}}=\left[ \begin{matrix}
1 & {} & 1 \\
2 & 1 & 3 \\
-1 & 1 & 1 \\
\end{matrix} \right]$
Thus, $ B={{A}^{-1}}$