Answer
The product of $ AB={{I}_{4}}$, product of $ BA={{I}_{4}}$ and $ B $ is the multiplicative inverse of $ A $ $ B={{A}^{-1}}$.
Work Step by Step
The given expression is
$ A=\left[ \begin{matrix}
1 & -2 & 1 & 0 \\
0 & 1 & -2 & 1 \\
0 & 0 & 1 & -2 \\
0 & 0 & 0 & 1 \\
\end{matrix} \right],B=\left[ \begin{matrix}
1 & 2 & 3 & 4 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 1 \\
\end{matrix} \right]$
Now, we will compute the matrix as $\left[ AB \right]$
$\begin{align}
& AB=\left[ \begin{matrix}
1 & -2 & 1 & 0 \\
0 & 1 & -2 & 1 \\
0 & 0 & 1 & -2 \\
0 & 0 & 0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 2 & 3 & 4 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{4}}
\end{align}$
Now, we will compute the matrix as $\left[ BA \right]$
$\begin{align}
& BA=\left[ \begin{matrix}
1 & 2 & 3 & 4 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & -2 & 1 & 0 \\
0 & 1 & -2 & 1 \\
0 & 0 & 1 & -2 \\
0 & 0 & 0 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{4}}
\end{align}$
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 & 2 & 3 & 4 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 1 \\
\end{matrix} \right]$ And $ A=\left[ \begin{matrix}
1 & -2 & 1 & 0 \\
0 & 1 & -2 & 1 \\
0 & 0 & 1 & -2 \\
0 & 0 & 0 & 1 \\
\end{matrix} \right]$
Determine the inverse of matrix A.
Now, reduce matrix to row form to get, $ A={{\left[ \begin{matrix}
1 & -2 & 1 & 0 \\
0 & 1 & -2 & 1 \\
0 & 0 & 1 & -2 \\
0 & 0 & 0 & 1 \\
\end{matrix} \right]}^{-1}}$
Augment with an identity matrix.
$ A=\left[ \begin{matrix}
1 & -2 & 1 & 0 & 1 & 0 & 0 & 0 \\
0 & 1 & -2 & 1 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & -2 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\
\end{matrix} \right]$
Swap matrix rows:
$\begin{align}
& {{R}_{3}}\to {{R}_{3}}+2\times {{R}_{4}} \\
& {{R}_{2}}\to {{R}_{2}}-1\times {{R}_{4}}
\end{align}$
$ A=\left[ \begin{matrix}
1 & -2 & 1 & 0 & 1 & 0 & 0 & 0 \\
0 & 1 & -2 & 0 & 0 & 1 & 0 & -1 \\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 2 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\
\end{matrix} \right]$
Again
$\begin{align}
& {{R}_{2}}\to {{R}_{2}}+2\times {{R}_{3}} \\
& {{R}_{1}}\to {{R}_{1}}-1\times {{R}_{3}} \\
& {{R}_{1}}\to {{R}_{1}}+2\times {{R}_{2}} \\
\end{align}$
$\left[ \begin{matrix}
1 & 0 & 0 & 0 & 1 & 2 & 3 & 4 \\
0 & 1 & 0 & 0 & 0 & 1 & 2 & 3 \\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 2 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\
\end{matrix} \right]$
Thus, ${{A}^{-1}}=\left[ \begin{matrix}
1 & 2 & 3 & 4 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 1 \\
\end{matrix} \right]$
Therefore, matrix $ B $ is the multiplicative inverse of matrix $ A $.