Answer
Matrix $ B $ is a multiplicative inverse of the matrix $ A $.
Work Step by Step
Find the product of $ AB $ as follows:
$\begin{align}
& AB=\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 2 & -7 \\
0 & -1 & 4 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 4 & 7 \\
0 & 1 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1\left( 1 \right)+0\left( 0 \right)+0\left( 0 \right) & 1\left( 0 \right)+0\left( 4 \right)+0\left( 1 \right) & 1\left( 0 \right)+0\left( 7 \right)+0\left( 2 \right) \\
0\left( 1 \right)+2\left( 0 \right)-7\left( 0 \right) & 0\left( 0 \right)+2\left( 4 \right)-7\left( 1 \right) & 0\left( 0 \right)+2\left( 7 \right)-7\left( 2 \right) \\
0\left( 1 \right)-1\left( 0 \right)+4\left( 0 \right) & 0\left( 0 \right)-1\left( 4 \right)+4\left( 1 \right) & 0\left( 0 \right)-1\left( 7 \right)+4\left( 2 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 8-7 & 14-14 \\
0 & -4+4 & -7+8 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right]
\end{align}$
Therefore, $ AB=\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right]$
Next we will find the product of $ BA $ as follows:
$\begin{align}
& BA=\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 4 & 7 \\
0 & 1 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 2 & -7 \\
0 & -1 & 4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1\left( 1 \right)+0\left( 0 \right)+0\left( 0 \right) & 1\left( 0 \right)+0\left( 2 \right)+0\left( -1 \right) & 1\left( 0 \right)+0\left( -7 \right)+0\left( 4 \right) \\
0\left( 1 \right)+4\left( 0 \right)+7\left( 0 \right) & 0\left( 0 \right)+4\left( 2 \right)+7\left( -1 \right) & 0\left( 0 \right)+4\left( -7 \right)+7\left( 4 \right) \\
0\left( 1 \right)+1\left( 0 \right)+2\left( 0 \right) & 0\left( 0 \right)+1\left( 2 \right)+2\left( -1 \right) & 0\left( 0 \right)+1\left( -7 \right)+2\left( 4 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 8-7 & 28-28 \\
0 & 2-2 & -7+8 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right]
\end{align}$
As $ AB=BA=I $, where $ I $ is an $3\times 3$ identity matrix, therefore, the given matrix $ B $ is a multiplicative inverse of the matrix $ A $.
