Answer
The product of $ AB=\left[ \begin{matrix}
8 & -16 \\
-2 & 7 \\
\end{matrix} \right]$, product of $ BA=\left[ \begin{matrix}
12 & 12 \\
1 & 3 \\
\end{matrix} \right]$ and $ B $ is not multiplicative inverse of $ A $ $ B\ne {{A}^{-1}}$
Work Step by Step
The given expression is
$ A=\left[ \begin{matrix}
-4 & 0 \\
1 & 3 \\
\end{matrix} \right],B=\left[ \begin{matrix}
-2 & 4 \\
0 & 1 \\
\end{matrix} \right]$
Now, we will compute the matrix as $\left[ AB \right]$
$\begin{align}
& AB=\left[ \begin{matrix}
-4 & 0 \\
1 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
-2 & 4 \\
0 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\left( -4 \right)\times 1+0\times 0 & \left( -4 \right)\times 4+0\times 1 \\
1\times \left( -2 \right)+3\times 0 & 1\times 4+3\times 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
8 & -16 \\
-2 & 7 \\
\end{matrix} \right]
\end{align}$
And, we will compute the matrix as $\left[ BA \right]$
$\begin{align}
& BA=\left[ \begin{matrix}
-2 & 4 \\
0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
-4 & 0 \\
1 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\left( -2 \right)\times \left( -4 \right)+4\times 1 & \left( -2 \right)\times 0+4\times 3 \\
0\times \left( -4 \right)+1\times 1 & 0\times 0+1\times 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
12 & 12 \\
1 & 3 \\
\end{matrix} \right]
\end{align}$
Since both the matrix $ AB=\left[ \begin{matrix}
8 & -16 \\
-2 & 7 \\
\end{matrix} \right]$ and $ BA=\left[ \begin{matrix}
12 & 12 \\
1 & 3 \\
\end{matrix} \right]$
Now, we will compute the matrix $ B={{A}^{-1}}$.
$ B=\left[ \begin{matrix}
-2 & 4 \\
0 & 1 \\
\end{matrix} \right]$
And
$ A=\left[ \begin{matrix}
-4 & 0 \\
1 & 3 \\
\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=0,c=1,d=3$
So, $\begin{align}
& {{A}^{-1}}=\frac{1}{\left| -12-0 \right|}\left[ \begin{matrix}
3 & -0 \\
-1 & -4 \\
\end{matrix} \right] \\
& =\frac{1}{-12}\left[ \begin{matrix}
3 & -0 \\
-1 & -4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{3}{-12} & \frac{0}{12} \\
\frac{1}{12} & \frac{4}{12} \\
\end{matrix} \right]
\end{align}$
Therefore, $ B\ne {{A}^{-1}}$
