Answer
The product of $ AB=\left[ \begin{matrix}
-3 & -4 \\
0 & 1 \\
\end{matrix} \right]$, product of $ BA=\left[ \begin{matrix}
-3 & 0 \\
-4 & 1 \\
\end{matrix} \right]$ and $ B $ is not the multiplicative inverse of $ A $ i.e. $ B\ne {{A}^{-1}}$
Work Step by Step
The given expression is:
$ A=\left[ \begin{matrix}
-2 & -1 \\
-1 & 1 \\
\end{matrix} \right],B=\left[ \begin{matrix}
1 & 1 \\
1 & 2 \\
\end{matrix} \right]$
Now, we will compute the matrix as $\left[ AB \right]$
$\begin{align}
& AB=\left[ \begin{matrix}
-2 & -1 \\
-1 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 1 \\
1 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\left( -2 \right)\times 1+\left( -1 \right)\times 1 & \left( -2 \right)\times 1+\left( -1 \right)\times 2 \\
\left( -1 \right)\times 1+1\times 1 & \left( -1 \right)\times 1+1\times 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-3 & -4 \\
0 & 1 \\
\end{matrix} \right]
\end{align}$
Now we will compute the matrix as $\left[ BA \right]$
$\begin{align}
& BA=\left[ \begin{matrix}
1 & 1 \\
1 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
-2 & -1 \\
-1 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1\times \left( -2 \right)+1\times \left( -1 \right) & 1\times \left( -1 \right)+1\times 1 \\
1\times \left( -2 \right)+2\times \left( -1 \right) & 1\times \left( -1 \right)+2\times 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-3 & 0 \\
-4 & 1 \\
\end{matrix} \right]
\end{align}$
Neither matrix $ AB $ nor matrix $ BA $ are equal to the identity matrix.
Now, we will compute the matrix $ B={{A}^{-1}}$
$ B=\left[ \begin{matrix}
1 & 1 \\
1 & 2 \\
\end{matrix} \right]$
And, $ A=\left[ \begin{matrix}
-2 & -1 \\
-1 & 1 \\
\end{matrix} \right]$
Determine the inverse of matrix $ A $.
Now, using the 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, $\begin{align}
& a=-2 \\
& b=-1 \\
& c=-1 \\
& d=1 \\
\end{align}$
So, $\begin{align}
& {{A}^{-1}}=\frac{1}{\left| -2-1 \right|}\left[ \begin{matrix}
1 & 1 \\
1 & -2 \\
\end{matrix} \right] \\
& =\frac{1}{-3}\left[ \begin{matrix}
1 & 1 \\
1 & -2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{1}{-3} & \frac{1}{-3} \\
\frac{1}{-3} & \frac{2}{3} \\
\end{matrix} \right]
\end{align}$
Therefore, no $ B $ is not the multiplicative inverse of $ A $.