Answer
The product of $ AB={{I}_{2}}$, product of $ BA={{I}_{2}}$ and $ B $ is the multiplicative inverse of $ A $ $ B={{A}^{-1}}$.
Work Step by Step
The given expression is
$ A=\left[ \begin{matrix}
4 & 5 \\
2 & 3 \\
\end{matrix} \right],B=\left[ \begin{matrix}
\frac{3}{2} & -\frac{5}{2} \\
-1 & 2 \\
\end{matrix} \right]$
Now, we will compute the matrix as $\left[ AB \right]$
$\begin{align}
& AB=\left[ \begin{matrix}
4 & 5 \\
2 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
\frac{3}{2} & -\frac{5}{2} \\
-1 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
4\times \frac{3}{2}+5\times \left( -1 \right) & 4\times \left( -\frac{5}{2} \right)+5\times 2 \\
2\times \frac{3}{2}+3\times \left( -1 \right) & 2\times \left( -\frac{5}{2} \right)+3\times 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{2}}
\end{align}$
Now, we will compute the matrix as $\left[ BA \right]$
$\begin{align}
& BA=\left[ \begin{matrix}
\frac{3}{2} & -\frac{5}{2} \\
-1 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
4 & 5 \\
2 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{3}{2}\times 4+\left( -\frac{5}{2} \right)\times 2 & \frac{3}{2}\times 5+\left( -\frac{5}{2} \right)\times 3 \\
\left( -1 \right)\times 4+2\times 2 & \left( -1 \right)\times 5+2\times 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{2}}
\end{align}$
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}
\frac{3}{2} & -\frac{5}{2} \\
-1 & 2 \\
\end{matrix} \right]$
And, $ A=\left[ \begin{matrix}
4 & 5 \\
2 & 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 values to get, $\begin{align}
& a=4 \\
& b=5 \\
& c=2 \\
& d=3 \\
\end{align}$
So, $\begin{align}
& {{A}^{-1}}=\frac{1}{\left| 12-10 \right|}\left[ \begin{matrix}
3 & -5 \\
-2 & 4 \\
\end{matrix} \right] \\
& =\frac{1}{2}\left[ \begin{matrix}
3 & -5 \\
-2 & 4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{3}{2} & \frac{-5}{2} \\
-1 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{3}{2} & \frac{-5}{2} \\
-1 & 2 \\
\end{matrix} \right]
\end{align}$
Therefore, $ B={{A}^{-1}}$.
