Answer
The product of $ AB={{I}_{3}}$, product of $ BA={{I}_{3}}$ and $ B $ is the multiplicative inverse of $ A $ $ B={{A}^{-1}}$.
The given expression is
$ A=\left[ \begin{matrix}
0 & 2 & 0 \\
3 & 3 & 2 \\
2 & 5 & 1 \\
\end{matrix} \right],B=\left[ \begin{matrix}
-3.5 & -1 & 2 \\
0.5 & 0 & 0 \\
4.5 & 2 & -3 \\
\end{matrix} \right]$
Work Step by Step
Now, we will compute the matrix as $\left[ AB \right]$
$\begin{align}
& AB=\left[ \begin{matrix}
0 & 2 & 0 \\
3 & 3 & 2 \\
2 & 5 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
-3.5 & -1 & 2 \\
0.5 & 0 & 0 \\
4.5 & 2 & -3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
0\times \left( -3.5 \right)+2\times 0.5+0\times 4.5 & 0\times \left( -1 \right)+2\times 0+0\times 2 & 0\times 2+2\times 0+0\times -3 \\
3\times \left( -3.5 \right)+3\times 0.5+2\times 4.5 & 3\times \left( -1 \right)+3\times 0+2\times 2 & 3\times 2+3\times 0+2\times \left( -3 \right) \\
2\times \left( -3.5 \right)+5\times 0.5+1\times 4.5 & 2\times \left( -1 \right)+5\times 0+1\times 2 & 2\times 2+5\times 0+1\times \left( -3 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{3}}
\end{align}$
Now, we will compute the matrix as $\left[ BA \right]$
$\begin{align}
& BA=\left[ \begin{matrix}
-3.5 & -1 & 2 \\
0.5 & 0 & 0 \\
4.5 & 2 & -3 \\
\end{matrix} \right]\left[ \begin{matrix}
0 & 2 & 0 \\
3 & 3 & 2 \\
2 & 5 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\left( -3.5 \right)\times 0+\left( -1 \right)\times 3+2\times 2 & \left( -3.5 \right)\times 2+\left( -1 \right)\times 3+2\times 5 & \left( -3.5 \right)\times 0+\left( -1 \right)\times 2+2\times 1 \\
0.5\times 0+0\times 3+0\times 2 & 0.5\times 2+0\times 3+0\times 5 & 0.5\times 0+0\times 2+0\times 1 \\
4.5\times 0+2\times 3+\left( -3 \right)\times 2 & 4.5\times 2+2\times 3+\left( -3 \right)\times 5 & 4.5\times 0+2\times 2+\left( -3 \right)\times 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{3}}
\end{align}$
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}
-3.5 & -1 & 2 \\
0.5 & 0 & 0 \\
4.5 & 2 & -3 \\
\end{matrix} \right]$
And
$ A=\left[ \begin{matrix}
0 & 2 & 0 \\
3 & 3 & 2 \\
2 & 5 & 1 \\
\end{matrix} \right]$
Determine the inverse of matrix A.
Now, Reduce matrix to row form:
$ A={{\left[ \begin{matrix}
0 & 2 & 0 \\
3 & 3 & 2 \\
2 & 5 & 1 \\
\end{matrix} \right]}^{-1}}$
Augment with an identity matrix, $ A=\left[ \begin{matrix}
0 & 2 & 0 & 1 & 0 & 0 \\
3 & 3 & 2 & 0 & 1 & 0 \\
2 & 5 & 1 & 0 & 0 & 1 \\
\end{matrix} \right]$
Swap matrix rows:
$\begin{align}
& {{R}_{1}}\leftrightarrow {{R}_{2}} \\
& {{R}_{3}}\to {{R}_{3}}-\frac{2}{3}\times {{R}_{1}} \\
& {{R}_{2}}\leftrightarrow {{R}_{3}} \\
& {{R}_{3}}\to {{R}_{3}}-\frac{2}{3}\times {{R}_{2}}
\end{align}$
$ A=\left[ \begin{matrix}
3 & 3 & 2 & 0 & 1 & 0 \\
0 & 3 & -\frac{1}{3} & 0 & -\frac{2}{3} & 1 \\
0 & 0 & \frac{2}{9} & 1 & \frac{4}{9} & -\frac{2}{3} \\
\end{matrix} \right]$
Again
$\begin{align}
& {{R}_{3}}\to \frac{9}{2}\times {{R}_{3}} \\
& {{R}_{2}}\to {{R}_{2}}+\frac{1}{3}\times {{R}_{3}} \\
& {{R}_{1}}\to {{R}_{1}}-2\times {{R}_{3}} \\
\end{align}$
$\left[ \begin{matrix}
3 & 3 & 0 & -9 & -3 & 6 \\
0 & 3 & 0 & \frac{3}{2} & 0 & 0 \\
0 & 0 & 1 & \frac{9}{2} & 2 & -3 \\
\end{matrix} \right]$
And, $\begin{align}
& {{R}_{2}}\to \frac{1}{3}\times {{R}_{2}} \\
& {{R}_{1}}\to {{R}_{1}}-3\times {{R}_{2}} \\
& {{R}_{1}}\to \frac{1}{3}\times {{R}_{1}} \\
\end{align}$
$\left[ \begin{matrix}
1 & 0 & 0 & -\frac{7}{2} & -1 & 2 \\
0 & 1 & 0 & \frac{1}{2} & 0 & 0 \\
0 & 0 & 1 & \frac{9}{2} & 2 & -3 \\
\end{matrix} \right]$
Thus, ${{A}^{-1}}=\left[ \begin{matrix}
-3.5 & -1 & 2 \\
0.5 & 0 & 0 \\
4.5 & 2 & -3 \\
\end{matrix} \right]$
Therefore, matrix $ B $ is the multiplicative inverse of matrix $ A $
