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