Answer
Yes
Work Step by Step
We are given the matrices:
$A=\begin{bmatrix}1&2&3\\1&3&4\\1&4&3\end{bmatrix}$
$B=\begin{bmatrix}\frac{7}{2}&-3&\frac{1}{2}\\-\frac{1}{2}&0&\frac{1}{2}\\-\frac{1}{2}&1&-\frac{1}{2}\end{bmatrix}$
In order to check if $B$ is the multiplicative inverse of $A$, we have to compute $AB$ and $BA$ and see if $AB=BA=I_3$.
Compute $AB$:
$AB=\begin{bmatrix}1&2&3\\1&3&4\\1&4&3\end{bmatrix}\begin{bmatrix}\frac{7}{2}&-3&\frac{1}{2}\\-\frac{1}{2}&0&\frac{1}{2}\\-\frac{1}{2}&1&-\frac{1}{2}\end{bmatrix}$
$=\begin{bmatrix}\frac{7}{2}-1-\frac{3}{2}&-3+0+3&\frac{1}{2}+1-\frac{3}{2}\\\frac{7}{2}-\frac{3}{2}-2&-3+0+4&\frac{1}{2}+\frac{3}{2}-2\\\frac{7}{2}-2-\frac{3}{2}&-3+0+3&\frac{1}{2}+2-\frac{3}{2}\end{bmatrix}$
$=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I_3$
Compute $BA$:
$BA=\begin{bmatrix}\frac{7}{2}&-3&\frac{1}{2}\\-\frac{1}{2}&0&\frac{1}{2}\\-\frac{1}{2}&1&-\frac{1}{2}\end{bmatrix}\begin{bmatrix}1&2&3\\1&3&4\\1&4&3\end{bmatrix}$
$=\begin{bmatrix}\frac{7}{2}-3+\frac{1}{2}&7-9+2&\frac{21}{2}-12+\frac{3}{2}\\-\frac{1}{2}+0+\frac{1}{2}&-1+0+2&-\frac{3}{2}+0+\frac{3}{2}\\-\frac{1}{2}+1-\frac{1}{2}&-1+3-2&-\frac{3}{2}+4-\frac{3}{2}\end{bmatrix}=I_3$
Because $AB=BA=I_3$, $B$ is the multiplicative inverse of $A$.
