Answer
See below
Work Step by Step
$\begin{bmatrix}
2 &1 &3\\1 &-1 &2\\3 & 3 &4
\end{bmatrix} \approx \begin{bmatrix}
1 &-1 &2\\2 &1 &3\\3 & 3 &4
\end{bmatrix} \approx \begin{bmatrix}
1 &-1 &2\\0 & 3 & -1\\0 & 6 &-2
\end{bmatrix} \approx \begin{bmatrix}
1 &-1 &2\\0 & 1 & -\frac{1}{3}\\0 & 6 &-2
\end{bmatrix} \approx \begin{bmatrix}
1 & 0 & \frac{5}{3}\\0 & 1 & -\frac{1}{3}\\0 & 0 & 0
\end{bmatrix} $
Since matrix does not have third singular point, hence $A^{-1}$ does not exist.
