For getting the inverse of the given matrix $A$ by row elementary operations we may write the given matrix as
$$
A=I A
$$
(i) $\because\left[\begin{array}{ccc}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$ A
Apply: $\mathrm{R}_2 \rightarrow \mathrm{R}_2+\mathrm{R}_1$
$$
\Rightarrow\left[\begin{array}{ccc}
2 & -1 & 3 \\
-3 & 2 & 4 \\
-3 & 2 & 3
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
1 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \mathrm{A}
$$
Apply: $\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_2$
$\Rightarrow\left[\begin{array}{ccc}2 & -1 & 3 \\ -3 & 2 & 4 \\ 0 & 0 & -1\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 1\end{array}\right] A$ Apply: $\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2$
$$
\Rightarrow\left[\begin{array}{ccc}
-1 & 1 & 7 \\
-3 & 2 & 4 \\
0 & 0 & -1
\end{array}\right]=\left[\begin{array}{ccc}
2 & 1 & 0 \\
1 & 1 & 0 \\
-1 & -1 & 1
\end{array}\right] \mathrm{A}
$$
Apply: $\mathrm{R}_2 \rightarrow \mathrm{R}_2-3 \mathrm{R}_1$
$$
\Rightarrow\left[\begin{array}{ccc}
-1 & 1 & 7 \\
0 & -1 & -17 \\
0 & 0 & -1
\end{array}\right]=\left[\begin{array}{ccc}
2 & 1 & 0 \\
-5 & -2 & 0 \\
-1 & -1 & 1
\end{array}\right] \mathrm{A}
$$
Apply: $\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2$ and $\mathrm{R}_3 \rightarrow-1 . \mathrm{R}_3$
$$
\Rightarrow\left[\begin{array}{ccc}
-1 & 0 & -10 \\
0 & -1 & -17 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
-3 & -1 & 0 \\
-5 & -2 & 0 \\
1 & 1 & -1
\end{array}\right] \mathrm{A}
$$
Apply: $\mathrm{R}_1 \rightarrow \mathrm{R}_1+10 \mathrm{R}_3$ and $\mathrm{R}_2 \rightarrow \mathrm{R}_2+17 \mathrm{R}_3$
$$
\Rightarrow\left[\begin{array}{ccc}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
7 & 9 & -10 \\
12 & 15 & -17 \\
1 & 1 & -1
\end{array}\right] \mathrm{A}
$$
Apply : $\mathrm{R}_1 \rightarrow 1 \mathrm{R}_1$ and $\mathrm{R}_2 \rightarrow-1 \mathrm{R}_2$
$$
\Rightarrow\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
-7 & -9 & 10 \\
-12 & -15 & 17 \\
1 & 1 & -1
\end{array}\right] \mathrm{A}
$$
Therefore, the inverse of $A$ is $\left[\begin{array}{ccc}-7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1\end{array}\right]$
(ii) $\therefore\left[\begin{array}{ccc}2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & -1\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \mathrm{A}$
Apply: $\mathrm{R}_2 \rightarrow \mathrm{R}_2+\mathrm{R}_3$ and $\mathrm{R}_1 \rightarrow \mathrm{R}_1-2 \mathrm{R}_3$
$$
\Rightarrow\left[\begin{array}{ccc}
0 & 1 & -1 \\
0 & -1 & 1 \\
1 & 1 & -1
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & -2 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{array}\right] \mathrm{A}
$$
Apply: $\mathrm{R}_2 \rightarrow \mathrm{R}_2+\mathrm{R}_1$
$$
\Rightarrow\left[\begin{array}{ccc}
0 & 1 & -1 \\
0 & 0 & 0 \\
1 & 1 & -1
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & -2 \\
1 & 1 & -1 \\
0 & 0 & 1
\end{array}\right] \mathrm{A}
$$
Since, second row of the matrix on LHS is containing all zeroes, so we can say that inverse of matrix A does not exist.