Solution:
We have $A=\left(\begin{array}{ccc}1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4\end{array}\right)$
To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:
$(A \mid I) \rightarrow\left(I \mid A^{-1}\right)$
Let’s proceed
$\begin{array}{c}
\left(\begin{array}{lll|ccc}
1 & 2 & -3 & 1 & 0 & 0 \\
2 & 3 & 2 & 0 & 1 & 0 \\
3 & -3 & -4 & 0 & 0 & 1
\end{array}\right) \\
\stackrel{R_{3} \rightarrow R_{3}-R_{2}}{\longrightarrow}\left(\begin{array}{lll|ccc}
1 & 2 & -3 & 1 & 0 & 0 \\
2 & 3 & 2 & 0 & 1 & 0 \\
1 & -6 & -3 & -1 & 1
\end{array}\right) \\
\stackrel{R_{2} \rightarrow R_{2}-R_{1}}{\longrightarrow} & \left.\begin{array}{lll|ccc}
1 & 2 & -3 & 1 & 0 & 0 \\
1 & 1 & 5 & -1 & 1 & 0 \\
1 & -6 & -3 & 0 & -1 & 1
\end{array}\right)
\end{array}$
$\stackrel{R_{3} \rightarrow R_{3}-R_{2}}{\longrightarrow}\left(\begin{array}{ccc|ccc}1 & 2 & -3 & 1 & 0 & 0 \\ 1 & 1 & 5 & -1 & 1 & 0 \\ 0 & -7 & -11 & 1 & -2 & 1\end{array}\right)$
$\stackrel{R_{2} \rightarrow R_{2}-R_{1}}{\longrightarrow}\left(\begin{array}{ccc|ccc}1 & 2 & -3 & 1 & 0 & 0 \\ 0 & -1 & 8 & -2 & 1 & 0 \\ 0 & -7 & -11 & 1 & -2 & 1\end{array}\right)$
$\stackrel{R_{2} \rightarrow-R_{2}}{\longrightarrow}\left(\begin{array}{ccc|ccc}1 & 2 & -3 & 1 & 0 & 0 \\ 0 & 1 & -8 & 2 & -1 & 0 \\ 0 & -7 & -11 & 1 & -2 & 1\end{array}\right)$
$\stackrel{R_{3} \rightarrow R_{3}+7 R_{2}}{\longrightarrow}\left(\begin{array}{ccc|ccc}1 & 2 & -3 & 1 & 0 & 0 \\ 0 & 1 & -8 & 2 & -1 & 0 \\ 0 & 0 & -67 & 15 & -9 & 1\end{array}\right)$
$\stackrel{R_{3} \rightarrow-\frac{1}{67} R_{3}}{\longrightarrow}\left(\begin{array}{ccc|ccc}1 & 2 & -3 & 1 & 0 & 0 \\ 0 & 1 & -8 & 2 & -1 & 0 \\ 0 & 0 & 1 & -\frac{15}{67} & \frac{9}{67} & -\frac{1}{67}\end{array}\right)$
$\stackrel{R_{2} \rightarrow R_{2}+8 R_{3}}{\longrightarrow}\left(\begin{array}{ccc|ccc}1 & 2 & -3 & 1 & 0 & 0 \\ 0 & 1 & 0 & \frac{14}{67} & \frac{5}{67} & -\frac{8}{67} \\ 0 & 0 & 1 & -\frac{15}{67} & \frac{9}{67} & -\frac{1}{67}\end{array}\right)$
$\stackrel{R_{1} \rightarrow R_{1}-2 R_{2}}{\longrightarrow}\left(\begin{array}{ccc|ccc}1 & 0 & -3 & \frac{39}{67} & -\frac{10}{67} & \frac{16}{67} \\ 0 & 1 & 0 & \frac{14}{67} & \frac{5}{67} & -\frac{8}{67} \\ 0 & 0 & 1 & -\frac{15}{67} & \frac{9}{67} & -\frac{1}{67}\end{array}\right)$
$\stackrel{R_{1} \rightarrow R_{1}+3 R_{3}}{\longrightarrow}\left(\begin{array}{ccc|ccc}1 & 0 & 0 & \frac{-6}{67} & \frac{17}{67} & \frac{13}{67} \\ 0 & 1 & 0 & \frac{14}{67} & \frac{5}{67} & -\frac{8}{67} \\ 0 & 0 & 1 & -\frac{15}{67} & \frac{9}{67} & -\frac{1}{67}\end{array}\right)$
Therefore we got the inverse of the matrix as
$A^{-1}=\left(\begin{array}{ccc}
\frac{-6}{67} & \frac{17}{67} & \frac{13}{67} \\
\frac{14}{67} & \frac{5}{67} & -\frac{8}{67} \\
-\frac{15}{67} & \frac{9}{67} & -\frac{1}{67}
\end{array}\right)$