If possible,using elementary row transformations,find the inverse of the following matrix:
$\left[\begin{array}{ccc}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right]$

(A) Let $A = \left[\begin{array}{ccc}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right]$. To find $A^{-1}$ using elementary row transformations,we write $A = IA$.
$\left[\begin{array}{ccc}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$
Applying $R_2 \rightarrow R_2 + 2R_1$ and $R_3 \rightarrow R_3 + R_1$:
$\left[\begin{array}{ccc}2 & -1 & 3 \\ -1 & 1 & 7 \\ -1 & 1 & 6\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 0 & 1\end{array}\right] A$
Applying $R_3 \rightarrow R_3 - R_2$:
$\left[\begin{array}{ccc}2 & -1 & 3 \\ -1 & 1 & 7 \\ 0 & 0 & -1\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \\ 2 & 1 & 0 \\ -1 & -1 & 1\end{array}\right] A$
Applying $R_1 \rightarrow R_1 + R_2$:
$\left[\begin{array}{ccc}1 & 0 & 10 \\ -1 & 1 & 7 \\ 0 & 0 & -1\end{array}\right] = \left[\begin{array}{ccc}3 & 1 & 0 \\ 2 & 1 & 0 \\ -1 & -1 & 1\end{array}\right] A$
Applying $R_2 \rightarrow R_2 + R_1$:
$\left[\begin{array}{ccc}1 & 0 & 10 \\ 0 & 1 & 17 \\ 0 & 0 & -1\end{array}\right] = \left[\begin{array}{ccc}3 & 1 & 0 \\ 5 & 2 & 0 \\ -1 & -1 & 1\end{array}\right] A$
Applying $R_1 \rightarrow R_1 + 10R_3$,$R_2 \rightarrow R_2 + 17R_3$,and $R_3 \rightarrow -R_3$:
$\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}-7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1\end{array}\right] A$
Thus,$A^{-1} = \left[\begin{array}{ccc}-7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1\end{array}\right]$.