Find the inverse of each of the following matrices by using elementary row transformations:

$A=\left[\begin{array}{ccc}1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0\end{array}\right]$

We know

$A=I A$

$\Rightarrow\left[\begin{array}{ccc}1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$

$\Rightarrow\left[\begin{array}{ccc}1 & 3 & -2 \\ 0 & 9 & -7 \\ 0 & -5 & 4\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1\end{array}\right] A$           [Applying $R_{2} \rightarrow R_{2}+3 R_{1}$ and $R_{3} \rightarrow R_{3}-2 R_{1}$ ]

$\Rightarrow\left[\begin{array}{ccc}1 & 3 & -2 \\ 0 & 1 & -\frac{7}{9} \\ 0 & -5 & 4\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ -2 & 0 & 1\end{array}\right] A$           $\left[\right.$ Applying $\left.R_{2} \rightarrow \frac{1}{9} R_{2}\right]$

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & \frac{1}{3} \\ 0 & 1 & -\frac{7}{9} \\ 0 & 0 & \frac{1}{9}\end{array}\right]=\left[\begin{array}{ccc}0 & -\frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ -\frac{1}{3} & \frac{5}{9} & 1\end{array}\right] A$             [Applying $R_{1} \rightarrow R_{1}-3 R_{2}$ and $R_{3} \rightarrow R_{3}+5 R_{2}$ ]

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & \frac{1}{3} \\ 0 & 1 & -\frac{7}{9} \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}0 & -\frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ -3 & 5 & 9\end{array}\right] A$            [Applying $R_{3} \rightarrow 9 R_{3}$ ]

$\Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9\end{array}\right] A$         [Applying $R_{2} \rightarrow R_{2}+\frac{7}{9} R_{3}$ and $R_{1} \rightarrow R_{1}-\frac{1}{3} R_{3}$ ]

$\Rightarrow A^{-1}=\left[\begin{array}{ccc}1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9\end{array}\right]$