Using elementary transformations, find the inverse of the matrx

$\left( \begin{array}{ccc}1 & 3& - 2\\ - 3 & 0 & - 1\\ 2& 1& 0\end{array} \right)$

$$\begin{gathered} \mathrm{The} \mathrm{given} \mathrm{matrix} \mathrm{is} \mathrm{A} = \left[ \begin{array}{ccc}1& 3& - 2\\ - 3 & 0& - 1\\ 2& 1& 0\end{array} \right]. \\ \mathrm{We} \mathrm{have} \mathrm{A} {\mathrm{A}}^{- 1} = \mathrm{I} \\ \mathrm{Thus}, \mathrm{A} = \mathrm{I} \mathrm{A} \\ \mathrm{Or}, \left[ \begin{array}{ccc}1& 3& - 2\\ - 3 & 0& - 1\\ 2& 1& 0\end{array} \right] =\left[ \begin{array}{ccc}1 & 0& 0 \\ 0 & 1& 0\\ 0 & 0& 1\end{array} \right] \mathrm{A} \\ \mathrm{Applying} {\mathrm{R}}_{2 } \to + {\mathrm{R}}_2 + 3 {\mathrm{R}}_1 \mathrm{and} {\mathrm{R}}_3 \to {\mathrm{R}}_3 - 2 {\mathrm{R}}_1 \\ \left[ \begin{array}{ccc} 1 & 3& - 2\\ 0 & 9& - 7\\ 0 & - 5 & 4 \end{array} \right] = \left[\begin{array}{ccc}1 & 0& 0 \\ 3 & 1& 0\\ - 2 & 0& 1\end{array} \right] \mathrm{A} \\ \mathrm{Now}, \mathrm{applying} {\mathrm{R}}_2 \to \frac{1}{9} {\mathrm{R}}_2 \\ \left[ \begin{array}{ccc} 1 & 3& - 2\\ 0 & 1& - \frac{7}{9}\\ 0 & - 5 & 4 \end{array} \right] = \left[\begin{array}{ccc}1 & 0& 0 \\ \frac{1}{3} & \frac{1}{9}& 0\\ - 2 & 0& 1\end{array} \right] \mathrm{A} \end{gathered}$$

$$\begin{gathered} \mathrm{Applying} {\mathrm{R}}_1 \to {\mathrm{R}}_1 - 3 {\mathrm{R}}_2 \mathrm{and} {\mathrm{R}}_3 \to {\mathrm{R}}_3 + 5 {\mathrm{R}}_2 \\ \left[ \begin{array}{ccc}1 & 0& \frac{1}{3}\\ 0 & 1& -\frac{7}{9}\\ 0 & 0& \frac{1}{9}\end{array} \right] = \left[\begin{array}{ccc}0& - \frac{1}{3} & 0 \\ \frac{1}{3}& \frac{1}{9}& 0\\ - \frac{1}{3} & \frac{5}{9}& 1\end{array} \right] \mathrm{A} \\ \mathrm{Applying} {\mathrm{R}}_3 \to 9 {\mathrm{R}}_3 \\ \left[ \begin{array}{ccc} 1 & 0& \frac{1}{3} \\ 0 & 1& -\frac{7}{9}\\ 0 & 0& 1\end{array} \right] = \left[\begin{array}{ccc}0& - \frac{1}{3} & 0 \\ \frac{1}{3}& \frac{1}{9}& 0\\ - 3 & 5& 9\end{array} \right] \mathrm{A} \\ \mathrm{Applying} {\mathrm{R}}_1 \to {\mathrm{R}}_1 - \frac{1}{3 } {\mathrm{R}}_3 \mathrm{and} {\mathrm{R}}_2 \to {\mathrm{R}}_2 + \frac{7}{9 } {\mathrm{R}}_3 \end{gathered}$$

$$\begin{gathered} \left[ \begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array} \right] = \left[\begin{array}{ccc} 1& - 2 & - 3 \\ - 2 & 4& 7\\ - 3 & 5& 9\end{array}\right] \mathrm{A} \Rightarrow \mathrm{I} = \left[\begin{array}{ccc} 1& - 2 & - 3 \\ - 2 & 4& 7\\ - 3 & 5& 9\end{array}\right] \mathrm{A} \\ \therefore {\mathrm{A}}^{-1} = \left[\begin{array}{ccc} 1& - 2 & - 3 \\ - 2 & 4& 7\\ - 3 & 5& 9\end{array}\right] \\ \mathrm{Hence}, \mathrm{inverse} \mathrm{of} \mathrm{the} \mathrm{matrix} \mathrm{A} \mathrm{is} \left[\begin{array}{ccc} 1& - 2 & - 3 \\ - 2 & 4& 7\\ - 3 & 5& 9\end{array}\right] . \end{gathered}$$