Question

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

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

Answer 1

$$\begin{gathered} A=\left[\begin{array}{ccc}1& 3& -2\\ -3& 0& -1\\ 2& 1& 0\end{array}\right] \\ \mathrm{We}\mathrm{know} \\ A=IA \\ \Rightarrow \left[\begin{array}{ccc}1& 3& -2\\ -3& 0& -1\\ 2& 1& 0\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A \\ \Rightarrow \left[\begin{array}{ccc}1& 3& -2\\ 0& 9& -7\\ 0& -5& 4\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ -2& 0& 1\end{array}\right]A\left[\mathrm{Applying}{R}_2\to R_2+3R_1\mathrm{and}{R}_3\to R_3-2R_1\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 3& -2\\ 0& 1& -\frac{7}{9}\\ 0& -5& 4\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ \frac{1}{3}& \frac{1}{9}& 0\\ -2& 0& 1\end{array}\right]A\left[\mathrm{Applying}{R}_2\to \frac{1}{9}{R}_2\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& \frac{1}{3}\\ 0& 1& -\frac{7}{9}\\ 0& 0& \frac{1}{9}\end{array}\right]=\left[\begin{array}{ccc}0& -\frac{1}{3}& 0\\ \frac{1}{3}& \frac{1}{9}& 0\\ -\frac{1}{3}& \frac{5}{9}& 1\end{array}\right]A\left[\mathrm{Applying}{R}_1\to R_1-3R_2\mathrm{and}{R}_3\to R_3+5R_2\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& \frac{1}{3}\\ 0& 1& -\frac{7}{9}\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}0& -\frac{1}{3}& 0\\ \frac{1}{3}& \frac{1}{9}& 0\\ -3& 5& 9\end{array}\right]A\left[\mathrm{Applying}{R}_3\to 9R_3\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& -2& -3\\ -2& 4& 7\\ -3& 5& 9\end{array}\right]A\left[\mathrm{Applying}{R}_2\to R_2+\frac{7}{9}{R}_3\mathrm{and}{R}_1\to R_1-\frac{1}{3}{R}_3\right] \\ \Rightarrow A^{-1}=\left[\begin{array}{ccc}1& -2& -3\\ -2& 4& 7\\ -3& 5& 9\end{array}\right] \end{gathered}$$