Question

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

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

Answer 1

$$\begin{gathered} A=\left[\begin{array}{ccc}1& 2& 0\\ 2& 3& -1\\ 1& -1& 3\end{array}\right] \\ \mathrm{We}\mathrm{know} \\ A=IA \\ \Rightarrow \left[\begin{array}{ccc}1& 2& 0\\ 2& 3& -1\\ 1& -1& 3\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A \\ \Rightarrow \left[\begin{array}{ccc}1& 2& 0\\ 0& -1& -1\\ 0& -3& 3\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ -1& 0& 1\end{array}\right]A\left[\mathrm{Applying}{R}_2\to R_2-2R_1\mathrm{and}{R}_3\to R_3-R_1\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 2& 0\\ 0& 1& 1\\ 0& -3& 3\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 2& -1& 0\\ -1& 0& 1\end{array}\right]A\left[\mathrm{Applying}{R}_2\to -R_2\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& -2\\ 0& 1& 1\\ 0& 0& 6\end{array}\right]=\left[\begin{array}{ccc}-3& 2& 0\\ 2& -1& 0\\ 5& -3& 1\end{array}\right]A\left[\mathrm{Applying}{R}_1\to R_1-2R_2\mathrm{and}{R}_3\to R_3+3R_2\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& -2\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}-3& 2& 0\\ 2& -1& 0\\ \frac{5}{6}& -\frac{1}{2}& \frac{1}{6}\end{array}\right]A\left[\mathrm{Applying}{R}_3\to \frac{1}{6}{R}_3\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}-\frac{4}{3}& 1& \frac{1}{3}\\ \frac{7}{6}& -\frac{1}{2}& \frac{-1}{6}\\ \frac{5}{6}& -\frac{1}{2}& \frac{1}{6}\end{array}\right]A\left[\mathrm{Applying}{R}_1\to R_1+2R_3\mathrm{and}{R}_2\to R_2-R_3\right] \\ \therefore A^{-1}=\left[\begin{array}{ccc}-\frac{4}{3}& 1& \frac{1}{3}\\ \frac{7}{6}& -\frac{1}{2}& \frac{-1}{6}\\ \frac{5}{6}& -\frac{1}{2}& \frac{1}{6}\end{array}\right] \end{gathered}$$