Question

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

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

Answer 1

$$\begin{gathered} \mathrm{Let}A=\left[\begin{array}{ccc}-1& 1& 2\\ 1& 2& 3\\ 3& 1& 1\end{array}\right]. \\ \mathrm{To}\mathrm{find}\mathrm{inverse},\mathrm{first}\mathrm{write}A=IA. \\ \mathrm{i}.\mathrm{e}.,\left[\begin{array}{ccc}-1& 1& 2\\ 1& 2& 3\\ 3& 1& 1\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& -1& -2\\ 1& 2& 3\\ 3& 1& 1\end{array}\right]=\left[\begin{array}{ccc}-1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A\left[\mathrm{Applying}{R}_1\to \left(-1\right)R_1\right] \\ \Rightarrow \left[\begin{array}{ccc}1& -1& -2\\ 0& 3& 5\\ 0& 4& 7\end{array}\right]=\left[\begin{array}{ccc}-1& 0& 0\\ 1& 1& 0\\ 3& 0& 1\end{array}\right]A\left[\mathrm{Applying}{R}_2\to R_2-R_1\mathrm{and}{R}_3\to R_3-3R_1\right] \\ \Rightarrow \left[\begin{array}{ccc}1& -1& -2\\ 0& 1& \frac{5}{3}\\ 0& 4& 7\end{array}\right]=\left[\begin{array}{ccc}-1& 0& 0\\ \frac{1}{3}& \frac{1}{3}& 0\\ 3& 0& 1\end{array}\right]A\left[\mathrm{Applying}{R}_2\to \frac{1}{3}{R}_2\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& \frac{-1}{3}\\ 0& 1& \frac{5}{3}\\ 0& 0& \frac{1}{3}\end{array}\right]=\left[\begin{array}{ccc}\frac{-2}{3}& \frac{1}{3}& 0\\ \frac{1}{3}& \frac{1}{3}& 0\\ \frac{5}{3}& -\frac{4}{3}& 1\end{array}\right]A\left[\mathrm{Applying}{R}_3\to R_3-4R_2\mathrm{and}{R}_1\to R_1+R_2\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& \frac{-1}{3}\\ 0& 1& \frac{5}{3}\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}\frac{-2}{3}& \frac{1}{3}& 0\\ \frac{1}{3}& \frac{1}{3}& 0\\ 5& -4& 3\end{array}\right]A\left[\mathrm{Applying}{R}_3\to 3R_3\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& -1& 1\\ -8& 7& -5\\ 5& -4& 3\end{array}\right]A\left[\mathrm{Applying}{R}_2\to R_2-\frac{5}{3}{R}_3\mathrm{and}{R}_1\to R_1+\frac{1}{3}{R}_3\right] \\ \mathrm{Hence},A^{-1}=\left[\begin{array}{ccc}1& -1& 1\\ -8& 7& -5\\ 5& -4& 3\end{array}\right]. \end{gathered}$$