Question

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

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

Answer 1

$$\begin{gathered} A=\left[\begin{array}{ccc}1& 1& 2\\ 3& 1& 1\\ 2& 3& 1\end{array}\right] \\ \mathrm{We}\mathrm{know} \\ A=IA \\ \Rightarrow \left[\begin{array}{ccc}1& 1& 2\\ 3& 1& 1\\ 2& 3& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A \\ \Rightarrow \left[\begin{array}{ccc}1& 1& 2\\ 0& -2& -5\\ 0& 1& -3\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ -3& 1& 0\\ -2& 0& 1\end{array}\right]A\left[\mathrm{Applying}{R}_2\to R_2-3R_1\mathrm{and}{R}_3\to R_3-2R_1\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 1& 2\\ 0& 1& \frac{5}{2}\\ 0& 1& -3\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ \frac{3}{2}& -\frac{1}{2}& 0\\ -2& 0& 1\end{array}\right]A\left[\mathrm{Applying}{R}_2\to \frac{-1}{2}{R}_2\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& -\frac{1}{2}\\ 0& 1& \frac{5}{2}\\ 0& 0& -\frac{11}{2}\end{array}\right]=\left[\begin{array}{ccc}-\frac{1}{2}& \frac{1}{2}& 0\\ \frac{3}{2}& -\frac{1}{2}& 0\\ -\frac{7}{2}& \frac{1}{2}& 1\end{array}\right]A\left[\mathrm{Applying}{R}_1\to R_1-R_2\mathrm{and}{R}_3\to R_3-R_2\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& -\frac{1}{2}\\ 0& 1& \frac{5}{2}\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}-\frac{1}{2}& \frac{1}{2}& 0\\ \frac{3}{2}& -\frac{1}{2}& 0\\ \frac{7}{11}& \frac{-1}{11}& \frac{-2}{11}\end{array}\right]A\left[\mathrm{Applying}{R}_3\to -\frac{2}{11}{R}_3\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}\frac{-2}{11}& \frac{5}{11}& \frac{-1}{11}\\ \frac{-1}{11}& \frac{-3}{11}& \frac{5}{11}\\ \frac{7}{11}& \frac{-1}{11}& \frac{-2}{11}\end{array}\right]A\left[\mathrm{Applying}{R}_2\to R_2-\frac{5}{2}{R}_3\mathrm{and}{R}_1\to R_1+\frac{1}{2}{R}_3\right] \\ \Rightarrow A^{-1}=\frac{1}{11}\left[\begin{array}{ccc}-2& 5& -1\\ -1& -3& 5\\ 7& -1& -2\end{array}\right] \end{gathered}$$