Question

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

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

Answer 1

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