Question

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

$\left[\begin{array}{ccc}3& -3& 4\\ 2& -3& 4\\ 0& -1& 1\end{array}\right]$

Answer 1

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