Question

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

$\left[\begin{array}{ccc}2& -1& 4\\ 4& 0& 7\\ 3& -2& 7\end{array}\right]$

Answer 1

$$\begin{gathered} A=\left[\begin{array}{ccc}2& -1& 4\\ 4& 0& 2\\ 3& -2& 7\end{array}\right] \\ \mathrm{We}\mathrm{know} \\ A=IA \\ \Rightarrow \left[\begin{array}{ccc}2& -1& 4\\ 4& 0& 2\\ 3& -2& 7\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& -\frac{1}{2}& 2\\ 4& 0& 2\\ 3& -2& 7\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2}& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A\left[\mathrm{Applying}{R}_1\to \frac{1}{2}{R}_1\right] \\ \Rightarrow \left[\begin{array}{ccc}1& -\frac{1}{2}& 2\\ 0& 2& -6\\ 0& -\frac{1}{2}& 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2}& 0& 0\\ -2& 1& 0\\ \frac{-3}{2}& 0& 1\end{array}\right]A\left[\mathrm{Applying}{R}_2\to R_2-4R_1\mathrm{and}{R}_3\to R_3-3R_1\right] \\ \Rightarrow \left[\begin{array}{ccc}1& -\frac{1}{2}& 2\\ 0& 1& -3\\ 0& -\frac{1}{2}& 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2}& 0& 0\\ -1& \frac{1}{2}& 0\\ \frac{-3}{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& -3\\ 0& 0& -\frac{1}{2}\end{array}\right]=\left[\begin{array}{ccc}0& \frac{1}{4}& 0\\ -1& \frac{1}{2}& 0\\ -2& \frac{1}{4}& 1\end{array}\right]A\left[\mathrm{Applying}{R}_1\to R_1+\frac{1}{2}{R}_2\mathrm{and}{R}_3\to R_3+\frac{1}{2}{R}_2\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& \frac{1}{2}\\ 0& 1& -3\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}0& \frac{1}{4}& 0\\ -1& \frac{1}{2}& 0\\ 4& -\frac{1}{2}& -2\end{array}\right]A\left[\mathrm{Applying}{R}_3\to -2R_3\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}-2& \frac{1}{2}& 1\\ 11& -1& -6\\ 4& -\frac{1}{2}& -2\end{array}\right]A\left[\mathrm{Applying}{R}_1\to R_1-\frac{1}{2}{R}_3\mathrm{and}{R}_2\to R_2+3R_3\right] \\ \Rightarrow A^{-1}=\left[\begin{array}{ccc}-2& \frac{1}{2}& 1\\ 11& -1& -6\\ 4& -\frac{1}{2}& -2\end{array}\right] \end{gathered}$$