Question

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

$\left[\begin{array}{ccc}2& 0& -1\\ 5& 1& 0\\ 0& 1& 3\end{array}\right]$

Answer 1

$$\begin{gathered} A=\left[\begin{array}{ccc}2& 0& -1\\ 5& 1& 0\\ 0& 1& 3\end{array}\right] \\ \mathrm{We}\mathrm{know} \\ A=IA \\ \Rightarrow \left[\begin{array}{ccc}2& 0& -1\\ 5& 1& 0\\ 0& 1& 3\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A \\ \Rightarrow \left[\begin{array}{ccc}1& 0& \frac{-1}{2}\\ 5& 1& 0\\ 0& 1& 3\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2}& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A\left[\mathrm{Applying}{R}_1\to \frac{1}{2}{R}_1\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& \frac{-1}{2}\\ 0& 1& \frac{5}{2}\\ 0& 1& 3\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2}& 0& 0\\ \frac{-5}{2}& 1& 0\\ 0& 0& 1\end{array}\right]A\left[\mathrm{Applying}{R}_2\to R_2-5R_1\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& \frac{-1}{2}\\ 0& 1& \frac{5}{2}\\ 0& 0& \frac{1}{2}\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2}& 0& 0\\ \frac{-5}{2}& 1& 0\\ \frac{5}{2}& -1& 1\end{array}\right]A\left[\mathrm{Applying}{R}_3\to R_3-R_2\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& \frac{-1}{2}\\ 0& 1& \frac{5}{2}\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2}& 0& 0\\ -\frac{5}{2}& 1& 0\\ 5& -2& 2\end{array}\right]A\left[\mathrm{Applying}{R}_3\to 2R_3\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right]A\left[\mathrm{Applying}{R}_1\to R_1+\frac{1}{2}{R}_3\mathrm{and}{R}_2\to R_2-\frac{5}{2}{R}_3\right] \\ \Rightarrow A^{-1}=\left[\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right] \end{gathered}$$