Question

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

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

Answer 1

$$\begin{gathered} A=\left[\begin{array}{ccc}2& 3& 1\\ 2& 4& 1\\ 3& 7& 2\end{array}\right] \\ \mathrm{We}\mathrm{know} \\ A=IA \\ \Rightarrow \left[\begin{array}{ccc}2& 3& 1\\ 2& 4& 1\\ 3& 7& 2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A \\ \Rightarrow \left[\begin{array}{ccc}1& \frac{3}{2}& \frac{1}{2}\\ 2& 4& 1\\ 3& 7& 2\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2}& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A\left[\mathrm{Applying}{R}_1\to \frac{1}{2}{R}_1\right] \\ \Rightarrow \left[\begin{array}{ccc}1& \frac{3}{2}& \frac{1}{2}\\ 0& 1& 0\\ 0& \frac{5}{2}& \frac{1}{2}\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2}& 0& 0\\ -1& 1& 0\\ \frac{-3}{2}& 0& 1\end{array}\right]A\left[\mathrm{Applying}{R}_2\to R_2-2R_1\mathrm{and}{R}_3\to R_3-3R_1\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& \frac{1}{2}\\ 0& 1& 0\\ 0& 0& \frac{1}{2}\end{array}\right]=\left[\begin{array}{ccc}2& \frac{-3}{2}& 0\\ -1& 1& 0\\ 1& \frac{-5}{2}& 1\end{array}\right]A\left[\mathrm{Applying}{R}_1\to R_1-\frac{3}{2}{R}_2\mathrm{and}{R}_3\to R_3-\frac{5}{2}{R}_2\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& \frac{1}{2}\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}2& \frac{-3}{2}& 0\\ -1& 1& 0\\ 2& -5& 2\end{array}\right]A\left[\mathrm{Applying}{R}_3\to 2R_3\right] \\ \Rightarrow \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 1& -1\\ -1& 1& 0\\ 2& -5& 2\end{array}\right]A\left[\mathrm{Applying}{R}_1\to R_1-\frac{1}{2}{R}_3\right] \\ \therefore A^{-1}=\left[\begin{array}{ccc}1& 1& -1\\ -1& 1& 0\\ 2& -5& 2\end{array}\right] \end{gathered}$$