Question

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

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

Answer 1

$$\begin{gathered} A=\left[71 \\ 4-3\right] \\ \mathrm{We}\mathrm{know} \\ A\mathit{=}IA \\ \Rightarrow \left[71 \\ 4-3\right]=\left[10 \\ 01\right]A \\ \Rightarrow \left[1\frac{1}{7} \\ 4-3\right]=\left[\frac{1}{7}0 \\ 01\right]A\mathit{}\left(\mathrm{Applying}{R}_1\to \frac{1}{7}{R}_1\right) \\ \Rightarrow \left[1\frac{1}{7} \\ 0-\frac{25}{7}\right]=\left[\frac{1}{7}0 \\ -\frac{4}{7}1\right]A\mathit{}\left(\mathrm{Applying}{R}_2\to R_2-4R_1\right) \\ \Rightarrow \left[1\frac{1}{7} \\ 01\right]=\left[\frac{1}{7}0 \\ \frac{4}{25}-\frac{7}{25}\right]A\mathit{}\left(\mathrm{Applying}{R}_2\to -\frac{7}{25}{R}_2\right) \\ \Rightarrow \left[10 \\ 01\right]=\left[\frac{3}{25}\frac{1}{25} \\ \frac{4}{25}-\frac{7}{25}\right]A\mathit{}\left(\mathrm{Applying}{R}_1\to R_1-\frac{1}{7}{R}_2\right) \\ \therefore A^{-1}=\frac{1}{25}\left[31 \\ 4-7\right] \end{gathered}$$