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

Solution:

$A=\left[\begin{array}{ll}7 & 1\end{array}\right.$

$4-3]$

We know

$A=I A$

$\Rightarrow\left[\begin{array}{ll}7 & 1\end{array}\right.$

$4-3]=\left[\begin{array}{ll}1 & 0\end{array}\right.$

$\begin{array}{ll}0 & 1] A\end{array}$

$4-3]=\left[\begin{array}{ll}\frac{1}{7} & 0\end{array}\right.$

$0 \quad 1] A$   (Applying $R_{1} \rightarrow \frac{1}{7} R_{1}$ )

$\Rightarrow\left[\begin{array}{ll}1 & \frac{1}{7}\end{array}\right.$

$\left.0-\frac{25}{7}\right]=\left[\begin{array}{ll}\frac{1}{7} & 0\end{array}\right.$

$\left.-\frac{4}{7} \quad 1\right] A$         (Applying $R_{2} \rightarrow R_{2}-4 R_{1}$ )

$\Rightarrow\left[\begin{array}{ll}1 & \frac{1}{7}\end{array}\right.$

$\left.\begin{array}{ll}0 & 1\end{array}\right]=\left[\begin{array}{ll}\frac{1}{7} & 0\end{array}\right.$

$\left.\frac{4}{25}-\frac{7}{25}\right] A \quad$ (Applying $R_{2} \rightarrow-\frac{7}{25} R_{2}$ )

$\Rightarrow\left[\begin{array}{ll}1 & 0\end{array}\right.$

$\left.\begin{array}{ll}0 & 1\end{array}\right]=\left[\begin{array}{ll}\frac{3}{25} & \frac{1}{25}\end{array}\right.$

$\left.\frac{4}{25}-\frac{7}{25}\right] A$     (Applying $R_{1} \rightarrow R_{1}-\frac{1}{7} R_{2}$ )

$\therefore A^{-1}=\frac{1}{25}\left[\begin{array}{ll}3 & 1\end{array}\right.$

$4-7]$