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

Solution:

$A=\left[\begin{array}{ll}1 & 6\end{array}\right.$

$\left.\begin{array}{ll}-3 & 5\end{array}\right]$

We know

$A=I A$

$\Rightarrow\left[\begin{array}{ll}1 & 6\end{array}\right.$

$-3 \quad 5]=\left[\begin{array}{ll}1 & 0\end{array}\right.$

$0 \quad$ 1] $A$

$\Rightarrow\left[\begin{array}{ll}1 & 6\end{array}\right.$

$-3+3 \quad 5+18]=\left[\begin{array}{ll}1 & 0\end{array}\right.$

$0+3 \quad 1+0] A$

$\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}+3 R_{1}\right]$

$\Rightarrow\left[\begin{array}{ll}1 & 6\end{array}\right.$

$0 \quad 23]=\left[\begin{array}{ll}1 & 0\end{array}\right.$

$\begin{array}{ll}3 & 1] A\end{array}$

$\Rightarrow\left[\begin{array}{lll}1 & 6-6\end{array}\right.$

$0 \quad 23]=\left[\begin{array}{ll}1-\frac{18}{23} & 0-\frac{6}{23}\end{array}\right.$

$3 \quad$ 1] $A \quad$ [Applying $\left.R_{1} \rightarrow R_{1}-\frac{6}{23} R_{2}\right]$

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

$\left.\begin{array}{ll}0 & 1\end{array}\right]=\left[\begin{array}{ll}5 / 23 & -6 / 23\end{array}\right.$

$3 / 23 \quad 1 / 23] A \quad\left[\right.$ Applying $\left.R_{2} \rightarrow \frac{1}{23} R_{2}\right]$

$\Rightarrow A^{-1}=\left[\begin{array}{ll}5 / 23 & -6 / 23\end{array}\right.$

$3 / 23 \quad 1 / 23]=\frac{1}{23}\left[\begin{array}{ll}5 & -6\end{array}\right.$

$\left.\begin{array}{ll}3 & 1\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{23}\left[\begin{array}{ll}5 & -6\end{array}\right.$

$3 \quad 1]$