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

Solution:

$A=\left[\begin{array}{ll}2 & 5\end{array}\right.$

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

We know

$A=I A$

$\Rightarrow\left[\begin{array}{ll}2-1 & 5-3\end{array}\right.$

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

$0 \quad 1] A \quad\left[\right.$ Applying $\left.R_{1} \rightarrow R_{1}-R_{2}\right]$

$\Rightarrow\left[\begin{array}{ll}1 & 2\end{array}\right.$

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

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

$\Rightarrow\left[\begin{array}{ll}1 & 2\end{array}\right.$

$1-1 \quad 3-2]=\left[\begin{array}{ll}1 & -1\end{array}\right.$

$\begin{array}{lll}0-1 & 1+1] A & \text { [Applying } R_{2} \rightarrow R_{2}-R_{1} \text { ] }\end{array}$

$\Rightarrow\left[\begin{array}{ll}1 & 2\end{array}\right.$

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

$\begin{array}{ll}-1 & 2] A\end{array}$

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

$\left.\begin{array}{ll}0 & 1\end{array}\right]=\left[\begin{array}{ll}1+2 & -1-4\end{array}\right.$

$\begin{array}{ll}-1 & 2] A\end{array}$                       [Applying $R_{1} \rightarrow R_{1}-2 R_{2}$ ]

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

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

$\begin{array}{ll}-1 & 2] A\end{array}$

$\Rightarrow A^{-1}=\left[\begin{array}{ll}3 & -5\end{array}\right.$

$\left.\begin{array}{ll}-1 & 2\end{array}\right]$