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

$A=\left[\begin{array}{ll}3 & 10\end{array}\right.$

$\left.\begin{array}{ll}2 & 7\end{array}\right]$

We know

$A=I A$

$\Rightarrow\left[\begin{array}{ll}3 & 10\end{array}\right.$

$2 \quad 7]=\left[\begin{array}{ll}1 & 0\end{array}\right.$

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

$\Rightarrow\left[\begin{array}{ll}3-2 & 10-7\end{array}\right.$

$2 \quad 7]=\left[\begin{array}{ll}1-0 & 0-1\end{array}\right.$

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

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

$2 \quad 7]=\left[\begin{array}{ll}1 & -1\end{array}\right.$

$0 \quad 1] A$

$2-2 \quad 7-6]=\left[\begin{array}{ll}1 & -1\end{array}\right.$

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

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

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

$\begin{array}{ll}-2 & 3] A\end{array}$

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

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

$0 \quad 1]=\left[\begin{array}{ll}1+6 & -1-9\end{array}\right.$

$\begin{array}{ll}-2 & 3] A\end{array}$ $\left[\right.$ Applying $\left.R_{1} \rightarrow R_{1}-3 R_{2}\right]$

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

$0 \quad 1]=\left[\begin{array}{ll}7 & -10\end{array}\right]$

$-2 \quad 3] A$

$\Rightarrow A^{-1}=\left[\begin{array}{ll}7 & -10\end{array}\right.$

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