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

$A=\left[\begin{array}{lll}2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2\end{array}\right]$

We know

$A=I A$

$\Rightarrow\left[\begin{array}{lll}2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A$

$\Rightarrow\left[\begin{array}{lll}1 & \frac{3}{2} & \frac{1}{2} \\ 2 & 4 & 1 \\ 3 & 7 & 2\end{array}\right]=\left[\begin{array}{lll}\frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] A \quad\left[\right.$ Applying $\left.R_{1} \rightarrow \frac{1}{2} R_{1}\right]$

$\Rightarrow\left[\begin{array}{lll}1 & \frac{3}{2} & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & \frac{5}{2} & \frac{1}{2}\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ -1 & 1 & 0 \\ \frac{-3}{2} & 0 & 1\end{array}\right] A$        [Applying $R_{2} \rightarrow R_{2}-2 R_{1}$ and $R_{3} \rightarrow R_{3}-3 R_{1}$ ]

$\Rightarrow\left[\begin{array}{lll}1 & 0 & \frac{1}{2} \\ 0 & 1 & 0 \\ 0 & 0 & \frac{1}{2}\end{array}\right]=\left[\begin{array}{ccc}2 & \frac{-3}{2} & 0 \\ -1 & 1 & 0 \\ 1 & \frac{-5}{2} & 1\end{array}\right] A$               [Applying $R_{1} \rightarrow R_{1}-\frac{3}{2} R_{2}$ and $R_{3} \rightarrow R_{3}-\frac{5}{2} R_{2}$ ]

$\Rightarrow\left[\begin{array}{lll}1 & 0 & \frac{1}{2} \\ 0 &  1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}2 & \frac{-3}{2} & 0 \\ -1 & 1 & 0 \\ 2 & -5 & 2\end{array}\right] A$    [Applying $R_{3} \rightarrow 2 R_{3}$ ]

$\Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & 1 & -1 \\ -1 & 1 & 0 \\ 2 & -5 & 2\end{array}\right] A$  [Applying $R_{1} \rightarrow R_{1}-\frac{1}{2} R_{3}$ ]

$\therefore A^{-1}=\left[\begin{array}{ccc}1 & 1 & -1 \\ -1 & 1 & 0 \\ 2 & -5 & 2\end{array}\right]$