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

$A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3\end{array}\right]$

We know

$A=I A$

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

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

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

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

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

$\Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right] A$             $\left[\right.$ Applying $R_{1} \rightarrow R_{1}+\frac{1}{2} R_{3}$ and $\left.R_{2} \rightarrow R_{2}-\frac{5}{2} R_{3}\right]$

$\Rightarrow A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$