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

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

We know

$A=I A$

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

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

$\Rightarrow\left[\begin{array}{ccc}1 & -\frac{1}{2} & 2 \\ 0 & 2 & -6 \\ 0 & -\frac{1}{2} & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ -2 & 1 & 0 \\ \frac{-3}{2} & 0 & 1\end{array}\right] A$   [Applying $R_{2} \rightarrow R_{2}-4 R_{1}$ and $R_{3} \rightarrow R_{3}-3 R_{1}$ ]

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

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & \frac{1}{2} \\ 0 & 1 & -3 \\ 0 & 0 & -\frac{1}{2}\end{array}\right]=\left[\begin{array}{ccc}0 & \frac{1}{4} & 0 \\ -1 & \frac{1}{2} & 0 \\ -2 & \frac{1}{4} & 1\end{array}\right] A$     [Applying $R_{1} \rightarrow R_{1}+\frac{1}{2} R_{2}$ and $R_{3} \rightarrow R_{3}+\frac{1}{2} R_{2}$ ]

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

$\Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-2 & \frac{1}{2} & 1 \\ 11 & -1 & -6 \\ 4 & -\frac{1}{2} & -2\end{array}\right] A$             [Applying $R_{1} \rightarrow R_{1}-\frac{1}{2} R_{3}$ and $R_{2} \rightarrow R_{2}+3 R_{3}$ ]

$\Rightarrow A^{-1}=\left[\begin{array}{ccc}-2 & \frac{1}{2} & 1 \\ 11 & -1 & -6 \\ 4 & -\frac{1}{2} & -2\end{array}\right]$