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

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

We know

$A=I A$

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

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

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

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & -\frac{11}{2}\end{array}\right]=\left[\begin{array}{ccc}-\frac{1}{2} & \frac{1}{2} & 0 \\ \frac{3}{2} & -\frac{1}{2} & 0 \\ -\frac{7}{2} & \frac{1}{2} & 1\end{array}\right] A$   [Applying $R_{1} \rightarrow R_{1}-R_{2}$ and $R_{3} \rightarrow R_{3}-R_{2}$ ]

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & -\frac{1}{2} \\ 0 & 1 & \frac{5}{2} \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-\frac{1}{2} & \frac{1}{2} & 0 \\ \frac{3}{2} & -\frac{1}{2} & 0 \\ \frac{7}{11} & \frac{-1}{11} & \frac{-2}{11}\end{array}\right] A$     [Applying $R_{3} \rightarrow-\frac{2}{11} R_{3}$ ]

$\Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{-2}{11} & \frac{5}{11} & \frac{-1}{11} \\ \frac{-1}{11} & \frac{-3}{11} & \frac{5}{11} \\ \frac{7}{11} & \frac{-1}{11} & \frac{-2}{11}\end{array}\right] A$              [Applying $R_{2} \rightarrow R_{2}-\frac{5}{2} R_{3}$ and $R_{1} \rightarrow R_{1}+\frac{1}{2} R_{3}$ ]

$\Rightarrow A^{-1}=\frac{1}{11}\left[\begin{array}{ccc}-2 & 5 & -1 \\ -1 & -3 & 5 \\ 7 & -1 & -2\end{array}\right]$