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

Solution:

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

We know

$A=I A$

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

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

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

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

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

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

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

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

$\Rightarrow\left[\begin{array}{ccc}-3 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & -1 \\ 4 & -3 & 1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}\right] A$           [Applying $R_{3} \rightarrow R_{3}+R_{2}$ ]

$\Rightarrow\left[\begin{array}{ccc}-3 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-\frac{3}{2} & \frac{3}{2} & -\frac{3}{2} \\ 4 & -3 & 1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}\right] A \quad$ [Applying $R_{1} \rightarrow R_{1}-R_{3}$ ]

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\ 4 & -3 & 1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}\right]$ A $\quad\left[\right.$ Applying $\left.R_{1} \rightarrow \frac{-1}{3} R_{1}\right]$

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\ 4 & -3 & 1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}\right] A$       $\left[\right.$ Applying $\left.R_{1} \rightarrow \frac{-1}{3} R_{1}\right]$

$\Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{2} & \frac{-1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & \frac{-3}{2} & \frac{1}{2}\end{array}\right] A$      [Applying $R_{2} \rightarrow-R_{2}$ ]