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

Question:

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

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

Solution:

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

We know

$A=I A$

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

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

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

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

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

$\Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}-\frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & -\frac{1}{2} & \frac{-1}{6} \\ \frac{5}{6} & -\frac{1}{2} & \frac{1}{6}\end{array}\right] A$       [Applying $R_{1} \rightarrow R_{1}+2 R_{3}$ and $R_{2} \rightarrow R_{2}-R_{3}$ ]

$\therefore A^{-1}=\left[\begin{array}{ccc}-\frac{4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & -\frac{1}{2} & \frac{-1}{6} \\ \frac{5}{6} & -\frac{1}{2} & \frac{1}{6}\end{array}\right]$