# 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:

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

Solution:

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

We know

$A=I A$

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

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

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

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

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

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

$\Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{15} & \frac{-2}{15} & \frac{1}{3} \\ -\frac{11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & -\frac{1}{6}\end{array}\right] A$                   [Applying $R_{2} \rightarrow R_{2}-R_{3}$ and $R_{1} \rightarrow R_{1}-2 R_{3}$ ]

$\Rightarrow A^{-1}=-\frac{1}{30}\left[\begin{array}{ccc}-2 & 4 & -10 \\ 11 & -7 & -5 \\ -5 & -5 & 5\end{array}\right]$