Find the inverse of the following matrices by using elementry row…

Question

Find the inverse of the following matrices by using elementry row transformation:$\begin{bmatrix} 2 & -1 & 3 \\ 1 & 2 &4 \\ 3 & 1 & 1 \end{bmatrix}$

✓

Answer

$\text{A}=\begin{bmatrix} 2 & -1 & 3 \\ 1 & 2 &4 \\ 3 & 1 & 1 \end{bmatrix}$We have A = IA
$\begin{bmatrix} 2 & -1 & 3 \\ 1 & 2 &4 \\ 3 & 1 & 1 \end{bmatrix}-\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\text{Applying R}_1\rightarrow\frac{\text{R}_1}{2}$
$\begin{bmatrix} 1 & \frac{-1}{2} & \frac{3}{2} \\ 1 & 2 &4 \\ 3 & 1 & 1 \end{bmatrix}=\begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\text{A}$
$\big[\text{Applying R}_2\rightarrow\text{R}_2-\text{R}_1, \text{R}_3\rightarrow\text{R}_3-3\text{R}_1\big]$
$\begin{bmatrix} 1 & \frac{-1}{2} & \frac{3}{2} \\ 0 & \frac{5}{2} & \frac{5}{2} \\ 0 & \frac{5}{2} & \frac{-7}{2} \end{bmatrix}=\begin{bmatrix} \frac{1}{2} & 0 & 0 \\ \frac{-1}{2} & 1 & 0 \\ \frac{-3}{2} & 0 & 1 \end{bmatrix}\text{A}$
$\text{Applying R}_2\rightarrow\Big(\frac{2}{5}\Big)\text{R}_2$
$\begin{bmatrix} 1 & \frac{-1}{2} & \frac{3}{2} \\ 0 & 1 & 1 \\ 0 & \frac{5}{2} & \frac{-7}{2} \end{bmatrix}=\begin{bmatrix} \frac{1}{2} & 0 & 0 \\ \frac{-1}{5} & \frac{2}{5} & 0 \\ \frac{-3}{2} & -1 & 1 \end{bmatrix}\text{A}$
$\text{Applying R}_1\rightarrow\text{R}_1+\frac{1}{2}\text{R}_2,\text{R}_3\rightarrow\text{R}_3-\frac{5}{2}\text{R}_2$
$\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & -6 \end{bmatrix}=\begin{bmatrix} \frac{2}{5} & \frac{1}{5} & 0 \\ \frac{-1}{5} & \frac{2}{5} & 0 \\ -1 & -1 & 1 \end{bmatrix}\text{A}$
$\text{Applying R}_3\rightarrow\frac{\text{R}_3}{-6}$
$\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} \frac{2}{5} & \frac{1}{5} & 0 \\ \frac{-1}{5} & \frac{2}{5} & 0 \\ \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \end{bmatrix}\text{A}$
$\text{Applying R}_2\rightarrow\text{R}_2-\text{R}_3,\text{R}_1\rightarrow\text{R}_1-2\text{R}_3$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} \frac{1}{15} & \frac{-2}{15} & \frac{-1}{3} \\ \frac{-11}{30} & \frac{7}{30} & \frac{1}{6} \\ \frac{1}{6} & \frac{1}{6} & \frac{-1}{6} \end{bmatrix}\text{A}\ \big[\because\ \text{I}=\text{A}^{-1}\text{A}\big]$