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

Question

Find the inverse of the following matrices by using elementry row transformation:$\begin{bmatrix}1 & 6 \\ -3 & 5 \end{bmatrix}$

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Answer

$\text{A}=\begin{bmatrix}1 & 6 \\ -3 & 5 \end{bmatrix}$We know $A = IA$
$\Rightarrow \begin{bmatrix}1 & 6 \\ -3 & 5 \end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\text{A}$
$\Rightarrow\begin{bmatrix}1 &0 \\ -3+3 & 5+18 \end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0+3 & 1+0 \end{bmatrix}\text{A}$
$[$Applying $R^2 → R^2 + 3R_1]$
$\Rightarrow\begin{bmatrix}1 & 6 \\ 0 & 23 \end{bmatrix}=\begin{bmatrix}1 & 0 \\ 3 & 1 \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_2\rightarrow\ \text{R}_1-\frac{6}{23}\text{R}_2\Big]$
$\Rightarrow\ \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} \frac{5}{23} & \frac{-6}{23} \\ \frac{-3}{23} & \frac{1}{23} \end{bmatrix}\text{A}$
$\Big[\text{Applying R}_2\rightarrow\ \frac{1}{23}\text{R}_2\Big]$
$\Rightarrow\text{A}^{-1}\begin{bmatrix}\frac{5}{23} & \frac{-6}{23} \\ \frac{3}{23} & \frac{1}{23} \end{bmatrix}=\frac{1}{23}\begin{bmatrix}5 & -6 \\ 3 & 1 \end{bmatrix}$
$\Rightarrow\text{A}^{-1}\begin{bmatrix}5 & -6 \\ 3 & 1 \end{bmatrix}$

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