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Adjoint and Inverse of a Matrix — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAdjoint and Inverse of a Matrix5 Marks
Question
Find the inverse of the following matrices by using elementry row transformation:$\begin{bmatrix}7 & 1 \\4 & -3 \end{bmatrix}$
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Answer

$\text{A}=\begin{bmatrix}7 & 1 \\4 & -3 \end{bmatrix}$We know A = IA
$\Rightarrow\ \begin{bmatrix}7 & 1 \\4 & -3 \end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\text{A}$
$\Rightarrow\ \begin{bmatrix}1 & \frac{1}{7} \\ 4 & -3 \end{bmatrix}=\begin{bmatrix}\frac{1}{7} & 0 \\0 & 1 \end{bmatrix}\text{A}$
$\Big(\text{Applying R}_1\rightarrow\ \frac{1}{7}\text{R}_1\Big)$
$\Rightarrow\ \begin{bmatrix}1 & \frac{1}{7} \\ 0 & -\frac{25}{7} \end{bmatrix}=\begin{bmatrix}\frac{1}{7} & 0 \\ -\frac{4}{7} & 1 \end{bmatrix}\text{A}$
$\Big(\text{Applying R}_2>\text{R}_2\ 4\text{R}_1\Big)$
$\Rightarrow\ \begin{bmatrix}1 & \frac{1}{7} \\ 0 & 1 \end{bmatrix}=\begin{bmatrix}\frac{1}{7} & 0 \\ \frac{4}{25} & -\frac{7}{25} \end{bmatrix}\text{A}$
$\Big(\text{Applying R}_2\rightarrow\ -\frac{7}{25}\text{R}_2\Big)$
$\Rightarrow\ \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix}\frac{3}{25} & \frac{1}{25} \\ \frac{4}{25} & -\frac{7}{25} \end{bmatrix}\text{A}$
$\Big(\text{Applying R}_1\rightarrow\text{R}_1-\frac{1}{7}\text{R}_2\Big)$
$\therefore\ \text{A}^{-1}=\frac{1}{25}\begin{bmatrix} 3 & 1 \\ 4 & -7 \end{bmatrix}$

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