#### Please solve RD Sharma class 12 Chapter 6. Adjoint and Inverse of Matrix excercise 6.2 question 1 maths textbook solution

$\frac{1}{25}\begin{bmatrix} 3 &1 \\ 4&-7 \end{bmatrix}$
Hint: Here we use the concept of elementary row operation

Given: $\begin{bmatrix} 7 &1 \\ 4&-3 \end{bmatrix}$

Solution: $A = IA$
$A = \begin{bmatrix} 1& 0\\ 0& 1 \end{bmatrix}A$
$\Rightarrow \begin{bmatrix} 7& 1\\ 4& -3 \end{bmatrix} = \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}A$
Applying    $R_1 \rightarrow \frac{1}{7}R_1$
$\Rightarrow \begin{bmatrix} 1 &\frac{1}{7} \\ 4&-3 \end{bmatrix} =\begin{bmatrix} \frac{1}{7} &0 \\ 0&1 \end{bmatrix}A$

Applying    $R_2\rightarrow R_2-4R_1$
$\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}A$
Applying     $R_2\rightarrow \frac{-7}{25}R_2$

$\Rightarrow \begin{bmatrix} 1 & \frac{1}{7}\\ 0&1 \end{bmatrix} =\begin{bmatrix} \frac{1}{7}&0 \\ \frac{4}{25}&\frac{-7}{25} \end{bmatrix}$
Applying    $R_1\rightarrow R_1-\frac{1}{7}R_1$
$\Rightarrow \begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix} =\begin{bmatrix} \frac{21}{75}&\frac{1}{25} \\ \frac{4}{25}&\frac{-7}{25} \end{bmatrix}A$

$\Rightarrow \begin{bmatrix} 1 & 0\\ 0&1 \end{bmatrix} =\begin{bmatrix} \frac{3}{25}&\frac{1}{25} \\ \frac{4}{25}&\frac{-7}{25} \end{bmatrix}A$

$\Rightarrow A^{-1} = \frac{1}{25}\begin{bmatrix} 3 & 1\\ 4& -7 \end{bmatrix}$