#### Please solve rd  sharma class 12 Chapter 6 Adjoint and Inverse of Matrix excercise 6.2 question 4  maths textbook solution

Answer:$A^{-1}=\left[\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}\right]$

Hint: Here, we use the concept of matrix inverse using elementary row operation

Given: $\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]$

Solution:             Let    $A = \left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]$

$A = IA$

$A=\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right], I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$

$\Rightarrow\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] A$

Applying $R_{1} \rightarrow R_{1}-R_{2}$

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

Applying $R_{2} \rightarrow R_{2}-R_{1}$

$\Rightarrow\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ -1 & 2 \end{array}\right] A$

Applying $R_{1} \rightarrow R_{1}-2R_{2}$

$\Rightarrow\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}\right] A$

Hence, $A^{-1}=\left[\begin{array}{cc} 3 & -5 \\ -1 & 2 \end{array}\right]$