#### Need solution for rd sharma maths class 12 chapter adjoint and inverse of a matrix exercise multiple choice question question 24

option (b) $\frac{1}{19}$

Given: $A=\begin{bmatrix} 2 &3 \\ 5& -2 \end{bmatrix} \& A^{-1}=KA$

Hint: simply multiply the matrices

\begin{aligned} &\text { Solution: } \mathrm{A}=\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]\\ &\therefore|\mathrm{A}|=-4-15=-19\\ &\text { So } \mathrm{A}^{-1} \text { exists }\\ &\text { We knows= that } A^{-1} \mathrm{~A}=\mathrm{I}\\ &\Rightarrow \mathrm{KA} \cdot \mathrm{A}=\mathrm{I} \quad\left(\therefore \mathrm{A}^{-1}=\mathrm{KA}\right)\\ &\Rightarrow \mathrm{K}\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\\ &\Rightarrow \mathrm{K}\left[\begin{array}{cc} 19 & 0 \\ 0 & 19 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\\ &\Rightarrow \mathrm{K}(19 \mathrm{I})=\mathrm{I}\\ &\Rightarrow 19 \mathrm{~K}=1\\ &\Rightarrow \mathrm{K}=\frac{1}{19} \end{aligned}

Hence, option (b) is correct