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#### Explain Solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and inverse of a Matrix Exercise very short answer type Question 19 maths textbook solution

VSQ : 19

Answer : $\frac{1}{19}$

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

Hint : Find $A^{-1}$ and put it in equation and then compare it.

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

\begin{aligned} &|A|=-4-15 \\ &=-19 \\ &|A| \neq 0 \\ &A^{-1} \text { exists } \\ &A^{-1}=\frac{\operatorname{adj} A}{|A|} \end{aligned}

\begin{aligned} &=\frac{-1}{19}\left[\begin{array}{cc} -2 & -3 \\ -5 & 2 \end{array}\right]\\ &=\frac{-1}{19}\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]\\ &\text { Now } \mathrm{A}^{-1}=K A\\ &\frac{1}{19}\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]=K\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]\\ &\text { On comparing }\\ &\mathrm{K}=\frac{1}{19} \end{aligned}