If $A =\begin{bmatrix}2&3\\5&-2\end{bmatrix}$ be such that $A^{-1} = KA$, then the value of K is:

$\frac{1}{19}$

The correct answer is Option (2) → $\frac{1}{19}$

$A=\begin{bmatrix}2&3\\5&-2\end{bmatrix}$

$\det(A)=2(-2)-15=-19$

$A^{-1}=\frac{1}{-19}\begin{bmatrix}-2&-3\\-5&2\end{bmatrix} =\frac{1}{19}\begin{bmatrix}2&3\\5&-2\end{bmatrix} =\frac{1}{19}A$

Given $A^{-1}=KA$ $\Rightarrow$ $K=\frac{1}{19}$

$K=\frac{1}{19}$