If $A=\begin{bmatrix}2 & 3 \\ 5 & -2\end{bmatrix}$ be such that $A^{-1}=\lambda A$. Then, the value of $\lambda$ is

$|A| = (2)(-2) - (3)(5) = -4 - 15 = -19$ $\text{adj }A = \begin{bmatrix}-2 & -3 \\ -5 & 2\end{bmatrix}$ $A^{-1} = \frac{1}{|A|}\text{adj }A = \frac{-1}{19}\begin{bmatrix}-2 & -3 \\ -5 & 2\end{bmatrix}$ $= \begin{bmatrix}\frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & \frac{-2}{19}\end{bmatrix}$ Given $A^{-1} = \lambda A$ $\Rightarrow \begin{bmatrix}\frac{2}{19} & \frac{3}{19} \\ \frac{5}{19} & \frac{-2}{19}\end{bmatrix} = \lambda \begin{bmatrix}2 & 3 \\ 5 & -2\end{bmatrix}$ Comparing, $\lambda = \frac{1}{19}$ $\boxed{\frac{1}{19}}$