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

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

For any matrix $A$, we have $A \cdot A^{-1} = I$ where $I$ is the identity matrix.

Substituting $A^{-1} = KA$:

$A \cdot (KA) = I$

$K(A \cdot A) = I$

$KA^2 = I$

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Calculate $A^2$:

$A^2 = A \times A$

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

For position (1,1): $(2)(2) + (3)(5) = 4 + 15 = 19$

For position (1,2): $(2)(3) + (3)(-2) = 6 - 6 = 0$

For position (2,1): $(5)(2) + (-2)(5) = 10 - 10 = 0$

For position (2,2): $(5)(3) + (-2)(-2) = 15 + 4 = 19$

$A^2 = \begin{bmatrix} 19 & 0 \\ 0 & 19 \end{bmatrix}$

$A^2 = 19\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$A^2 = 19I$

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From $KA^2 = I$ and $A^2 = 19I$:

$K(19I) = I$

$19KI = I$

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

Therefore, the value of $K$ is $\frac{1}{19}$.