If A = […] be such that A^-1 = KA , then the value of K is:

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

Using \(A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A\):

    \(|A| = (2)(-2) - (3)(5) = -19\)

    \( \displaystyle \mathrm{adj}\,A = \begin{bmatrix}-2&-3\\-5&2\end{bmatrix} \)

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

    \( \displaystyle A^{-1} = \frac{1}{19}A \)

So \(K = \frac{1}{19}\).

Shortcut: by the Cayley-Hamilton theorem for a 2 × 2 matrix, \(A^2 - (\mathrm{tr}\,A)A + |A|I = O\). Here \(\mathrm{tr}\,A = 0\) and \(|A| = -19\):

    \(A^2 = 19I\)

    \( \displaystyle A\left(\frac{1}{19}A\right) = I \)

So \(A^{-1} = \frac{1}{19}A\).