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

The correct answer is Option (3) → $\frac{1}{11}$

$A=\begin{bmatrix} 3 & 1 \\2 & -3 \end{bmatrix}$

$A^2=\begin{bmatrix} 3 & 1 \\2 & -3 \end{bmatrix}\begin{bmatrix} 3 & 1 \\2 & -3 \end{bmatrix}=\begin{bmatrix} 11 & 0 \\0& 11 \end{bmatrix}$

We know that $A^{-1}=KA$

So $A^{-1}A=KAA$

$⇒I=KA^2$

So $\begin{bmatrix} 1 & 0 \\0& 1 \end{bmatrix}=K\begin{bmatrix} 11 & 0 \\0& 11 \end{bmatrix}$

$⇒K=\frac{1}{11}$