If $A =\begin{bmatrix}2&3\\5&-2\end{bmatrix}$ be such that $A^{-1} = KA$, then the value of K is:
Answer & explanation
Correct answer: option 2
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 $A^{-1}=KA$ $\Rightarrow$ $K=\frac{1}{19}$
$K=\frac{1}{19}$