If $A=\begin{bmatrix}K&4\\4&K\end{bmatrix}$ and $|A^3|=729$, then value of $K^8$ is

$5^8$

The correct answer is Option (3) → $5^8$

$A=\begin{bmatrix}K&4\\4&K\end{bmatrix}$

$|A^3|=\begin{bmatrix}K&4\\4&K\end{bmatrix}^3=(K^2-16)^3=729$

$|A|=K^2-16$

$|A^3|=(|A|)^3=(K^2-16)^3=729$

$=K^2-16=\sqrt[3]{729}$

$=K^2-16=9$

$=K^2=25$

$=K=5$

$∴K^8=5^8$