If $A=\begin{bmatrix}K&4\\4&K\end{bmatrix}$ and $|A^3|=729$, then value of $K^8$ is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $5^8$
$A=\begin{bmatrix}K&4\\4&K\end{bmatrix}$
$|A^3|=\begin{bmatrix}K&4\\4&K\end{bmatrix}^3=(K^2-16)^3=729$
$|A|=K^2-16$
$|A^3|=(|A|)^3=(K^2-16)^3=729$
$=K^2-16=\sqrt[3]{729}$
$=K^2-16=9$
$=K^2=25$
$=K=5$
$∴K^8=5^8$