if A =

$2 \mathrm{~A}$ adj $(2 \mathrm{~A})=12 \mathrm{AlI}$

$\Rightarrow \mathrm{A} \operatorname{adj}(2 \mathrm{~A})=-4 \mathrm{I} \quad \ldots .(\mathrm{i})$

Now, $E=\left|A^{4}\right|+\left|A^{10}-(\operatorname{adj}(2 A))^{10}\right|$

$=(-2)^{4}+\frac{\left|\mathrm{A}^{20}-\mathrm{A}^{10}(\operatorname{adj} 2 \mathrm{~A})^{10}\right|}{|\mathrm{A}|^{10}}$

$=16+\frac{\left|\mathrm{A}^{20}-(\mathrm{A} \operatorname{adj}(2 \mathrm{~A}))^{10}\right|}{|\mathrm{A}|^{10}}$

$=16+\frac{\left|A^{20}-2^{10} I\right|}{2^{10}}$ (from

Now, characteristic roots of $\mathrm{A}$ are 2 and $-1 .$ So, characteristic roots of $\mathrm{A}^{20}$ are $2^{10}$ and $1 .$

Hence, $\left(\mathrm{A}^{20}-2^{10} \mathrm{I}\right)\left(\mathrm{A}^{20}-\mathrm{I}\right)=0$

$\Rightarrow \mid \mathrm{A}^{20}-2^{10} \mathrm{Il}=0\left(\right.$ as $\left.\mathrm{A}^{20} \neq \mathrm{I}\right)$

$\Rightarrow \mathrm{E}=16 \mathrm{Ans} .$