# If A =

Question:

If $\mathrm{A}=\left[\begin{array}{cc}2 & 3 \\ 0 & -1\end{array}\right]$, then the value of $\operatorname{det}\left(\mathrm{A}^{4}\right)+\operatorname{det}\left(\mathrm{A}^{10}-(\operatorname{Adj}(2 \mathrm{~A}))^{10}\right)$ is equal to

Solution:

$2 \mathrm{~A}$ adj $(2 \mathrm{~A})=|2 \mathrm{~A}| \mathrm{I}$

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

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

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

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

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

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

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

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

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

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