If $A=\left[\begin{array}{cc}2 & 3 \ 0 & -1\end{array} ight]$, then the value of $\operatorname{de

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

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(16) : $$ \begin{gathered} A=\left[\begin{array}{cc} 2 & 3 \ 0 & -1 \end{array}ight], A^{2}=\left[\begin{array}{ll} 4 & 3 \ 0 & 1 \end{array}ight], A^{3}=\left[\begin{array}{cc} 8 & 9 \ 0 & -1 \end{array}ight], \ldots, \ A^{n}=\left[\begin{array}{cc} 2^{n} & 2^{n}-(-1)^{n} \ 0 & (-1)^{n} \end{array}ight] \end{gathered} $$ Also, $2 A \cdot(\operatorname{adj}(2 A))=|2 A| I$\ $\Rightarrow \quad 2 A \cdot \operatorname{adj}(2 A)=22(-2) I$\ $\Rightarrow \quad A \operatorname{adj}(2 A)=-22 I$\ Now, $\left|A^{10}-(\operatorname{adj}(2 A))^{10}ight|$\ $=\frac{1}{|A|^{10}}\left|A^{20}-A^{10}(\operatorname{adj}(2 A))^{10}ight|$ $$\begin{equation*} =\frac{1}{|A|^{10}}\left|A^{20}-2^{20} Iight| 	ag{i} \end{equation*}$$ $\because \quad A^{20}-220 I$\ $=\left[\begin{array}{cc}2^{20} & 2^{20}-1 \ 0 & 1\end{array}ight]-\left[\begin{array}{cc}2^{20} & 0 \ 0 & 2^{20}\end{array}ight]=\left[\begin{array}{ll}0 & 2^{20}-1 \ 0 & 1-2^{20}\end{array}ight]$\ $\Rightarrow \quad\left|A^{20}-220 Iight|=0$\ From (i), $\left|A^{10}-(\operatorname{adj}(2 A))^{10}ight|=0$ Hence, $\left|A^{4}ight|+\left|A^{10}-(\operatorname{adj}(2 A))^{10}ight|$\ $=|A|^{4}+0=(-2)^{4}=16$\