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

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Solution

## Step 1: Determine the value of $\mathrm{det}\left({A}^{4}\right)$It is given that, $A=\left[\begin{array}{cc}2& 3\\ 0& -1\end{array}\right]$.So, ${A}^{2}=A×A$.$⇒{A}^{2}=\left[\begin{array}{cc}2& 3\\ 0& -1\end{array}\right]×\left[\begin{array}{cc}2& 3\\ 0& -1\end{array}\right]\phantom{\rule{0ex}{0ex}}⇒{A}^{2}=\left[\begin{array}{cc}\left(2×2\right)+\left(3×0\right)& \left(2×3\right)+\left(-1×3\right)\\ \left(0×2\right)+\left(-1×0\right)& \left(3×0\right)+\left[\left(-1\right)×\left(-1\right)\right]\end{array}\right]\phantom{\rule{0ex}{0ex}}⇒{A}^{2}=\left[\begin{array}{cc}4& 3\\ 0& 1\end{array}\right]$Thus, ${A}^{4}={A}^{2}×{A}^{2}$$⇒{A}^{4}=\left[\begin{array}{cc}4& 3\\ 0& 1\end{array}\right]×\left[\begin{array}{cc}4& 3\\ 0& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}⇒{A}^{4}=\left[\begin{array}{cc}\left(4×4\right)+\left(3×0\right)& \left(4×3\right)+\left(3×1\right)\\ \left(0×4\right)+\left(1×0\right)& \left(0×3\right)+\left[1×1\right]\end{array}\right]\phantom{\rule{0ex}{0ex}}⇒{A}^{4}=\left[\begin{array}{cc}16& 15\\ 0& 1\end{array}\right]$Hence, $\mathrm{det}\left({A}^{4}\right)=\left|\begin{array}{cc}16& 15\\ 0& 1\end{array}\right|$$⇒\mathrm{det}\left({A}^{4}\right)=\left(16×1\right)-\left(0×15\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{det}\left({A}^{4}\right)=16$Step 2: Calculate the value of ${A}^{10}$It is calculated that, ${A}^{2}=\left[\begin{array}{cc}4& 3\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}{2}^{2}& {2}^{2}-1\\ 0& {\left(-1\right)}^{2}\end{array}\right]$ and ${A}^{4}=\left[\begin{array}{cc}16& 15\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}{2}^{4}& {2}^{4}-1\\ 0& {\left(-1\right)}^{4}\end{array}\right]$.Thus it can be written that, ${A}^{n}=\left[\begin{array}{cc}{2}^{n}& {2}^{n}-1\\ 0& {\left(-1\right)}^{n}\end{array}\right]$.So, ${A}^{10}=\left[\begin{array}{cc}{2}^{10}& {2}^{10}-1\\ 0& {\left(-1\right)}^{10}\end{array}\right]=\left[\begin{array}{cc}1024& 1023\\ 0& 1\end{array}\right]$.Step 3: Determine the value of the given expressionThe given expression: $\mathrm{det}\left({A}^{4}\right)+\mathrm{det}\left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]$$⇒\mathrm{det}\left({A}^{4}\right)+\mathrm{det}\left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]=16+\mathrm{det}\left[\left[\begin{array}{cc}1024& 1023\\ 0& 1\end{array}\right]-\mathrm{Adj}\left({2}^{10}×\left[\begin{array}{cc}1024& 1023\\ 0& 1\end{array}\right]\right)\right]\phantom{\rule{0ex}{0ex}}⇒\mathrm{det}\left({A}^{4}\right)+\mathrm{det}\left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]=16+\mathrm{det}\left[\left[\begin{array}{cc}1024& 1023\\ 0& 1\end{array}\right]-\mathrm{Adj}\left[\begin{array}{cc}1048576& 1047552\\ 0& 1024\end{array}\right]\right]\phantom{\rule{0ex}{0ex}}⇒\mathrm{det}\left({A}^{4}\right)+\mathrm{det}\left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]=16+\mathrm{det}\left[\left[\begin{array}{cc}1024& 1023\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}1024& -1047552\\ 0& 1048576\end{array}\right]\right]\phantom{\rule{0ex}{0ex}}⇒\mathrm{det}\left({A}^{4}\right)+\mathrm{det}\left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]=16+\left|\begin{array}{cc}0& 1046529\\ 0& -1048575\end{array}\right|\phantom{\rule{0ex}{0ex}}⇒\mathrm{det}\left({A}^{4}\right)+\mathrm{det}\left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]=16+0\phantom{\rule{0ex}{0ex}}⇒\mathrm{det}\left({A}^{4}\right)+\mathrm{det}\left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]=16$Hence, the value of $\mathrm{det}\left({A}^{4}\right)+\mathrm{det}\left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]$ is equal to $16$.

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