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

Step 1: Determine the value of $\det \left(A^4\right)$

It is given that, $A=\left[\begin{array}{cc}2& 3\\ 0& -1\end{array}\right]$.

So, $A^2=A\times A$.

$$\begin{gathered} \Rightarrow A^2=\left[\begin{array}{cc}2& 3\\ 0& -1\end{array}\right]\times \left[\begin{array}{cc}2& 3\\ 0& -1\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}\left(2\times 2\right)+\left(3\times 0\right)& \left(2\times 3\right)+\left(-1\times 3\right)\\ \left(0\times 2\right)+\left(-1\times 0\right)& \left(3\times 0\right)+\left[\left(-1\right)\times \left(-1\right)\right]\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}4& 3\\ 0& 1\end{array}\right] \end{gathered}$$

Thus, $A^4=A^2\times A^2$

$$\begin{gathered} \Rightarrow A^4=\left[\begin{array}{cc}4& 3\\ 0& 1\end{array}\right]\times \left[\begin{array}{cc}4& 3\\ 0& 1\end{array}\right] \\ \Rightarrow A^4=\left[\begin{array}{cc}\left(4\times 4\right)+\left(3\times 0\right)& \left(4\times 3\right)+\left(3\times 1\right)\\ \left(0\times 4\right)+\left(1\times 0\right)& \left(0\times 3\right)+\left[1\times 1\right]\end{array}\right] \\ \Rightarrow A^4=\left[\begin{array}{cc}16& 15\\ 0& 1\end{array}\right] \end{gathered}$$

Hence, $\det \left(A^4\right)=\left|\begin{array}{cc}16& 15\\ 0& 1\end{array}\right|$

$$\begin{gathered} \Rightarrow \det \left(A^4\right)=\left(16\times 1\right)-\left(0\times 15\right) \\ \Rightarrow \det \left(A^4\right)=16 \end{gathered}$$

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 expression

The given expression: $\det \left(A^4\right)+\det \left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]$

$$\begin{gathered} \Rightarrow \det \left(A^4\right)+\det \left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]=16+\det \left[\left[\begin{array}{cc}1024& 1023\\ 0& 1\end{array}\right]-\mathrm{Adj}\left(2^{10}\times \left[\begin{array}{cc}1024& 1023\\ 0& 1\end{array}\right]\right)\right] \\ \Rightarrow \det \left(A^4\right)+\det \left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]=16+\det \left[\left[\begin{array}{cc}1024& 1023\\ 0& 1\end{array}\right]-\mathrm{Adj}\left[\begin{array}{cc}1048576& 1047552\\ 0& 1024\end{array}\right]\right] \\ \Rightarrow \det \left(A^4\right)+\det \left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]=16+\det \left[\left[\begin{array}{cc}1024& 1023\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}1024& -1047552\\ 0& 1048576\end{array}\right]\right] \\ \Rightarrow \det \left(A^4\right)+\det \left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]=16+\left|\begin{array}{cc}0& 1046529\\ 0& -1048575\end{array}\right| \\ \Rightarrow \det \left(A^4\right)+\det \left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]=16+0 \\ \Rightarrow \det \left(A^4\right)+\det \left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]=16 \end{gathered}$$

Hence, the value of $\det \left(A^4\right)+\det \left[{\left(A\right)}^{10}-\mathrm{Adj}{\left(2A\right)}^{10}\right]$ is equal to $16$.