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If $A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$, then $\operatorname{det}\left(A^6+B^6\right)=$
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$-212$
Given,
$A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$
$\begin{aligned} & \Rightarrow A^2=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \\ & \Rightarrow A^2=\left[\begin{array}{ll}1+0 & 0+0 \\ 2+2 & 0+1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]\end{aligned}$
$\Rightarrow A^3=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \quad\left[A^2 \cdot A=A^3\right]$
$\Rightarrow A^3 \doteq\left[\begin{array}{cc}1+0 & 0+0 \\ 4+2 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 6 & 1\end{array}\right]$
Similarly, $A^6=\left[\begin{array}{cc}1 & 0 \\ 12 & 1\end{array}\right] \quad\left[A^n=A_{21}=n \times 2\right]$
Now, $B=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$
$\begin{aligned} & \Rightarrow B^2=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right] \\ & \Rightarrow B^2=\left[\begin{array}{ll}1+0 & 3+3 \\ 0+0 & 0+1\end{array}\right]=\left[\begin{array}{ll}1 & 6 \\ 0 & 1\end{array}\right] \\ & \Rightarrow B^3=\left[\begin{array}{ll}1 & 6 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 9 \\ 0 & 1\end{array}\right]\end{aligned}$
Similarly, $B^6=\left[\begin{array}{cc}1 & 18 \\ 0 & 1\end{array}\right]$ $\left[B^n=B_{12}=n \times 3\right]$
$\therefore \quad\left(A^6+B^6\right)=\left[\begin{array}{cc}1 & 0 \\ 12 & 1\end{array}\right]+\left[\begin{array}{cc}1 & 18 \\ 0 & 1\end{array}\right]$
$A^6+B^6=\left[\begin{array}{cc}2 & 18 \\ 12 & 2\end{array}\right]$
$\operatorname{Det}\left(A^6+B^6\right)=4-216=-212$
$A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$
$\begin{aligned} & \Rightarrow A^2=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \\ & \Rightarrow A^2=\left[\begin{array}{ll}1+0 & 0+0 \\ 2+2 & 0+1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]\end{aligned}$
$\Rightarrow A^3=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \quad\left[A^2 \cdot A=A^3\right]$
$\Rightarrow A^3 \doteq\left[\begin{array}{cc}1+0 & 0+0 \\ 4+2 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 6 & 1\end{array}\right]$
Similarly, $A^6=\left[\begin{array}{cc}1 & 0 \\ 12 & 1\end{array}\right] \quad\left[A^n=A_{21}=n \times 2\right]$
Now, $B=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$
$\begin{aligned} & \Rightarrow B^2=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right] \\ & \Rightarrow B^2=\left[\begin{array}{ll}1+0 & 3+3 \\ 0+0 & 0+1\end{array}\right]=\left[\begin{array}{ll}1 & 6 \\ 0 & 1\end{array}\right] \\ & \Rightarrow B^3=\left[\begin{array}{ll}1 & 6 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 9 \\ 0 & 1\end{array}\right]\end{aligned}$
Similarly, $B^6=\left[\begin{array}{cc}1 & 18 \\ 0 & 1\end{array}\right]$ $\left[B^n=B_{12}=n \times 3\right]$
$\therefore \quad\left(A^6+B^6\right)=\left[\begin{array}{cc}1 & 0 \\ 12 & 1\end{array}\right]+\left[\begin{array}{cc}1 & 18 \\ 0 & 1\end{array}\right]$
$A^6+B^6=\left[\begin{array}{cc}2 & 18 \\ 12 & 2\end{array}\right]$
$\operatorname{Det}\left(A^6+B^6\right)=4-216=-212$
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