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(C) Given $A = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{bmatrix}$.We observe $A^2 = \begin{bmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 &

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$3^{32}$

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(C) Given $A = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{bmatrix}$.We observe $A^2 = \begin{bmatrix} 0 & 0 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{bmatrix}$ and $A^3 = I$.Since $49 = 3 	imes 16 + 1$,$A^{49} = A^1 = A$.Since $98 = 3 	imes 32 + 2$,$A^{98} = A^2$.Thus,$B_{0} = A + 2A^2 = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 2 \ 2 & 0 & 0 \ 0 & 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \ 2 & 0 & 1 \ 1 & 2 & 0 \end{bmatrix}$.Calculating $\det(B_{0}) = 0(0-2) - 1(0-1) + 2(4-0) = 0 + 1 + 8 = 9$.We know $\det(	ext{Adj}(M)) = (\det(M))^{k-1}$ where $k$ is the order of the matrix. Here $k=3$,so $\det(	ext{Adj}(M)) = (\det(M))^2$.For $B_{n} = 	ext{Adj}(B_{n-1})$,$\det(B_{n}) = (\det(B_{n-1}))^2$.Thus,$\det(B_{4}) = (\det(B_{0}))^{2^4} = (\det(B_{0}))^{16}$.$\det(B_{4}) = 9^{16} = (3^2)^{16} = 3^{32}$.

Let the matrix $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$ and the matrix $B_{0} = A^{49} + 2A^{98}$. If $B_{n} = \text{Adj}(B_{n-1})$ for all $n \geq 1$,then $\det(B_{4})$ is equal to:

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