Let the matrix A=pmatrix0&1&00&0&11&0&0pmatrix and B_0=A^49+2A^98.

Question

Let the matrix $A=\begin{pmatrix}0&1&0\0&0&1\1&0&0\end{pmatrix}$ and $B_0=A^{49}+2A^{98}$. If $B_n=\operatorname{Adj}(B_{n-1})$ for all $n>1$, then $\det(B_4)$ is equal to:

Accepted Answer

$A^3=I$ (direct check), so $A^{49}=A^{48}\cdot A=A$ and $A^{98}=A^{96}\cdot A^2=A^2$.

$B_0=A+2A^2=\begin{pmatrix}0&1&2\2&0&1\1&2&0\end{pmatrix}$, and $\det(B_0)=9$.

For $2	imes2$... here $3	imes3$: $\det(B_n)=\det(\operatorname{Adj}(B_{n-1}))=\det(B_{n-1})^2$.

So $\det(B_1)=9^2$, $\det(B_2)=9^4$, $\det(B_3)=9^8$, $\det(B_4)=9^{16}=3^{32}$.

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