Let the matrix A= [0, 1, 0; 0, 0, 1; 1, 0, 0] and the matrix B_0=A^49+2 A^98. If B_n= Adj (B_n-1) for all n ≥ 1, then det ( B_4) is equal to:

Let the matrix $\displaystyle A=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right]$ and the matrix $\displaystyle B_0=A^{49}+2 A^{98}$. If $\displaystyle B_n=\operatorname{Adj}\left(B_{n-1}\right)$ for all $\displaystyle \mathrm{n} \geq 1$, then $\displaystyle \operatorname{det}\left(\mathrm{B}_4\right)$ is equal to :