Let the matrix $$A=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right]$$ and the matrix $B_{0}=A^{49}+2 A^{98}$. If $B_{n}=A d j\left(B_{n-1}\right)$ for all $n \geq 1$, then $\operatorname{det}\left(B_{4}\right)$ is equal to :

<p>$$A = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right]$$</p> <p>$$ \Rightarrow {A^2} = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] \times \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] = \left[ {\matrix{ 0 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr } } \right]$$</p> <p>$$ \Rightarrow {A^3} = \left[ {\matrix{ 0 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr } } \right]\left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] = l$$</p> <p>Now ${B_0} = {A^{49}} + 2{A^{98}} = {({A^3})^{16}}\,.\,A + 2{({A^3})^{32}}\,.\,{A^2}$</p> <p>$${B_0} = A + 2{A^2} = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right] + \left[ {\matrix{ 0 & 0 & 2 \cr 2 & 0 & 0 \cr 0 & 2 & 0 \cr } } \right] = \left[ {\matrix{ 0 & 1 & 2 \cr 2 & 0 & 1 \cr 1 & 2 & 0 \cr } } \right]$$</p> <p>$|{B_0}| = 9$</p> <p>Since, ${B_n} = Adj\,|{B_{n - 1}}| \Rightarrow |{B_n}| = |{B_{n - 1}}{|^2}$</p> <p>Hence $|{B_4}| = |{B_3}{|^2} = |{B_2}{|^4} = |{B_1}{|^8} = |{B_0}{|^{16}}$</p> <p>$= |{3^2}{|^{16}} = {3^{32}}$</p>