Let the matrix [A=[[arraylll 0 & 1 & 0 0 & 0 & 1 1 & 0 & 0 array]]] and the matrix B_0=A^4…

Question

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 :

Accepted Answer

$$A = \left[ {\begin{pmatrix} 0 & 1 & 0  \
   0 & 0 & 1  \
   1 & 0 & 0  \

\end{pmatrix} } \right]$$

$$ \Rightarrow {A^2} = \left[ {\begin{pmatrix} 0 & 1 & 0  \
   0 & 0 & 1  \
   1 & 0 & 0  \

\end{pmatrix} } \right] \times \left[ {\begin{pmatrix} 0 & 1 & 0  \
   0 & 0 & 1  \
   1 & 0 & 0  \

\end{pmatrix} } \right] = \left[ {\begin{pmatrix} 0 & 0 & 1  \
   1 & 0 & 0  \
   0 & 1 & 0  \

\end{pmatrix} } \right]$$

$$ \Rightarrow {A^3} = \left[ {\begin{pmatrix} 0 & 0 & 1  \
   1 & 0 & 0  \
   0 & 1 & 0  \

\end{pmatrix} } \right]\left[ {\begin{pmatrix} 0 & 1 & 0  \
   0 & 0 & 1  \
   1 & 0 & 0  \

\end{pmatrix} } \right] = \left[ {\begin{pmatrix} 1 & 0 & 0  \
   0 & 1 & 0  \
   0 & 0 & 1  \

\end{pmatrix} } \right] = l$$

Now $${B_0} = {A^{49}} + 2{A^{98}} = {({A^3})^{16}}\,.\,A + 2{({A^3})^{32}}\,.\,{A^2}$$

$${B_0} = A + 2{A^2} = \left[ {\begin{pmatrix} 0 & 1 & 0  \
   0 & 0 & 1  \
   1 & 0 & 0  \

\end{pmatrix} } \right] + \left[ {\begin{pmatrix} 0 & 0 & 2  \
   2 & 0 & 0  \
   0 & 2 & 0  \

\end{pmatrix} } \right] = \left[ {\begin{pmatrix} 0 & 1 & 2  \
   2 & 0 & 1  \
   1 & 2 & 0  \

\end{pmatrix} } \right]$$

$|{B_0}| = 9$

Since, $${B_n} = Adj\,|{B_{n - 1}}| \Rightarrow |{B_n}| = |{B_{n - 1}}{|^2}$$

Hence $$|{B_4}| = |{B_3}{|^2} = |{B_2}{|^4} = |{B_1}{|^8} = |{B_0}{|^{16}}$$

$= |{3^2}{|^{16}} = {3^{32}}$

Solutions · 1