Question

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

• A. $3^{28}$
• B. $3^{30}$
• C. $3^{32}$
• D. $3^{36}$

Answer 1

Answer: (C)

$3^{32}$

Answer 2

The correct answer is (C) : $3^{32}$
$A^2=\begin{bmatrix} 0 & 1 & 0 \\\ 1 & 0 & 0 \\\ 0 & 0 & 1 \\\ \end{bmatrix}\begin{bmatrix} 0 & 1 & 0 \\\ 1 & 0 & 0 \\\ 0 & 0 & 1 \\\ \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \\\ \end{bmatrix}=I$
$B_0=(A^2)^{49}+2(A^2)^{24}A⇒B_0=2A+I$
$B_0=\begin{bmatrix} 1 & 2 & 0 \\\ 2 & 1 & 0 \\\ 0 & 0 & 3 \\\ \end{bmatrix}⇒|B_0|=-9$
$B_4=adj\ B_3=adj(adj\ B_2)=adj(adj(adj\ B_1)=adj(adj/adj(adj\ B_0)$
$|B_4|=|B_0|^{(3-1)^4}$
$=|B_0|^{16}=(-9)^{16}$
$=(-9)^{16}=3^{32}$