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

The Correct Option is C

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