Let the matrix $A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and the matrix $B_0 =…

Let the matrix $A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and the matrix $B_0 = A^{49} + 2A^{98}$. If $B_n = \text{Adj}(B_{n-1})$ for all $n \geq 1$, then $\text{det}(B_4)$ is equal to.
  1. 328
  2. 330
  3. 332
  4. 336

Solution

$A=\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ $A^2=\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=I$ i.e. $A^{98}=I$; $A^{49}=A$ $\therefore B_{0}=A^{98}+2A^{49}=I+2A=\begin{bmatrix} 1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ Given $B_{n}=\text{adj}(B_{n-1})$ $\therefore B_{4}=\text{adj}(B_{3})=\text{adj}(\text{adj}(B_{2}))=\text{adj}(\text{adj}(\text{adj}(B_{1})))$ $=\text{adj}(\text{adj}(\text{adj}(\text{adj}(B_{0}))))$ $\therefore |B_{4}|=|B_{0}|^{2^{4}}=|B_{0}|^{16}$ Now $|B_{0}|=\begin{vmatrix} 1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 3 \end{vmatrix}=-9$ Hence, $|B_{4}|=|B_{0}|^{16}=9^{16}=3^{32}$

Asked in: JEE Main 2022 (28 Jul Shift 1)

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