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Let the matrix A=[ … ] and the matrix B_0=A^49+2A^98. If B_n= Adj(B_n-1) for all n≥ 1, then det(B_4) is equal to

JEE Main 2022 — Mathematics Algebra

2022mcqhard

Let the matrix A=[010100001]A=[\begin{matrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{matrix}]A=[010​100​001​] and the matrix B0=A49+2A98{B}_{0}={A}^{49}+2{A}^{98}B0​=A49+2A98. If Bn=Adj(Bn−1){B}_{n}=\mathrm{Adj}({B}_{n-1})Bn​=Adj(Bn−1​) for all n≥1n\geq 1n≥1, then det(B4)det({B}_{4})det(B4​) is equal to

Official previous-year question

Held on 28 Jul 2022 · Verified 6 Jul 2026.

Options

  1. A

    328{3}^{28}328

  2. B

    330{3}^{30}330

  3. C

    332{3}^{32}332

  4. D

    336{3}^{36}336

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