JEE Main 2022 · 28 Jul · Shift 1 · Mathematics Q14

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MathematicsSingle correct+4 / -1Question 14 of 90

Let the matrix $A=\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}$ and the matrix $B_0=A^{49}+2A^{98}$ . If $B_n=\operatorname{Adj}(B_{n-1})$ for all $n\ge1$ , then $\det(B_4)$ is equal to :

3^{28}

3^{30}

3^{32}

3^{36}

Answer and worked solution

Official answer: C

$A^3=I$ , so $B_0=A+2A^2$ .

Eigenvalues $1+2=3$ and $\omega+2\omega^2$ , $\omega^2+2\omega$ (product 3). So $\det B_0=9$ .

Adjugates. For $3\times3$ , $\det(\operatorname{adj}B)=(\det B)^2$ , so $\det B_4=9^{16}=3^{32}$ .

(C)

Topics:Adjoint and inverse Special matrices

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JEE Main 2022 · 28 Jul · Shift 1, Mathematics question 14