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  • Let the matrix <img alt="\mathrm{A=\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right]}" src="/latex-image/?%5Cmathrm%7BA%3D%5Cleft%5B%5Cbeg

Let the matrix \mathrm{A=\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right]} and the matrix \mathrm{\mathrm{B}_{0}=\mathrm{A}^{49}+2 \mathrm{~A}^{98}}. If \mathrm{\mathrm{B}_{\mathrm{n}}=\operatorname{Adj}\left(\mathrm{B}_{\mathrm{n}-1}\right)} for all \mathrm{\mathrm{n} \geq 1}, then \mathrm{\operatorname{det}\left(B_{4}\right)} is equal to:

Option: 1

3^{28}


Option: 2

3^{30}


Option: 3

3^{32}


Option: 4

3^{36}


Answers (1)

best_answer

\begin{aligned} \mathrm{A^{2}} &=\left[\begin{array}{lll} 0& 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right]\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right] \\ &=\left[\begin{array}{lll} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] \\ & \mathrm{ a \leftrightarrow R_{2}} \end{aligned}

\begin{aligned} &-\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]\\ & \mathrm{R_{2} \leftrightarrow R_{3}}\\ &\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I\\ \end{aligned}

\begin{aligned} & \mathrm{B_{0}=A^{49}+2 A^{98}}\\ & \mathrm{=A+2 I}\\ & \mathrm{B_{n}=\operatorname{Ad} J\left(B_{n-1}\right)}\\ & \mathrm{=\left|B_{0}\right|^{(n-1)^{4}}}\\ & \mathrm{=\left|B_{0}\right|^{\prime}}\\ \end{aligned}

\begin{aligned} &\mathrm{B_{0}}=\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right]+\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]\\ &=\left[\begin{array}{lll} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{array}\right]\\ & \mathrm{=2(4-0)-1(0-1)=9}\\ & \mathrm{B_{4} = (9)^{16}=3^{3^{2}}} \end{aligned}

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HARSH KANKARIA

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