Let the matrix A=010001100 and the matrix B0=A49+2A98. If Bn=AdjBn-1 for all n≥1,then detB4 is equal to :

A=010001100⇒A2=010001100010001100=001100010 ⇒A3=001100010010001100=100010001A3=I Given that B0=A49+2A98=A316A+2A332A2 ∴B0=A+2A2=010001100+2001100010=012201120 ⇒B0=9 Now Bn=AdjBn-1 ⇒B1=AdjB0,B2=AdjB1=AdjAdjB0 ⇒B4=AdjB3=AdjAdjB2=AdjAdjAdjAdjB0 ∴B4=B0n-14=924=916=332

Let the matrix A=010001100 and the matrix B0=A49+2A98. If Bn=AdjBn-1 for all n≥1,then detB4 is equal to :

A=010001100⇒A2=010001100010001100=001100010 ⇒A3=001100010010001100=100010001A3=I Given that B0=A49+2A98=A316A+2A332A2 ∴B0=A+2A2=010001100+2001100010=012201120 ⇒B0=9 Now Bn=AdjBn-1 ⇒B1=AdjB0,B2=AdjB1=AdjAdjB0 ⇒B4=AdjB3=AdjAdjB2=AdjAdjAdjAdjB0 ∴B4=B0n-14=924=916=332

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Let the matrix $A = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}]$ and the matrix $B_{0} = A^{49} + 2 A^{98} . I f B_{n} = A d j (B_{n - 1}) f o r a l l n \geq 1,$
then det $(B_{4})$ is equal to :

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3³⁶

$3^{28}$

$3^{30}$

$3^{32}$

answer is C.

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Detailed Solution

$A = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}] [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}] \Rightarrow A^{3} = [\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

$A^{3} = I G i v e n t h a t B_{0} = A^{49} + 2 A^{98} = {(A^{3})}^{16} A + 2 {(A^{3})}^{32} A^{2} ∴ B_{0} = A + 2 A^{2} = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}] + 2 [\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}] = [\begin{matrix} 0 & 1 & 2 \\ 2 & 0 & 1 \\ 1 & 2 & 0 \end{matrix}] \Rightarrow |B_{0}| = 9 N o w B_{n} = A d j (B_{n - 1}) \Rightarrow B_{1} = A d j (B_{0}), B_{2} = A d j (B_{1}) = A d j (A d j B_{0}) \Rightarrow B_{4} = A d j (B_{3}) = A d j (A d j B_{2}) = A d j (A d j (A d j (A d j B_{0}))) ∴ |B_{4}| = {|B_{0}|}^{{(n - 1)}^{4}} = 9^{2^{4}} = 9^{16} = 3^{32}$