Let the matrix A = [ cases 0 & 1 & 0 1 & 0 & 0 0 & 0 & 1 cases ] and the matrix B _ 0 = A ^ 49 + 2 A ^ 98 . If B _ n = Adj( B _ n - 1 ) for all n ≥ 1 , then det ( B _ 4 ) is equal to

Correct answer: C. 3 ^ 32 Given A = [ cases 0 & 1 & 0 1 & 0 & 0 0 & 0 & 1 cases ] Now A ^ 2 = [ cases 0 & 1 & 0 1 & 0 & 0 0 & 0 & 1 cases ][ cases 0 & 1 & 0 1 & 0 & 0 0 & 0 & 1 cases ] = [ cases 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 cases ] = I i.e. A ^ 98 = I; A ^ 49 = A ∴ B _ 0 = A ^ 98 + 2 A ^ 49 = I + 2A = [ cases 1 & 2 & 0 2 & 1 & 0 0 & 0 & 3 cases ] Given B _ n = adj ( B _ n - 1 ) ∴ B _ 4 = adj ( B _ 3 ) = adj (adj B _ 2 ) = adj (adj (adj B _ 1 )) = adj (adj (adj (adj B _ 0 ))) ∴ | B _ 4 | = | B _ 0 | ^ 2 ^ 4 = | B _ 0 | ^ 16 Now | B _ 0 | = [ cases 1 & 2 & 0 2 & 1 & 0 0 & 0 & 3 cases ] = - 9 Hence, | B _ 4 | = | B _ 0 | ^ 16 = 9 ^ 16 = 3 ^ 32

JEE Main 202228 JulyMorning Shift

Exam: JEE Main
Year: 2022
Date: 28 July
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2022 · 28 July · Morning Shift · Mathematics · Matrices

Let the matrix A = [ cases 0 & 1 & 0 1 & 0 & 0 0 & 0 & 1 cases ]...

Let the matrix

A = \left[

\begin{cases}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{cases}

\right] and the matrix

{B}_{0} = {A}^{49} + 2{A}^{98}

. If

{B}_{n} = Adj\left({B}_{n - 1}\right)

for all

n \ge 1

, then

det

\left({B}_{4}\right)

is equal to

Options

A ${3}^{28}$
B ${3}^{30}$
C ${3}^{32}$
D ${3}^{36}$

Correct answer

C. ${3}^{32}$

Step-by-step solution

Given $A =$ \left[ $\begin{cases}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{cases}$ \right] $Now$ {A}^{2} = \left[ $\begin{cases}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{cases}$ \right]\left[ $\begin{cases}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{cases}$ \right] = \left[ $\begin{cases}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}$ \right] = I $i.e.$ {A}^{98} = I; {A}^{49} = A $$ ∴ {B}_{0} = {A}^{98} + 2{A}^{49} = I + 2A = \left[ $\begin{cases}1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 3\end{cases}$ \right] $Given$ {B}_{n} = adj \left({B}_{n - 1}\right) $$ ∴ {B}_{4} = adj \left({B}_{3}\right) = adj \left(adj {B}_{2}\right) = adj \left(adj \left(adj {B}_{1}\right)\right) $$ = adj \left(adj \left(adj \left(adj {B}_{0}\right)\right)\right) $$ ∴ \left|{B}_{4}\right| = {\left|{B}_{0}\right|}^{{2}^{4}} = {\left|{B}_{0}\right|}^{16} $Now \left|{B}_{0}\right| = \left[$ \begin{cases}1 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 3\end{cases} $\right] = - 9$
Hence, \left|{B}_{4}\right| = {\left|{B}_{0}\right|}^{16} = {9}^{16} = {3}^{32}

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