Let S = \ ( cases - 1 & a 0 & b cases ); a, b ∈ \ 1, 2, 3, 100\ \ and let T _ n = \ A ∈ S: A ^ n(n + 1) = I\ . Then the number of elements in _ n = 1 ^ 100 T _ n is _____.

Correct answer: 0 Given, A = [ cases - 1 & a 0 & b cases ] A ^ 2 = [ cases - 1 & a 0 & b cases ][ cases - 1 & a 0 & b cases ] = [ cases 1 & - a + ab 0 & b ^ 2 cases ] ∴ T _ n = \ A ∈ S; A ^ n(a + 1) = I\ ∴ b must be equal to 1 ∴ In this case A ^ 2 will become identity matrix and a can take any value from 1 to 100 ∴ Total number of common element will be 100 .

JEE Main 202224 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 24 June
Shift: Evening Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2022 · 24 June · Evening Shift · Mathematics · Matrices

Let S = \ ( cases - 1 & a 0 & b cases ); a, b ∈ \ 1, 2, 3, 100\ \...

Let

S = \left\{\left(

\begin{cases}- 1 & a \\ 0 & b\end{cases}

\right); a, b \in \left\{1, 2, 3, \ldots 100\right\}\right\}

and let

{T}_{n} = \left\{A \in S: {A}^{n\left(n + 1\right)} = I\right\}

. Then the number of elements in

\mathop{\cap }_{n = 1}^{100}{T}_{n}

is _____.

Correct answer

Step-by-step solution

Given,

$A = \left[$ \begin{cases}- 1 & a \\ 0 & b\end{cases} $\right]$

${A}^{2} = \left[$ \begin{cases}- 1 & a \\ 0 & b\end{cases} $\right]\left[$ \begin{cases}- 1 & a \\ 0 & b\end{cases} $\right]$ $= \left[$ \begin{cases}1 & - a + ab \\ 0 & {b}^{2}\end{cases} $\right]$

$∴ {T}_{n} = \left\{A \in S; {A}^{n\left(a + 1\right)} = I\right\}$

$∴ b$ must be equal to $1$

$∴$ In this case ${A}^{2}$ will become identity matrix and $a$ can take any value from $1$ to $100$

$∴$ Total number of common element will be $100$ .

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