Let ( S=\ m ∈ Z: A^ m^2 +A^ m =3 I -A^ -6 \ ), where ( A=[ array cc 2 & -1 1 & 0 array ]). Then ( n(S) ) is equal to ______.

Correct answer: 0 A= [ array cc 2 & -1 1 & 0 array ] Now finding characteristic equation aligned & | array cc 2-λ & -1 1 & -λ array |=0 & (2-λ)(-λ)-(-1)(1)=-2 λ+λ^2+1=0 & λ^2-2 λ+1=0 & (λ-1)^2=0 & λ=1 aligned Since A satisfies (A-I)^2=0 aligned & A=I+N where & N=A-I & N=[ array ll 1 & -1 1 & -1 array ] & N^2=0 & A^m=(I+N)^m=I+m N & A^m · A^m=(I+m N)(I+m N)=I+2 m N+m^2 N^2 aligned Since N^2=0 A^ m^2 =I+2 m N Now putting in given condition aligned & I+m^2 N+I+m N=3 I-A^ -6 & A^ -1 =[ array cc 0 & 1 -1 & 2 array ] & A^ -6 =(A^ -1 )^6=I+(-6) N aligned Putting in (i) aligned & (m^2+m) N=I⁻(I-6 N) & (m^2+m) N=6 N aligned Since N ≠ 0 aligned & m^2+m=6 & m^2+m-6=0 & (m-2)(m+3)=0 & m=2,-3 aligned Number of elements in S is 2

JEE Main 202529 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2025 · 29 January · Morning Shift · Mathematics · Matrices

Let ( S=\ m ∈ Z: A^ m^2 +A^ m =3 I -A^ -6 \ ), where ( A=[ array cc...

Let (

\mathrm{S}=\left\{\mathrm{m} \in \mathbf{Z}: \mathrm{A}^{\mathrm{m}^2}+\mathrm{A}^{\mathrm{m}}=3

\mathrm{I}

-\mathrm{A}^{-6}\right\}

), where (

\mathrm{A}=\left[

\begin{array}{cc}2 & -1 \\ 1 & 0\end{array}

\right]). Then (

\mathrm{n}(\mathrm{S})$) is equal to ______.

Correct answer

Step-by-step solution

\left[

\begin{array}{cc} 2 & -1 \\ 1 & 0 \end{array}

\right]

Now finding characteristic equation

\begin{aligned} & \left|\begin{array}{cc} 2-\lambda & -1 \\ 1 & -\lambda \end{array}\right|=0 \\ & \Rightarrow(2-\lambda)(-\lambda)-(-1)(1)=-2 \lambda+\lambda^2+1=0 \\ & \Rightarrow \lambda^2-2 \lambda+1=0 \\ & \Rightarrow(\lambda-1)^2=0 \\ & \Rightarrow \lambda=1 \end{aligned}

Since

A

satisfies

(A-I)^2=0

\begin{aligned} & \therefore \quad A=I+N \text{ where } \\ & N=A-I \\ & N=\left[\begin{array}{ll} 1 & -1 \\ 1 & -1 \end{array}\right] \\ & N^2=0 \\ & A^m=(I+N)^m=I+m N \\ & A^m \cdot A^m=(I+m N)(I+m N)=I+2 m N+m^2 N^2 \end{aligned}

\text{ Since } N^2=0

\Rightarrow \quad A^{m^2}=I+2 m N

Now putting in given condition

\begin{aligned} & I+m^2 N+I+m N=3 I-A^{-6} \\ & A^{-1}=\left[\begin{array}{cc} 0 & 1 \\ -1 & 2 \end{array}\right] \\ & A^{-6}=\left(A^{-1}\right)^6=I+(-6) N \end{aligned}

\therefore \quad

Putting in (i)

\begin{aligned} & \left(m^2+m\right) N=I⁻(I-6 N) \\ & \left(m^2+m\right) N=6 N \end{aligned}

Since

N \neq 0

\begin{aligned} & \Rightarrow m^2+m=6 \\ & \Rightarrow m^2+m-6=0 \\ & \Rightarrow(m-2)(m+3)=0 \\ & \Rightarrow m=2,-3 \end{aligned}

\therefore \quad

Number of elements in

S

is 2

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