Let A=[ array ccc θ & 0 & - θ 0 & 1 & 0 θ & 0 & θ array ] . If for some θ ∈(0, π) , A^2=A^T , then the sum of the diagonal elements of the matrix (A+ I )^3+(A- I )^3-6 ~A is equal to _____ .

Correct answer: 0 A is orthogonal matrix aligned & A^T=A^ -1 & A^2=A^ -1 ( A^2=A^ T ) & A^3=I & let B=(A+I)^3+(A-I)^3-6 A & =2(A^3+3 A)-6 A & =2 A^3 & B=2 I=[ array lll 2 & 0 & 0 0 & 2 & 0 0 & 0 & 2 array ] aligned Now sum of diagonal elements =2+2+2=6

JEE Main 20254 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2025 · 4 April · Morning Shift · Mathematics · Matrices

Let A=[ array ccc θ & 0 & - θ 0 & 1 & 0 θ & 0 & θ array ] . If for...

Let

A=\left[

\begin{array}{ccc}\cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta\end{array}

\right]

. If for some

\theta \in(0, \pi)

A^2=A^T

, then the sum of the diagonal elements of the matrix

(\mathrm{A}+

\mathrm{I}

)^3+(\mathrm{A}-

\mathrm{I}

)^3-6 \mathrm{~A}

is equal to _____ .

Correct answer

Step-by-step solution

\because \mathrm{A}

is orthogonal matrix<br/>

\begin{aligned}<br/>& \therefore A^T=A^{-1} \\ & \Rightarrow A^2=A^{-1} \left(\because \mathrm{A}^2=\mathrm{A}^{\mathrm{T}}\right) \\ & \Rightarrow A^3=I \\ & \text{ let } B=(A+I)^3+(A-I)^3-6 A \\ & =2\left(A^3+3 A\right)-6 A \\ & =2 A^3 \\ & B=2 I=\left[\begin{array}{lll}<br/>2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2<br/>\end{array}\right]<br/>\end{aligned}

<br/>Now sum of diagonal elements

=2+2+2=6

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