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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Matrices

JEE Main 2025 · 2 April · Evening Shift · Mathematics · Matrices

Let A be a 3 × 3 real matrix such that A^2(A-2 I)- 4(~A- I )=O ,...

Let AA be a 3×33 \times 3 real matrix such that A2(A−2I)−A^2(A-2 I)- 4( A−4(\mathrm{~A}-\mathrm{I})=O)=\mathrm{O}, where I and O are the identity and null matrices, respectively. If A5=αA2+βA+γIA^5=\alpha A^2+\beta A+\gamma I, where α,β\alpha, \beta and γ\gamma are real constants, then α+β+γ\alpha+\beta+\gamma is equal to:

Options

  1. A1212
  2. B2020
  3. C7676
  4. D44

Correct answer

A. 1212

Step-by-step solution

A3−2A2−4A+4I=0A3=2A2+4A−4IA4=2A3+4A2−4A=2(2A2+4A−4I)+4A2−4AA4=8A2+4A−8IA5=8A3+4A2−8A=8(2A2+4A−4I)+4A2−8AA5=20A2+24A−32I∴α=20,β=24,γ=−32∴α+β+γ=12\begin{aligned} & A^3-2 A^2-4 A+4 I=0 \\ & A^3=2 A^2+4 A-4 I \\ & A^4=2 A^3+4 A^2-4 A \\ & =2\left(2 A^2+4 A-4 I\right)+4 A^2-4 A \\ & A^4=8 A^2+4 A-8 I \\ & A^5=8 A^3+4 A^2-8 A \\ & =8\left(2 A^2+4 A-4 I\right)+4 A^2-8 A \\ & A^5=20 A^2+24 A-32 I \\ & \therefore \alpha=20, \beta=24, \gamma=-32 \\ & \therefore \alpha+\beta+\gamma=12\end{aligned}
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