Mathematics · Matrices

JEE Main 2025 — 2 April, Evening Shift — Question 40

Let $A$ be a $3 \times 3$ real matrix such that $A^{2}(A-2 I)-$ $4(A-I)=0$ , where $I$ and $O$ are the identity and

null matrices, respectively. If $A^{5}=\alpha A^{2}+\beta A+\gamma$ l, where $\alpha, \beta$ and $\gamma$ are real constants, then $\alpha+\beta+\gamma$ is equal to:

Option A:
4
Option B:
20
Option C:
12
Correct
Option D:
76

Answer: C

Step-by-step solution

$A^{2}(A-2 I)-4(A-I)=0$

$A^{3}-2 A^{2}-4 A+4 I=0$

Multiply by $A$

$A^{4}=2 A^{3}+4 A^{2}-4 A$

$A^{4}=2\left(2 A^{2}+4 A-4 I\right)+4 A^{2}-4 A$

$A^{4}=8 A^{2}+4 A-8 I$

Multiply again by $A$

$\Rightarrow A^{5}=8 A^{3}+4 A^{2}-8 A$

$\Rightarrow A^{5}=8\left(2 A^{2}+4 A-4 I\right)+4 A^{2}-8 A$

$\Rightarrow A^{5}=20 A^{2}+24 A-321$

Comparing with $A^{5}=\alpha A^{2}+\beta A+\gamma l$

$\alpha=20, \beta=24, \gamma=-32$

$\therefore \alpha+\beta+\gamma=20+24-32$

$=44-32$

$=12$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Evening Shift
Subject: Mathematics
Chapter: Matrices
Topic: Characteristic Equation & roots,application of cayley - hamilton theorem