JEE Main 2025 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 2 April 2025, Shift 2, Maths Q3

Question 3 of 75 in this shift, Maths question 3 of 25, Section A.

Let

A

be a

3\times3

real matrix such that

A^2(A-2I) - 4(A-I) = O

, where

I

and

O

are the identity and null matrices, respectively. If

A^5 = \alpha A^2 + \beta A + \gamma I

, where

\alpha, \beta

, and

\gamma

are real constants, then

\alpha+\beta+\gamma

is equal to :

(1) $4$
(2) $12$ Official answer
(3) $20$
(4) $76$

Official answer

Option 2

NTA final key.

Chapter: Matrices
Topic: Matrix Multiplication
Idea tested: Cayley-Hamilton Theorem Application

Same idea in other shifts

Asked 4× in all

2 Apr 2026, Shift 1 · Q4Let $A=\begin{bmatrix}1&2\\1&\alpha\end{bmatrix}$ and $B=\begin{bmatrix}3&3\\\beta&2\end{bmatrix}$ . If $A^2-4A+I=O$ and $B^2-5B-6I=O$ ,…HardOther
8 Apr 2024, Shift 1 · Q4Let $A=\begin{bmatrix}2&a&0\\1&3&1\\0&5&b\end{bmatrix}$ . If $A^3=4A^2-A-21I$ , where $I$ is the identity matrix of order $3\times3$ , then…MediumSingle correct
9 Apr 2024, Shift 2 · Q4Let $B=\begin{bmatrix}1&3\\1&5\end{bmatrix}$ and $A$ be a $2\times2$ matrix such that $AB^{-1}=A^{-1}$ . If $BCB^{-1}=A$ and…HardSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.