JEE Main 2022 · MathsNumerical answerNumerical valueHardMulti-step

JEE Main 26 June 2022, Shift 2, Maths Q24

Question 24 of 90 in this shift, Maths question 24 of 30, Section B.

Let

X = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}

Y = \alpha I + \beta X + \gamma X^2

and

Z = \alpha^2 I - \alpha\beta X + (\beta^2 - \alpha\gamma)X^2

\alpha, \beta, \gamma \in \mathbb{R}

. If

Y^{-1} = \begin{bmatrix} \frac{1}{5} & -\frac{2}{5} & \frac{1}{5} \\ 0 & \frac{1}{5} & -\frac{2}{5} \\ 0 & 0 & \frac{1}{5} \end{bmatrix}

, then

(\alpha - \beta + \gamma)^2

is equal to ______.

Official answer

100

NTA final key.

Chapter: Matrices
Topic: Matrix Multiplication
Idea tested: Inverse of (I+A) for Nilpotent Matrices

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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 108 of the 174 JEE Main shifts held from 2017 to 2026.