JEE Main 2019MathematicsMatricesMediumMCQ

JEE Main 2019 — Matrices Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

Let A=cosα-sinαsinαcosα, a∈R  such that A32=0-110. Then, a value of α is:

Choose an option

Show full solutionCorrect option: C
Correct answer
Cπ64

Step-by-step explanation

∵ A=cosα-sinαsinαcosα 

⇒A2=cos2α-sin2αsin2αcos2α

⇒A3 =cos3α-sin3αsin3αcos3α

Similarly, A32 =cos32α-sin32αsin32αcos32α
Given, 

A32 =0-110

⇒ cos32α-sin32αsin32αcos32α=0-110

On comparing cos32α=0 & sin32α=1, we get,
α=π64

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About this question

This is a previous-year question from JEE Main 2019, covering the Matrices chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.