JEE Main 2019MathematicsMatricesMediumMCQ

JEE Main 2019 — Matrices Question with Solution

JEE Main 2019 (09 Jan Shift 2)

Question

If A=ete-tcos te-tsin⁡ tet-e-tcos⁡t-e-tsin⁡t-e-tsin⁡t+e-tcos⁡tet2e-tsin⁡t-2e-tcos⁡t, then A is:

Choose an option

Show full solutionCorrect option: D
Correct answer
DInvertible for all t∈R

Step-by-step explanation

Since given matrix A is invertible ⇒A≠0
 

A=e-t1cos tsin t1-cos t-sin t-sin t+cos t12 sin t-2 cos t

R2→R2-R1

R3→R3-R1

=e-t1cos tsin t0-2 cos t-sin t-2 sin t+cos t02 sin t-cos t-2 cos t-sin t

=e-t2cost+sint2+2sint-cost2

=5e-t

A=5 e-t≠0 ∀ t∈R

Hence, given matrix is always invertible.

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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.