If A = e t e - t c o s   t e - t sin ⁡   t e t - e - t cos ⁡ t - e - t sin ⁡ t -…

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:
  1. Invertible only if t=π
  2. Not invertible for any t∈R
  3. Invertible only if t=π2
  4. Invertible for all t∈R

Solution

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.

Asked in: JEE Main 2019 (09 Jan Shift 2)

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