The set of all values of t ∈ ℝ , for which the matrix e t e - t sin t - 2 cos t e - t - 2 sin t…

The set of all values of t, for which the matrix ete-tsint-2coste-t-2sint-costete-t2sint+coste-tsint-2costete-tcoste-tsint is invertible, is

  1. 2k+1π2,k
  2. kπ+π4,k
  3. kπ,k

Solution

Let

A=ete-tsint-2coste-t-2sint-costete-t2sint+coste-tsint-2costete-tcoste-tsint

Given matrix A is invertible if

A0

ete-tsint-2coste-t-2sint-costete-t2sint+coste-tsint-2costete-tcoste-tsint0

et·e-t·e-t1sint-2cost-2sint-cost12sint+costsint-2cost1costsint0

Applying R1R1-R2 then R2R2-R3, ee get

e-t0-sint-3cost-3sint+cost02sint-2cost1costsint0

By expanding we have,

e-t×2sintcost+6cos2t+6sin2t-2sintcost0

e-t×60 for t

Asked in: JEE Main 2023 (29 Jan Shift 2)

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