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JEE Main20199 Jan 2019Evening ShiftMathematicsMatricesActual

If A = e t e - t c o s t e - t sin ⁡ t e t - e - t cos ⁡ t - e - t sin ⁡ t - e - t sin ⁡ t + e - t cos ⁡ t e t 2 e - t sin ⁡ t - 2 e - t cos ⁡ t , then A is:

Options

  1. AInvertible only if t = π
  2. BNot invertible for any t ∈ R
  3. CInvertible only if t = π 2
  4. DInvertible for all t ∈ R

Correct answer

D. Invertible for all t ∈ R

Step-by-step solution

Since given matrix A is invertible ⇒ A ≠ 0 A = e - t 1 cos   t sin   t 1 - cos   t - sin   t - sin   t + cos   t 1 2   sin   t - 2   cos   t R 2 → R 2 - R 1 R 3 → R 3 - R 1 = e - t 1 cos   t sin   t 0 - 2   cos   t - sin   t - 2   sin   t + cos   t 0 2   sin   t - cos   t - 2   cos   t - sin   t = e - t 2 cos t + sin t 2 + 2 sin t - cos t 2 = 5 e - t A = 5   e - t &#88

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