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JEE Main 20199 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
9 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Matrices

JEE Main 2019 · 9 January · Evening Shift · Mathematics · Matrices

If A = [ cases e ^ t & e ^ - t t & e ^ - t sin t e ^ t & - e ^ - t...

If A = \left[\begin{cases}{e}^{t} & {e}^{- t}\cos 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} & {2e}^{- t}sin t & - 2{e}^{- t}cos t\end{cases}\right], thenA$ is:

Options

  1. AInvertible only if t=πt = \pi
  2. BNot invertible for any t∈Rt \in R
  3. CInvertible only if t=π2t = \dfrac{\pi}{2}
  4. DInvertible for all t∈Rt \in R

Correct answer

D. Invertible for all t∈Rt \in R

Step-by-step solution

Since given matrix AA is invertible ⇒∣A∣≠0⇒ \left|A\right| \ne 0

\left|A\right| = {e}^{- t}\left|{1cos⁡tsin⁡t1−cos⁡t−sin⁡t−sin⁡t+cos⁡t12sin⁡t−2cos⁡t\begin{cases}1 & \cos t & \sin t \\ 1 & - \cos t - \sin t & - \sin t + \cos t \\ 1 & 2\sin t & - 2\cos t\end{cases}\right|</p><p></p><p>{R}_{2} \to {R}_{2} - {R}_{1}</p><p></p><p>{R}_{3} \to {R}_{3} - {R}_{1}</p><p></p><p>= {e}^{- t}\left|{1cos⁡tsin⁡t0−2cos⁡t−sin⁡t−2sin⁡t+cos⁡t02sin⁡t−cos⁡t−2cos⁡t−sin⁡t\begin{cases}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\end{cases}\right|</p><p></p><p>= {e}^{- t}\left\{{\left(2\cos t + \sin t\right)}^{2} + {\left(2\sin t - \cos t\right)}^{2}\right\}</p><p></p><p>= 5{e}^{- t}</p>∣A∣=5e−t≠0∀t∈R</p>\left|A\right| = 5{e}^{- t} \ne 0 \forall t \in R

Hence, given matrix is always invertible.

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