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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Matrices

NDA 2018 · Mathematics · Matrices

What is the adjoint of the matrix (-θ)- (-θ) - (-θ) (-θ) ?

What is the adjoint of the matrix (cos⁡(−θ)−sin⁡(−θ)−sin⁡(−θ)cos⁡(−θ))?\binom{\cos (-\theta)-\sin (-\theta)}{-\sin (-\theta) \cos (-\theta)} ?

Options

  1. A(cos⁡θ−sin⁡θ−sin⁡θcos⁡θ)\binom{\cos \theta-\sin \theta}{-\sin \theta \cos \theta}
  2. B(cos⁡θsin⁡θsin⁡θcos⁡θ)\binom{\cos \theta \sin \theta}{\sin \theta \cos \theta}
  3. C(cos⁡θsin⁡θ−sin⁡θcos⁡θ)\binom{\cos \theta \sin \theta}{-\sin \theta \cos \theta}
  4. D(cos⁡θ−sin⁡θsin⁡θcos⁡θ)\binom{\cos \theta -\sin \theta}{\sin \theta \cos \theta}

Correct answer

A. (cos⁡θ−sin⁡θ−sin⁡θcos⁡θ)\binom{\cos \theta-\sin \theta}{-\sin \theta \cos \theta}

Step-by-step solution

A=[cos⁡θsin⁡θsin⁡θcos⁡θ](∵cos⁡(−θ)=cos⁡θ and sin⁡(−θ)=−sin⁡θ)adj⁡A=[cos⁡θ−sin⁡θ−sin⁡θcos⁡θ]\begin{aligned} & A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right] \\ & \quad(\because \cos (-\theta)=\cos \theta \text{ and } \sin (-\theta)=-\sin \theta) \\ & \operatorname{adj} A=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\end{aligned}
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