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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Matrices

NDA 2025 · Mathematics · Matrices

If f(θ) = pmatrix θ & θ - θ & θ pmatrix , then what is \ f(π)\ ^2...

If f(θ)=(cos⁡θsin⁡θ−sin⁡θcos⁡θ)f(\theta) = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}, then what is {f(π)}2\{f(\pi)\}^2 equal to?

Options

  1. A(−100−1)\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}
  2. B(1111)\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}
  3. C(−1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}
  4. D(1001)\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

Correct answer

D. (1001)\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

Step-by-step solution

Given f(θ)=(cos⁡θsin⁡θ−sin⁡θcos⁡θ)f(\theta) = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}

Substituting θ=π\theta = \pi:

f(π)=(cos⁡πsin⁡π−sin⁡πcos⁡π)=(−100−1)f(\pi) = \begin{pmatrix} \cos\pi & \sin\pi \\ -\sin\pi & \cos\pi \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}

Squaring the matrix:

{f(π)}2=(−100−1)(−100−1)=(1001)\{f(\pi)\}^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
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