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NDA 1 2025 Maths, question 26

Topic: Matrices and determinants

Q26

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

  1. (a) [−100−1]\begin{bmatrix} - 1 & 0 \\ 0 & - 1 \end{bmatrix}
  2. (b) [1111]\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}
  3. (c) [−1001]\begin{bmatrix} - 1 & 0 \\ 0 & 1 \end{bmatrix}
  4. (d) [1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}Correct answer

Solution.

Answer: (d) [1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

  1. Since cos⁡π=−1\cos\pi = - 1 and sin⁡π=0\sin\pi = 0, f(π)=−I2f(\pi) = - I_{2}.
  2. Squaring gives {f(π)}2=(−I2)2=I2\{ f(\pi)\}^{2} = \left( - I_{2} \right)^{2} = I_{2}.