Let \(A=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}\) and \(I=\begin{bmatrix}1&0\\0&1\end{bmatrix}\). If \(A^{T}+A=I\), then
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\theta=2n\pi+\displaystyle\frac{\pi}{3},\ n\in \mathrm{Z}\)
Taking the transpose (rows become columns):
\( \displaystyle A^{T} = \begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix} \)
\( \displaystyle A^{T} + A = \begin{bmatrix}2\cos\theta&0\\0&2\cos\theta\end{bmatrix} \)
Equating with I:
\(2\cos\theta = 1\)
\(\cos\theta = \frac{1}{2}\)
The general solution of \(\cos\theta = \cos\frac{\pi}{3}\) is:
\( \displaystyle \theta = 2n\pi \pm \frac{\pi}{3},\ n \in \mathrm{Z} \)
Every value \(2n\pi + \frac{\pi}{3}\) in Option (1) satisfies the condition, and it is the only option that does.
Why the other options are incorrect
Option (2) \(\theta=n\pi,\ n\in \mathrm{Z}\) — Incorrect. Then \(\cos\theta = \pm 1\), not \(\frac{1}{2}\).
Option (3) \(\theta=(2n+1)\displaystyle\frac{\pi}{2},\ n\in \mathrm{Z}\) — Incorrect. Then \(\cos\theta = 0\), not \(\frac{1}{2}\).
Option (4) \(\theta=2n\pi+\displaystyle\frac{\pi}{6},\ n\in \mathrm{Z}\) — Incorrect. Then \(\cos\theta = \frac{\sqrt{3}}{2}\), not \(\frac{1}{2}\).
Note: It seems that there is a mistake in the original NTA paper, which prints Option (1) as \(\theta = 2n\pi + \frac{\pi}{3},\ n \in \mathrm{Z}\) when the complete general solution is \(\theta = 2n\pi \pm \frac{\pi}{3},\ n \in \mathrm{Z}\); the values \(2n\pi - \frac{\pi}{3}\) also satisfy \(A^{T} + A = I\). The question can still be solved, because every value in Option (1) satisfies the condition and no value in the other options does.