What is Q equal to ?

Consider the following for the next three (03) items that follow : Let A=\begin{pmatrix}0&\sin^{2}\theta&\cos^{2}\theta\\ \cos^{2}\theta&0&\sin^{2}\theta\\ \sin^{2}\theta&\cos^{2}\theta&0\end{pmatrix} and A=P+Q where P is symmetric matrix and Q is skew-symmetric matrix.

  1. A. \begin{pmatrix}0&1/2&1/2\\ 1/2&0&1/2\\ 1/2&1/2&0\end{pmatrix}
  2. B. \begin{pmatrix}0&1&1\\ 1&0&1\\ 1&1&0\end{pmatrix}
  3. C. \cos~2\theta\begin{pmatrix}0&-1&1\\ 1&0&-1\\ -1&1&0\end{pmatrix}
  4. D. \cos~2\theta\begin{pmatrix}0&-1/2&1/2\\ 1/2&0&-1/2\\ -1/2&1/2&0\end{pmatrix} ✓

Correct Answer: D. \cos~2\theta\begin{pmatrix}0&-1/2&1/2\\ 1/2&0&-1/2\\ -1/2&1/2&0\end{pmatrix}

Explanation

The skew-symmetric part is Q = \frac{1}{2}(A - A^T). Calculating the (1,2) entry: A_{12} - A_{21} = \sin^2\theta - \cos^2\theta = -\cos 2\theta. This implies Q_{12} = -\frac{1}{2}\cos 2\theta. Repeating this for other elements and factoring out \cos 2\theta yields Q = \cos 2\theta \begin{pmatrix}0&-1/2&1/2\\ 1/2&0&-1/2\\ -1/2&1/2&0\end{pmatrix}.

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