Let \(A(\theta) = \begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\), where \(0 \lt \theta \lt \displaystyle\frac{\pi}{2}\). Consider the following statements:
(A) \((A(\theta))' = A(-\theta)\)
(B) \(A(\theta)\,A(-\theta) = I\)
(C) \(A(\theta)\) is a symmetric matrix.
(D) \((A(\theta))^2 = A(2\theta)\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
Since \(\cos(-\theta) = \cos\theta\) and \(\sin(-\theta) = -\sin\theta\),
\( \displaystyle A(-\theta) = \begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix} \)
Multiplying row by column and using the addition formulas,
\( \displaystyle A(\theta)A(\theta) = \begin{bmatrix}\cos^2\theta - \sin^2\theta&2\sin\theta\cos\theta\\-2\sin\theta\cos\theta&\cos^2\theta - \sin^2\theta\end{bmatrix} = \begin{bmatrix}\cos 2\theta&\sin 2\theta\\-\sin 2\theta&\cos 2\theta\end{bmatrix} \)
\( \displaystyle A(\theta)A(-\theta) = \begin{bmatrix}\cos^2\theta + \sin^2\theta&0\\0&\sin^2\theta + \cos^2\theta\end{bmatrix} = I \)
So \(A(-\theta)\) is the transpose of \(A(\theta)\), \(A(\theta)A(-\theta) = I\), \((A(\theta))^2 = A(2\theta)\), and \(a_{12} = \sin\theta\) while \(a_{21} = -\sin\theta\).
Checking the statements
(A) \((A(\theta))' = A(-\theta)\) — Correct. Interchanging rows and columns of \(A(\theta)\) gives exactly \(A(-\theta)\).
(B) \(A(\theta)\,A(-\theta) = I\) — Correct. Each diagonal element is \(\cos^2\theta + \sin^2\theta = 1\), and the others are 0.
(C) \(A(\theta)\) is a symmetric matrix. — Incorrect. \(a_{12} = \sin\theta\) and \(a_{21} = -\sin\theta\) differ, since \(\sin\theta \ne 0\) here.
(D) \((A(\theta))^2 = A(2\theta)\) — Correct. The square has elements \(\cos 2\theta\) and \(\pm\sin 2\theta\) in the same places.