Let \(A = \begin{bmatrix}\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{bmatrix}\), where \(\theta\) is a real number. Consider the following statements:
(A) A is a symmetric matrix.
(B) \(A^2 = I\)
(C) \(A^{-1} = A\)
(D) \(A + A' = O\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (C) only
Both non-diagonal elements of A are \(\sin\theta\), so \(A' = A\). Multiplying A by itself:
\( \displaystyle A^2 = \begin{bmatrix}\cos^2\theta + \sin^2\theta&\cos\theta\sin\theta - \sin\theta\cos\theta\\\sin\theta\cos\theta - \cos\theta\sin\theta&\sin^2\theta + \cos^2\theta\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix} \)
Since \(A \cdot A = I\), A is its own inverse. Also \(A + A' = 2A\), and A is never the zero matrix because \(\cos\theta\) and \(\sin\theta\) are never both 0.
So \(A' = A\), \(A^2 = I\), \(A^{-1} = A\) and \(A + A' = 2A \ne O\).
Checking the statements
(A) A is a symmetric matrix. — Correct. \(a_{12} = a_{21} = \sin\theta\).
(B) \(A^2 = I\) — Correct. Each diagonal element of \(A^2\) is \(\sin^2\theta + \cos^2\theta = 1\) and the others are 0.
(C) \(A^{-1} = A\) — Correct. \(A \cdot A = I\).
(D) \(A + A' = O\) — Incorrect. \(A + A' = 2A\); for \(\theta = 0\) it is \(\begin{bmatrix}2&0\\0&-2\end{bmatrix}\).