If \(A = \begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix}\), then \(A^{-1}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}\)
The determinant of A:
\( |A| = \cos\theta \times \cos\theta - \sin\theta \times (-\sin\theta) \)
\( = \cos^2\theta + \sin^2\theta = 1 \)
Since \(|A| \ne 0\), the inverse exists. For a 2 × 2 matrix the adjoint swaps the diagonal entries and changes the signs of the other two:
\( \displaystyle \mathrm{adj}\,A = \begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix} \)
\( \displaystyle A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A = \begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix} \)
Check: the (1, 1) entry of \(AA^{-1}\) is \(\cos^2\theta + \sin^2\theta = 1\) and the (1, 2) entry is \(-\cos\theta\sin\theta + \sin\theta\cos\theta = 0\).