Let \(A = \begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}\), then \((A^{-1})^T\) equals
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}\)
The determinant is
\(|A| = \cos^2\theta + \sin^2\theta = 1\)
For a 2 × 2 matrix, the adjoint swaps the diagonal entries and changes the signs of the other two:
\(\displaystyle A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A = \begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\)
Taking the transpose:
\(\displaystyle (A^{-1})^T = \begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix} = A\)
Shortcut: \(AA^T = I\), so A is an orthogonal matrix: \(A^{-1} = A^T\), and hence \((A^{-1})^T = (A^T)^T = A\).
Why the other options are incorrect
Option (2) \(\begin{bmatrix}\cos\theta&-\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\) — Incorrect. Its (2, 1) entry is \(-\sin\theta\); in \((A^{-1})^T\) it is \(\sin\theta\).
Option (3) \(\begin{bmatrix}-\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{bmatrix}\) — Incorrect. Its diagonal entries are \(-\cos\theta\); in \((A^{-1})^T\) they are \(\cos\theta\).
Option (4) \(\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\) — Incorrect. This is \(A^{-1}\) itself; the transpose has not been taken.