Discussion
If
\[ A = \begin{pmatrix}\cos\theta&\sin\theta&0\\-\sin\theta&\cos\theta&0\\0&0&1\end{pmatrix} \]
is the matrix of orthogonal transformation of a rigid body with one fixed, then \(A^{-1}\) is given by :
Answer: A
The inverse is the transpose of A, option A.
A is orthogonal (its rows are orthonormal), so \(A^{-1} = A^{T}\). Transposing swaps the off-diagonal entries:
\[ A^{T} = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
As a check, \(AA^{T} = I\) because \(\cos^{2}\theta + \sin^{2}\theta = 1\). It is the rotation by \(-\theta\) about the z-axis, undoing the rotation by \(\theta\). Options C and D rotate about the x-axis instead, and option B permutes the rows.
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