Question

If $F(\alpha)=\begin{bmatrix}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix}$, where $\alpha \in R$, then $[F(\alpha)]^{-1} =$

• A. $F(-\alpha)$
• B. $F(2\alpha)$
• C. $F(\alpha)$
• D. $F(3\alpha)$

Hint:

Conceptually, if $F(\alpha)$ rotates an object by an angle $\alpha$, the inverse operation must simply rotate it back by the same amount in the opposite direction, which is $-\alpha$. Hence, the inverse is intuitively $F(-\alpha)$!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a $3 \times 3$ transformation matrix $F(\alpha)$ and need to find its inverse, $[F(\alpha)]^{-1}$, in terms of the original function $F$.

Step 2: Key Formula or Approach:
Notice that $F(\alpha)$ represents a standard 2D rotation matrix embedded in 3D space (specifically, a rotation about the Z-axis by angle $\alpha$).
Rotation matrices are orthogonal matrices. A key property of any orthogonal matrix $A$ is that its inverse is simply its transpose: $A^{-1} = A^T$.
Alternatively, we can compute the adjoint and determinant, but the transpose method is much faster.

Step 3: Detailed Explanation:
Let's find the transpose of $F(\alpha)$ by swapping its rows and columns:
$$[F(\alpha)]^T = \begin{bmatrix}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix}$$ Because $F(\alpha)$ is orthogonal, $[F(\alpha)]^{-1} = [F(\alpha)]^T$.
Now, let's evaluate the matrix $F(-\alpha)$ by substituting $-\alpha$ into the original function:
$$F(-\alpha) = \begin{bmatrix}\cos(-\alpha) & -\sin(-\alpha) & 0 \\ \sin(-\alpha) & \cos(-\alpha) & 0 \\ 0 & 0 & 1\end{bmatrix}$$ Recall the even/odd trigonometric identities: $\cos(-\alpha) = \cos \alpha$ and $\sin(-\alpha) = -\sin \alpha$.
Substitute these identities into the matrix:
$$F(-\alpha) = \begin{bmatrix}\cos \alpha & -(-\sin \alpha) & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix}$$ $$F(-\alpha) = \begin{bmatrix}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix}$$ Notice that this resulting matrix is exactly identical to $[F(\alpha)]^T$.
Therefore, $[F(\alpha)]^{-1} = F(-\alpha)$.

Step 4: Final Answer:
The inverse matrix is $F(-\alpha)$, corresponding to option (A).