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: Recognise the matrix.
$F(\alpha)$ is a rotation matrix about the Z-axis by angle $\alpha$. We need its inverse.
Step 2: Use a property of rotations.
Geometrically, undoing a rotation by $\alpha$ is just rotating by $-\alpha$. So we expect $[F(\alpha)]^{-1}=F(-\alpha)$. Let us confirm.
Step 3: Write $F(-\alpha)$.
Replace $\alpha$ by $-\alpha$ and use $\cos(-\alpha)=\cos\alpha$, $\sin(-\alpha)=-\sin\alpha$. The off-diagonal signs swap.
Step 4: Multiply $F(\alpha)F(-\alpha)$.
The top-left entry gives $\cos\alpha\cos\alpha+\sin\alpha\sin\alpha=\cos^2\alpha+\sin^2\alpha=1$, and similarly the cross terms cancel.
Step 5: See the identity appear.
The product comes out to the identity matrix $I$, confirming $F(\alpha)F(-\alpha)=I$.
Step 6: Conclude.
So $[F(\alpha)]^{-1}=F(-\alpha)$, which is option (1). \[ \boxed{[F(\alpha)]^{-1}=F(-\alpha)} \]