Question

If $A(\alpha)=\begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$, then $[A^2(\alpha)]^{-1} =$

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

Hint:

For any standard rotation matrix matrix $A(\alpha)$, powers and inverses track linearly with the angle multiplier: $A^n(\alpha) = A(n\alpha)$. Therefore, $[A^2(\alpha)]^{-1} = A(2\alpha)^{-1} = A(-2\alpha)$ directly!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a parametric $2 \times 2$ rotation matrix $A(\alpha)$. We need to calculate the inverse of the squared matrix, $[A^2(\alpha)]^{-1}$, and express the final matrix in terms of the function $A$.

Step 2: Key Formula or Approach:
By using the properties of rotation matrices or basic matrix algebra laws, we know that squaring the matrix corresponds to doubling the rotation angle: $$ A^2(\alpha) = A(2\alpha) $$ Furthermore, taking the inverse of a rotation matrix corresponds to rotating in the opposite direction, which is equivalent to changing the sign of the angle parameter: $$ [A(\theta)]^{-1} = A(-\theta) $$

Step 3: Detailed Explanation: Let's verify this step by step using explicit matrix multiplication to find $A^2(\alpha)$: $$ A^2(\alpha) = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} $$ Evaluating the row-by-column products: $$ = \begin{bmatrix} \cos^2\alpha - \sin^2\alpha & \cos\alpha\sin\alpha + \sin\alpha\cos\alpha \\ -\sin\alpha\cos\alpha - \cos\alpha\sin\alpha & -\sin^2\alpha + \cos^2\alpha \end{bmatrix} $$ Applying standard double-angle trigonometric identities ($\cos^2\alpha - \sin^2\alpha = \cos 2\alpha$ and $2\sin\alpha\cos\alpha = \sin 2\alpha$): $$ A^2(\alpha) = \begin{bmatrix} \cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \end{bmatrix} = A(2\alpha) $$ Now, we find the inverse of this matrix. The determinant of this matrix is: $$ |A(2\alpha)| = \cos^2 2\alpha - (-\sin^2 2\alpha) = \cos^2 2\alpha + \sin^2 2\alpha = 1 $$ Using the standard inverse shortcut formula for a $2 \times 2$ matrix (swap main diagonal entries and invert signs of off-diagonal entries): $$ [A^2(\alpha)]^{-1} = \frac{1}{1} \begin{bmatrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{bmatrix} $$ Using the properties of even and odd functions where $\cos(-2\alpha) = \cos 2\alpha$ and $\sin(-2\alpha) = -\sin 2\alpha$: $$ = \begin{bmatrix} \cos(-2\alpha) & \sin(-2\alpha) \\ -\sin(-2\alpha) & \cos(-2\alpha) \end{bmatrix} = A(-2\alpha) $$

Step 4: Final Answer: The resulting matrix is represented by $A(-2\alpha)$, which corresponds to option (C).