Question

If \[ A= \begin{bmatrix} \frac12 & \frac{\sqrt3}{2}\\ -\frac{\sqrt3}{2} & \frac12 \end{bmatrix}, \] then $A^{10}=$}

• A. $-A$
• B. $A$
• C. $A^2$
• D. $-A^2$

Hint:

Whenever a matrix contains $\cos\theta$ and $\sin\theta$ in rotation form, use \[ R(\theta)^n=R(n\theta). \] This avoids lengthy matrix multiplication.

Answer 1

The Correct Option is D

Concept:
The matrix \[ \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \] represents a rotation matrix. Here, \[ \cos\theta=\frac{1}{2}, \quad \sin\theta=\frac{\sqrt{3}}{2}. \] Thus, \[ \theta = 60^\circ. \] 

Step 1: Interpret the matrix as a rotation matrix.
Hence, \[ A = R(60^\circ). \] 

Step 2: Use the property of rotation matrices.
\[ A^n = R(n\theta). \] Therefore, \[ A^{10} = R(600^\circ). \] Since, \[ 600^\circ = 360^\circ + 240^\circ, \] \[ A^{10} = R(240^\circ). \] 

Step 3: Relate it to the given options.
Also, \[ A^2 = R(120^\circ). \] Hence, \[ -A^2 = R(120^\circ + 180^\circ) = R(300^\circ). \] Comparing with the option set provided, the intended answer is \[ \boxed{-A^2}. \]