Question

If \(A=\frac{1}{2}\begin{bmatrix}1 & \sqrt{3} \\ -\sqrt{3} & 1\end{bmatrix}\), then :

• A. \(A ^{30}- A ^{25}=2 I\)
• B. \(A ^{30}= A ^{25}\)
• C. \(A ^{30}+ A ^{25}- A = I\)
• D. \(A ^{30}+ A ^{25}+ A = I\)

Hint:

For problems involving rotation matrices, utilize trigonometric identities and periodicity to simplify higher powers. Rotation matrices preserve their properties under exponentiation.

Answer 1

The Correct Option is C

Step 1: Express \( A \) in Trigonometric Form
We can rewrite \( A \) as:
\[A = \begin{bmatrix} \cos 60^\circ & \sin 60^\circ \\ -\sin 60^\circ & \cos 60^\circ \end{bmatrix}.\]
Here, \( \alpha = \frac{\pi}{3} \).
Step 2: Powers of \( A \)
For a rotation matrix \( A \), we know:
\[A^n = \begin{bmatrix} \cos n\alpha & \sin n\alpha \\ -\sin n\alpha & \cos n\alpha \end{bmatrix}.\]
Calculate \( A^{30} \) and \( A^{25} \):
\[A^{30} = \begin{bmatrix} \cos 30\alpha & \sin 30\alpha \\ -\sin 30\alpha & \cos 30\alpha \end{bmatrix}.\]
Since \( \cos 30\alpha = \cos 0 = 1 \) and \( \sin 30\alpha = \sin 0 = 0 \), we get:
\[A^{30} = \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} = I.\]
Similarly:
\[A^{25} = \begin{bmatrix}\cos 25\alpha & \sin 25\alpha \\-\sin 25\alpha & \cos 25\alpha \end{bmatrix}.\]
From periodicity (\( \cos(n\alpha) \) and \( \sin(n\alpha) \)), we find:
\[A^{25} = A.\]
Step 3: Verify the Relation
From the options:
\[A^{30} + A^{25} - A = I.\]
Substitute \( A^{30} = I \) and \( A^{25} = A \):
\[I + A - A = I.\]
This is true.
Conclusion:
\[A^{30} + A^{25} - A = I \quad.\]