If `A (1/2)[(1,sqrt3),(-sqrt3,1)] `, then

Options

A. ` A^(30) – A^(25) = 2I`
B. `A^(30) = A^(25)`
C. `A^(30) + A^(25) – A = I`
D. `A^(30) + A^(25) + A = I`

Answer

To solve the problem, we need to analyze the matrix \( A \) given by:

\[ A = \frac{1}{2} \begin{pmatrix} 1 & \sqrt{3} \\ -\sqrt{3} & 1 \end{pmatrix} \]

### Step 1: Identify the Matrix
First, we can rewrite the matrix \( A \):

\[ A = \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} \]

### Step 2: Recognize the Rotation Matrix
Notice that this matrix resembles a rotation matrix. The standard form of a 2D rotation matrix is:

\[ R(\theta) = \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix} \]

Comparing, we can identify that:

\[ \cos(\theta) = \frac{1}{2}, \quad \sin(\theta) = \frac{\sqrt{3}}{2} \]

This corresponds to \( \theta = \frac{\pi}{3} \) or \( 60^\circ \).

### Step 3: Calculate \( A^2 \)
To find \( A^2 \), we can use the property of rotation matrices:

\[ A^2 = R(2\theta) = R\left(2 \cdot \frac{\pi}{3}\right) = R\left(\frac{2\pi}{3}\right) \]

Calculating \( A^2 \):

\[ A^2 = \begin{pmatrix} \cos\left(\frac{2\pi}{3}\right) & -\sin\left(\frac{2\pi}{3}\right) \\ \sin\left(\frac{2\pi}{3}\right) & \cos\left(\frac{2\pi}{3}\right) \end{pmatrix} \]

Using the values:

\[ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}, \quad \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

Thus,

\[ A^2 = \begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix} \]

### Step 4: Generalize \( A^n \)
Using the rotation property, we can generalize:

\[ A^n = R(n \cdot \frac{\pi}{3}) = \begin{pmatrix} \cos\left(n \cdot \frac{\pi}{3}\right) & -\sin\left(n \cdot \frac{\pi}{3}\right) \\ \sin\left(n \cdot \frac{\pi}{3}\right) & \cos\left(n \cdot \frac{\pi}{3}\right) \end{pmatrix} \]

### Step 5: Find \( A^{30} \) and \( A^{25} \)
For \( A^{30} \):

\[ A^{30} = R\left(30 \cdot \frac{\pi}{3}\right) = R(10\pi) = I \quad (\text{Identity matrix}) \]

For \( A^{25} \):

\[ A^{25} = R\left(25 \cdot \frac{\pi}{3}\right) = R\left(\frac{25\pi}{3}\right) = R\left(8\pi + \frac{\pi}{3}\right) = R\left(\frac{\pi}{3}\right) \]

Thus,

\[ A^{25} = A \]

### Step 6: Solve the Equation
We need to solve \( A^{25} - A = 0 \):

\[ A - A = 0 \quad \Rightarrow \quad 0 = 0 \]

### Conclusion
Thus, the final result is:

\[ A^{30} + A^{25} - A = I \]