Let `A=({:(cosalpha,-,sinalpha),(sinalpha,,cosalpha):}),(alphaepsilonR)` such that
`A^(32)=({:(0,-1),(1,0):})`. Then a value of `alpha` is:

Options

A. 0
B. `pi/16`
C. `pi/32`
D. `pi/64`

Explanation

To solve the problem step by step, we need to find the value of \( \alpha \) such that the matrix \( A^{32} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \).

### Step 1: Define the matrix \( A \)
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \]

### Step 2: Calculate \( A^2 \)
To find \( A^{32} \), we first need to calculate \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \cdot \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \]
Using matrix multiplication:
- The (1,1) entry: \( \cos^2 \alpha + \sin^2 \alpha = 1 \)
- The (1,2) entry: \( -\cos \alpha \sin \alpha + \sin \alpha \cos \alpha = 0 \)
- The (2,1) entry: \( \sin \alpha \cos \alpha + \cos \alpha \sin \alpha = 0 \)
- The (2,2) entry: \( \sin^2 \alpha + \cos^2 \alpha = 1 \)

Thus,
\[ A^2 = \begin{pmatrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{pmatrix} \]

### Step 3: Generalize \( A^n \)
From the pattern observed, we can generalize:
\[ A^n = \begin{pmatrix} \cos n\alpha & -\sin n\alpha \\ \sin n\alpha & \cos n\alpha \end{pmatrix} \]

### Step 4: Calculate \( A^{32} \)
Using the general formula:
\[ A^{32} = \begin{pmatrix} \cos 32\alpha & -\sin 32\alpha \\ \sin 32\alpha & \cos 32\alpha \end{pmatrix} \]
We know that:
\[ A^{32} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \]

### Step 5: Set up equations
From the equality of matrices, we have:
1. \( \cos 32\alpha = 0 \)
2. \( -\sin 32\alpha = -1 \) or \( \sin 32\alpha = 1 \)

### Step 6: Solve \( \cos 32\alpha = 0 \)
The cosine function is zero at:
\[ 32\alpha = \frac{\pi}{2} + k\pi \quad (k \in \mathbb{Z}) \]
This gives:
\[ \alpha = \frac{\pi}{64} + \frac{k\pi}{32} \]

### Step 7: Solve \( \sin 32\alpha = 1 \)
The sine function equals 1 at:
\[ 32\alpha = \frac{\pi}{2} + 2m\pi \quad (m \in \mathbb{Z}) \]
This also gives:
\[ \alpha = \frac{\pi}{64} + \frac{m\pi}{16} \]

### Step 8: Find a specific solution
The simplest solution occurs when \( k = 0 \) and \( m = 0 \):
\[ \alpha = \frac{\pi}{64} \]

Thus, the value of \( \alpha \) is:
\[ \boxed{\frac{\pi}{64}} \]