Let `A = [{:(cos alpha,- sin alpha ), ( sin alpha , cos alpha ):}], (a in R)` such that ` A ^(32) = [{:(0,-1), (1, 0):}]` is the value of `alpha ` is `(pi)/(2^(k)),` then the value of k is

Answer

To solve the problem, we need to find the value of \( k \) such that \( \alpha = \frac{\pi}{2^k} \) and \( A^{32} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \).

### Step-by-Step Solution:

1. **Understanding the Matrix A**:
The matrix \( A \) is defined as:
\[ A = \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \]
This is a rotation matrix that rotates vectors by an angle \( \alpha \).

2. **Finding \( A^2 \)**:
To find \( A^{32} \), we first compute \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \cdot \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \]
Using matrix multiplication:
\[ A^2 = \begin{pmatrix} \cos^2 \alpha - \sin^2 \alpha & -2 \sin \alpha \cos \alpha \\ 2 \sin \alpha \cos \alpha & \cos^2 \alpha - \sin^2 \alpha \end{pmatrix} \]
This simplifies to:
\[ A^2 = \begin{pmatrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \end{pmatrix} \]

3. **Using Induction to Generalize**:
We can use mathematical induction to show that:
\[ A^n = \begin{pmatrix} \cos(n\alpha) & -\sin(n\alpha) \\ \sin(n\alpha) & \cos(n\alpha) \end{pmatrix} \]
- **Base Case**: For \( n = 1 \), it holds true.
- **Inductive Step**: Assume true for \( n = k \), then for \( n = k + 1 \):
\[ A^{k+1} = A^k \cdot A = \begin{pmatrix} \cos(k\alpha) & -\sin(k\alpha) \\ \sin(k\alpha) & \cos(k\alpha) \end{pmatrix} \cdot \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} \]
This results in:
\[ A^{k+1} = \begin{pmatrix} \cos((k+1)\alpha) & -\sin((k+1)\alpha) \\ \sin((k+1)\alpha) & \cos((k+1)\alpha) \end{pmatrix} \]
Thus, by induction, it holds for all \( n \).

4. **Finding \( A^{32} \)**:
From the induction result:
\[ A^{32} = \begin{pmatrix} \cos(32\alpha) & -\sin(32\alpha) \\ \sin(32\alpha) & \cos(32\alpha) \end{pmatrix} \]
We know from the problem statement:
\[ A^{32} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \]
This implies:
\[ \cos(32\alpha) = 0 \quad \text{and} \quad \sin(32\alpha) = 1 \]

5. **Solving for \( \alpha \)**:
The equation \( \cos(32\alpha) = 0 \) occurs when:
\[ 32\alpha = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
The simplest case is when \( n = 0 \):
\[ 32\alpha = \frac{\pi}{2} \implies \alpha = \frac{\pi}{64} \]

6. **Finding \( k \)**:
We have \( \alpha = \frac{\pi}{2^k} \). Setting this equal to our result:
\[ \frac{\pi}{64} = \frac{\pi}{2^k} \]
This gives:
\[ 2^k = 64 \implies k = 6 \]

### Final Answer:
Thus, the value of \( k \) is \( \boxed{6} \).