The least positive integer n such that `[{:(cos""(pi)/(4), sin""(pi)/(4)) ,(- sin ""(pi)/(4) , cos ""(pi)/(4)):}]^(n)`
is an identity matrix of order 2 is

Options

A. 4
B. 8
C. 12
D. 16

Answer

To find the least positive integer \( n \) such that the matrix

\[ A = \begin{pmatrix} \cos\left(\frac{\pi}{4}\right) & \sin\left(\frac{\pi}{4}\right) \\ -\sin\left(\frac{\pi}{4}\right) & \cos\left(\frac{\pi}{4}\right) \end{pmatrix}^n \]

is an identity matrix of order 2, we will follow these steps:

### Step 1: Identify the Matrix Elements
First, we calculate the values of \( \cos\left(\frac{\pi}{4}\right) \) and \( \sin\left(\frac{\pi}{4}\right) \):

\[ \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \quad \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \]

Thus, the matrix \( A \) becomes:

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

### Step 2: Matrix Power Calculation
We know that the \( n \)-th power of a rotation matrix can be expressed as:

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

where \( \theta = \frac{\pi}{4} \). Therefore,

\[ A^n = \begin{pmatrix} \cos\left(n \cdot \frac{\pi}{4}\right) & \sin\left(n \cdot \frac{\pi}{4}\right) \\ -\sin\left(n \cdot \frac{\pi}{4}\right) & \cos\left(n \cdot \frac{\pi}{4}\right) \end{pmatrix} \]

### Step 3: Set Equal to Identity Matrix
For \( A^n \) to be the identity matrix \( I_2 \), we need:

\[ A^n = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

This gives us the following equations:

1. \( \cos\left(n \cdot \frac{\pi}{4}\right) = 1 \)
2. \( \sin\left(n \cdot \frac{\pi}{4}\right) = 0 \)

### Step 4: Solve the Equations
From the first equation, \( \cos\left(n \cdot \frac{\pi}{4}\right) = 1 \) occurs when:

\[ n \cdot \frac{\pi}{4} = 2k\pi \quad \text{for } k \in \mathbb{Z} \]

This simplifies to:

\[ n = 8k \]

From the second equation, \( \sin\left(n \cdot \frac{\pi}{4}\right) = 0 \) occurs when:

\[ n \cdot \frac{\pi}{4} = m\pi \quad \text{for } m \in \mathbb{Z} \]

This simplifies to:

\[ n = 4m \]

### Step 5: Find the Least Positive Integer \( n \)
To satisfy both conditions, we need \( n \) to be a common multiple of 8 and 4. The least positive integer satisfying both equations is:

\[ n = 8 \]

### Conclusion
Thus, the least positive integer \( n \) such that \( A^n \) is the identity matrix of order 2 is:

\[ \boxed{8} \]