The least positive integral value of k for which `[(cos.(2pi)/(7),-sin.(2pi)/(7)),(sin.(2pi)/(7),cos.(2pi)/(7))]^(k)=[(1,0),(0,1)]` is

Options

A. 0
B. 3
C. 7
D. 14

Answer

To solve the problem, we need to find the least positive integral value of \( k \) such that:

\[ \begin{pmatrix} \cos\left(\frac{2\pi}{7}\right) & -\sin\left(\frac{2\pi}{7}\right) \\ \sin\left(\frac{2\pi}{7}\right) & \cos\left(\frac{2\pi}{7}\right) \end{pmatrix}^k = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

This matrix represents a rotation by an angle \( \theta = \frac{2\pi}{7} \). The matrix raised to the power \( k \) represents a rotation by \( k\theta \).

### Step 1: Set up the equation
We need to find \( k \) such that:

\[ k\theta = 2n\pi \]

for some integer \( n \). This means that the angle \( k\theta \) must be a multiple of \( 2\pi \).

### Step 2: Substitute \( \theta \)
Substituting \( \theta = \frac{2\pi}{7} \):

\[ k \cdot \frac{2\pi}{7} = 2n\pi \]

### Step 3: Simplify the equation
Dividing both sides by \( 2\pi \):

\[ \frac{k}{7} = n \]

### Step 4: Rearranging for \( k \)
From the equation above, we can express \( k \) in terms of \( n \):

\[ k = 7n \]

### Step 5: Find the least positive integral value of \( k \)
To find the least positive integral value of \( k \), we can set \( n = 1 \):

\[ k = 7 \cdot 1 = 7 \]

### Conclusion
Thus, the least positive integral value of \( k \) for which the original matrix raised to the power \( k \) equals the identity matrix is:

\[ \boxed{7} \]