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 of \( \frac{2\pi}{7} \). When we raise this matrix to the power of \( k \), it represents a rotation by an angle of \( k \cdot \frac{2\pi}{7} \).

### Step 1: Set up the equation for rotation
We want the resulting rotation to equal the identity matrix, which corresponds to a rotation of \( 0 \) radians (or \( 2\pi \) radians, or any multiple of \( 2\pi \)). Therefore, we need:

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

for some integer \( n \).

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

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

This implies:

\[ k = 7n \]

### Step 3: Find the least positive integral value of \( k \)
To find the least positive integral value of \( k \), we can set \( n = 1 \):

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

### Step 4: Verify the solution
Now we need to check if \( k = 7 \) satisfies the original equation:

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

This is indeed a multiple of \( 2\pi \), confirming that the rotation results in the identity matrix.

### Conclusion
Thus, the least positive integral value of \( k \) is:

\[ \boxed{7} \]