<p>If \(\begin{bmatrix} \cos\dfrac{2\pi}{7} & -\sin\dfrac{2\pi}{7} \\ \sin\dfrac{2\pi}{7} & \cos\dfrac{2\pi}{7} \end{bmatrix}^k = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), then the least positive integral value of \(k\) is</p>
Step-by-Step Solution
Key Concept: This rotation matrix raised to power k must equal the identity matrix, which happens when the total rotation angle equals a multiple of 2π. Since each matrix represents a rotation by 2π/7, we need 7 rotations to complete one full cycle.
<p><strong>Step 1:</strong> Recognize that the given matrix is a 2D rotation matrix with rotation angle θ = 2π/7.</p><p><strong>Step 2:</strong> A rotation matrix R(θ) raised to power k gives R(kθ). The matrix R(kθ)ⁿ equals the identity matrix when kθ = 2πn for some positive integer n.</p><p><strong>Step 3:</strong> Set up the equation: k · (2π/7) = 2πn, which simplifies to k/7 = n.</p><p><strong>Step 4:</strong> For the least positive integral value of k, take n = 1, giving k = 7.</p><p><strong>Step 5:</strong> Verify: The matrix represents a rotation by 2π/7. Applying it 7 times rotates by 7 × (2π/7) = 2π, which returns to the identity position.</p><p>∴ Answer: C (k = 7)</p>
Correct Answer: C