Let α be a root of the equation x<sup>2</sup> + x + 1 = 0 and the matrix A = 1/√3 [[1, 1, 1], [1, α, α<sup>2</sup>], [1, α<sup>2</sup>, α<sup>4</sup>]], then the matrix A<sup>31</sup> is equal to:
Step-by-Step Solution
Key Concept: The matrix A is a unitary matrix, meaning A*A^T = I. Given the properties of the roots of unity, A^3 = I, so A^31 = (A^3)^10 * A = A.
Given x<sup>2</sup> + x + 1 = 0, the roots are ω and ω<sup>2</sup>. Let α = ω. Then A = 1/√3 [[1, 1, 1], [1, ω, ω<sup>2</sup>], [1, ω<sup>2</sup>, ω<sup>4</sup>]]. Since ω<sup>4</sup> = ω, A = 1/√3 [[1, 1, 1], [1, ω, ω<sup>2</sup>], [1, ω<sup>2</sup>, ω]]. It can be shown that A<sup>3</sup> = I. Thus A<sup>31</sup> = (A<sup>3</sup>)<sup>10</sup> * A = I * A = A.
Correct Answer: 4