Let α be a root of the equation x^2 + x + 1 = 0 and the matrix A = 1/sqrt(3) * [[1, 1, 1], [1, α, α^2], [1, α^2, α^4]], then the matrix A^31 is equal to:
Step-by-Step Solution
Key Concept: The matrix A is a unitary matrix, meaning A*A^H = I. Given \alpha is a root of x^2+x+1=0, \alpha^3=1 and 1+\alpha+\alpha^2=0. Calculate A^2, A^3, and identify the cycle.
Given \alpha^2 + \alpha + 1 = 0, we have \alpha^3 = 1. The matrix A = 1/sqrt(3) * [[1, 1, 1], [1, \alpha, \alpha^2], [1, \alpha^2, \alpha^4]]. Since \alpha^4 = \alpha, A = 1/sqrt(3) * [[1, 1, 1], [1, \alpha, \alpha^2], [1, \alpha^2, \alpha]]. Calculating A^2 = A*A, it can be shown that A^3 = I. Therefore, A^31 = (A^3)^10 * A = I^10 * A = A.
Correct Answer: 1