Let A be a 3x3 matrix such that A^2 = I. If A is not equal to I and A is not equal to -I, then which of the following is true?
Step-by-Step Solution
Key Concept: Since A^2 = I, det(A^2) = det(I) = 1, so (det(A))^2 = 1, implying det(A) = 1 or -1. For a 3x3 matrix, if A^2 = I and A is not I or -I, the eigenvalues must be 1, 1, -1 or 1, -1, -1. The product of eigenvalues is the determinant.
Given A^2 = I, det(A^2) = det(I) = 1, so (det(A))^2 = 1, which means det(A) = 1 or -1. The minimal polynomial of A divides x^2 - 1 = (x-1)(x+1). The eigenvalues of A can only be 1 or -1. Since A is 3x3, the possible sets of eigenvalues are {1, 1, 1}, {1, 1, -1}, {1, -1, -1}, {-1, -1, -1}. If eigenvalues are {1, 1, 1}, A = I. If eigenvalues are {-1, -1, -1}, A = -I. Since A is neither, the eigenvalues must be {1, 1, -1} or {1, -1, -1}. In both cases, the product of eigenvalues (the determinant) is -1.
Correct Answer: 3