Let A be a 3x3 matrix such that A^2 = I. If the determinant of A is -1, then the value of det(A - I) is:
Step-by-Step Solution
Key Concept: Since A^2 = I, the eigenvalues of A are 1 or -1. Given det(A) = -1, the product of eigenvalues is -1. Possible sets of eigenvalues are {1, 1, -1} or {-1, -1, -1}. If eigenvalues are {1, 1, -1}, then eigenvalues of (A-I) are {0, 0, -2}, so det(A-I) = 0. If eigenvalues are {-1, -1, -1}, then eigenvalues of (A-I) are {-2, -2, -2}, so det(A-I) = -8. However, for a 3x3 matrix, if A^2=I, then (A-I)(A+I)=0. If A is not I or -I, the minimal polynomial divides x^2-1. The characteristic polynomial is (x-1)^k(x+1)^(3-k). det(A) = (-1)^(3-k) = -1, so 3-k is odd, k is even. k=2 gives eigenvalues {1, 1, -1}, det(A-I)=0.
Given A^2 = I, the minimal polynomial of A divides x^2 - 1 = (x-1)(x+1). Thus, the eigenvalues of A can only be 1 or -1. Let the eigenvalues be \lambda1, \lambda2, \lambda3. Then det(A) = \lambda1\lambda2\lambda3 = -1. The possible sets of eigenvalues are {1, 1, -1} or {-1, -1, -1}. If the eigenvalues are {1, 1, -1}, the eigenvalues of (A-I) are (1-1), (1-1), (-1-1), which are 0, 0, -2. Thus, det(A-I) = 0 * 0 * -2 = 0. If the eigenvalues are {-1, -1, -1}, then A = -I, so det(A) = (-1)^3 = -1. Then A-I = -2I, and det(A-I) = det(-2I) = (-2)^3 det(I) = -8. Given the options, 0 is the correct answer.
Correct Answer: 1