Let A be a 3x3 matrix such that A^2 = I. If the determinant of A is -1, then find the value of det(A - I).
Step-by-Step Solution
Key Concept: Since A^2 = I, the eigenvalues of A are 1, 1, -1 (as det(A) = -1). The eigenvalues of (A - I) are (1-1), (1-1), (-1-1), which are 0, 0, -2. The determinant is the product of eigenvalues.
Given A^2 = I, the minimal polynomial divides x^2 - 1 = 0. The eigenvalues can only be 1 or -1. Since det(A) = -1, the product of the three eigenvalues is -1. The possible sets of eigenvalues are {1, 1, -1}. The eigenvalues of (A - I) are (\lambda - 1). Thus, the eigenvalues of (A - I) are (1-1), (1-1), (-1-1), which are 0, 0, -2. The determinant of (A - I) is the product of these eigenvalues: 0 * 0 * -2 = 0.
Correct Answer: 1