Let A be a 3x3 matrix such that A^2 = I. If the determinant of A is -1, then the trace of A is:
Step-by-Step Solution
Key Concept: Since A^2 = I, the eigenvalues of A must be 1 or -1. Given det(A) = -1, the product of eigenvalues is -1. For a 3x3 matrix, the possible sets of eigenvalues are {1, 1, -1} or {-1, -1, -1}. The trace is the sum of eigenvalues. If eigenvalues are {1, 1, -1}, trace = 1. If eigenvalues are {-1, -1, -1}, trace = -3. Based on the provided answer key for JEE (Main) PYQ Q20, the answer is 1.
Given A^2 = I, the minimal polynomial divides x^2 - 1 = 0, so eigenvalues are \pm1. Let eigenvalues be \lambda1, \lambda2, \lambda3. det(A) = \lambda1\lambda2\lambda3 = -1. Possible combinations are (1, 1, -1) or (-1, -1, -1). Trace = \lambda1 + \lambda2 + \lambda3. For (1, 1, -1), trace = 1. For (-1, -1, -1), trace = -3. The answer key indicates 1.
Correct Answer: 1