Problem 6 - Olympiad

Given A^2 = 2A - I. Rearranging: A^2 - 2A + I = 0. This can be written as (A - I)^2 = 0. Taking determinant on both sides: |(A - I)^2| = |0| => |A - I|^2 = 0 => |A - I| = 0. But this doesn't directly give |A|. Alternatively, multiply both sides of A^2 = 2A - I by A^{-1}: A = 2I - A^{-1}. Rearranging: A + A^{-1} = 2I. Taking determinant: |A + A^{-1}| = |2I| = 2^3 = 8. But |A + A^{-1}| is not simply related. Better approach: From A^2 - 2A + I = 0, we have (A - I)^2 = 0. This means A - I is nilpotent. For a 3×3 nilpotent matrix, the determinant is 0, but that applies to A - I, not A. Taking determinant of both sides of A^2 = 2A - I: |A|^2 = |2A - I|. This is not straightforward. Let me use the characteristic polynomial approach. If A satisfies A^2 - 2A + I = 0, then the minimal polynomial divides λ^2 - 2λ + 1 = (λ-1)^2. So the only eigenvalue is 1. Therefore |A| = product of eigenvalues = 1·1·1 = 1. So |A| = 1.