Problem 2 - Olympiad

Since A + A^T = I and A is 2×2, trace(A) = 1/2 (because trace(A^T) = trace(A), so 2·trace(A) = 1). Given |A| = 1/2. The characteristic polynomial is |A - λI| = λ^2 - (trace A)·λ + |A| = λ^2 - (1/2)λ + 1/2. However, careful reading: the question asks for |A - λI| as a polynomial in λ. For a 2×2 matrix, |A - λI| = λ^2 - (tr A)λ + |A|. With tr A = 1/2 and |A| = 1/2, we get λ^2 - (1/2)λ + 1/2. But none of the options match this directly. Re-examining: if A + A^T = I, then taking determinants: |A + A^T| = |I| = 1. For 2×2, |A + A^T| = |A|(1 + tr(A^{-1}A^T)). This approach is messy. Let me reconsider: The problem likely expects recognizing that the characteristic polynomial is λ^2 - (tr A)λ + |A|. With tr A = 1/2, |A| = 1/2, the answer is λ^2 - (1/2)λ + 1/2, which corresponds to choice A if we multiply by 2: 2λ^2 - λ + 1. Hmm. Let me reinterpret: Actually, for a 2×2 matrix, |A - λI| = λ^2 - (tr A)λ + |A|. With tr A = 1/2 and |A| = 1/2, this equals λ^2 - 0.5λ + 0.5. Multiplying through by 2 gives 2λ^2 - λ + 1. None match exactly. Let me reconsider the trace: From A + A^T = I, taking trace: tr(A) + tr(A^T) = tr(I) => 2·tr(A) = 2 => tr(A) = 1. Yes! tr(I) for 2×2 is 2. So tr(A) = 1. Then |A - λI| = λ^2 - 1·λ + 1/2 = λ^2 - λ + 1/2. This matches choice A.