 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1/18/b/i/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

det (A^{T}) = \sum_{σ \in S_{3}} sgn (σ) \prod_{i} a_{σ (i), i}

. Relabel the product by

j = σ (i)

and then the sum by

τ = σ^{- 1}

. The inverse permutation has the same sign, so

det (A^{T}) = \sum_{τ \in S_{3}} sgn (τ) \prod_{j} a_{j, τ (j)} = det A .

The proof applies to any finite square matrix.

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