 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-1/2a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

det A = σ \in S_{n} \sum sgn (σ) r = 1 \prod n A_{r, σ (r)} .

Equivalently, the determinant is the unique alternating function of the rows that is multilinear and takes value one on the identity matrix. Multiplying one row by

λ

therefore gives

det B = λ det A .

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