 Solution

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

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

Let the columns be

C_{1}, \dots, C_{n}

, and replace

C_{k}

C_{k} + λ C_{l}

, where

k \neq = l

det (C_{1}, \dots, C_{k} + λ C_{l}, \dots, C_{n}) = det A + λ det (C_{1}, \dots, C_{l}, \dots, C_{l}, \dots, C_{n}) .

The second determinant vanishes because it has two equal columns. Hence

det D = det A .

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