 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-1/2a/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

(n \cdot x) n_{i} = n_{i} n_{j} x_{j}

, so the matrix and, for

α \neq = 0

A_{ij} = δ_{ij} + (α - 1) n_{i} n_{j}, B_{ij} = δ_{ij} + (α^{- 1} - 1) n_{i} n_{j} .

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