 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ia/paper-1/6b/a/ii/solution

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

The product identity always holds, since

\prod_{i} μ_{i} = det (A^{T} A) = det (A^{T}) det A = (det A)^{2} = (\prod_{i} λ_{i})^{2} .

Therefore

i = 1 \prod n μ_{i} = i = 1 \prod n λ_{i}^{2} .

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