 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-1/8a/d/i/solution

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

e^{x A}

commutes with

C

and is orthogonal, so

e^{x A} C (e^{x A})^{T} = C .

Since

C^{n} = diag (1, (- 1)^{n}, (- 1)^{n})

n = 1 \sum N (e^{x A} C (e^{x A})^{T})^{n} = diag (N, \frac{( - 1 ) ^{N} - 1}{2}, \frac{( - 1 ) ^{N} - 1}{2}) .

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