 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-1/8a/c/solution

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

Write

A = S diag (λ_{1}, \dots, λ_{n}) S^{- 1}

. The power series gives

e^{A} = S diag (e^{λ_{1}}, \dots, e^{λ_{n}}) S^{- 1}

. Therefore

det (e^{A}) = i \prod e^{λ_{i}} = e^{\sum_{i} λ_{i}} = e^{tr A} .

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