 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-3/17b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Expansion of

det (x I - A_{n})

along the last row gives

D_{n} = (x - α_{n}) D_{n - 1} - β_{n} D_{n - 2},

with

D_{0} = 1

D_{1} = x - α_{1}

. This is exactly the recurrence defining

P_{n}

, so induction gives

P_{n} = det (x I - A_{n})

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