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Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 1 / 1F / Solution

The Leibniz definition is

det A = \sum_{π \in S_{n}} sgn (π) \prod_{i = 1}^{n} A_{i, π (i)} .

The adjugate is the transpose of the cofactor matrix. Cofactor expansion gives

A adj (A) = adj (A) A = (det A) I .

For the displayed tridiagonal matrix, expansion along the last row gives

D_{n} = 2 D_{n - 1} - D_{n - 2}

, with

D_{1} = 2, D_{2} = 3

. Induction yields

det A_{n} = D_{n} = n + 1.

Solved by gpt-5.6-sol high.

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