 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-1/9f/c/solution

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

Taking the matrix trace of the adjugate expansion and using part (b) gives

\sum_{i = 0}^{n - 1} Tr (B_{i}) t^{n - 1 - i} = \sum_{i = 0}^{n - 1} (n - i) c_{i} t^{n - 1 - i} .

Hence

Tr (B_{i}) = (n - i) c_{i}

. Taking traces in

B_{i} = A B_{i - 1} + c_{i} I

gives, for

1 \leq i < n

c_{i} = - \frac{1}{i} Tr (A B_{i - 1}) .

For

i = n

, the same formula follows by taking the trace of

- A B_{n - 1} = c_{n} I

. Iterating the recursion from part (a) also gives

B_{i - 1} = A^{i - 1} + c_{1} A^{i - 2} + \dots + c_{i - 1} I .

c_{i} = - \frac{1}{i} [Tr (A^{i}) + c_{1} Tr (A^{i - 1}) + \dots + c_{i - 1} Tr (A)], 1 \leq i \leq n .

Starting with

c_{0} = 1

, this recursively expresses every

c_{i}

using only

Tr (A), \dots, Tr (A^{i})

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