 Solution

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

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

The adjugate matrix is the transpose of the cofactor matrix, and it satisfies

M adj (M) = adj (M) M = (det M) I .

Put

M = t I - A

and substitute the two given expansions into this identity:

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

t

gives

B_{0} = c_{0} I = I, B_{i} = A B_{i - 1} + c_{i} I (1 \leq i \leq n - 1), - A B_{n - 1} = c_{n} I .

Here

c_{0} = 1

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