 Solution

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

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

Let

M (t_{1}, \dots, t_{n}) = diag (t_{1}, \dots, t_{n}) - A

. Expanding the determinant along row

i

shows that

\frac{\partial p}{\partial t _{i}} = C_{ii} (M),

where

C_{ii}

is the corresponding cofactor. Since a diagonal entry of the adjugate matrix is that same cofactor, the multivariable chain rule along the diagonal

t_{1} = \dots = t_{n} = t

yields

\frac{d}{d t} det (t I - A) = \sum_{i = 1}^{n} \frac{\partial p}{\partial t _{i}}_{(t, \dots, t)} = \sum_{i = 1}^{n} [adj (t I - A)]_{ii} .

Therefore

\frac{d}{d t} det (t I - A) = Tr (adj (t I - A)) .

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