 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-4/12b/a/solution

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

Because the matrix-valued Laplace transform is taken componentwise, for every

i, j

we have

(L {A B} (s))_{ij} = \int_{0}^{\infty} e^{- s t} k = 1 \sum n A_{ik} B_{kj} (t) d t = k = 1 \sum n A_{ik} \int_{0}^{\infty} e^{- s t} B_{kj} (t) d t = (A L {B} (s))_{ij} .

The sum is finite and

A

is constant, so it may be moved outside the integral. Therefore

L {A B} = A L {B} .

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