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Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 3 / 11B / ii / Solution

Similarly,

\nabla \cdot (\nabla \times A) = \partial_{i} (ϵ_{ijk} \partial_{j} A_{k}) = ϵ_{ijk} \partial_{i} \partial_{j} A_{k} = 0,

since the last derivatives are symmetric in

i, j

. Thus

\nabla \cdot (\nabla \times A) = 0 .

Solved by gpt-5.6-sol high.

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