 Hadamard jump condition

Suppose an affine deformation on each side of a plane with unit normal $N$ is continuous across it. Tangential derivatives agree, so $(A-B)t=0$ for every $t\perp N$. Decomposition of an arbitrary vector into its tangent component and $(x\cdot N)N$ then gives $(A-B)x=((A-B)N)(N\cdot x)$, proving the rank-one form. Conversely that form makes the two affine maps agree on the plane when their constants agree there.

Orientation preservation on both sides also requires $1-N\cdot A^{-1}a>0$: the matrix determinant lemma gives $\det B=\det A(1-N\cdot A^{-1}a)$. This inequality is additional geometric information supplied by the determinant signs, not by continuity alone.