Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 1 / 1 / b / Solution

The cross product $\mathbf{b}\times \mathbf{c}$ is orthogonal to both factors. Division by the nonzero scalar $D$ preserves that perpendicularity, giving $\boxed{\mathbf{A}\cdot \mathbf{b}=\mathbf{A}\cdot \mathbf{c}=0}$. Cyclic permutation gives the other off-diagonal pairings needed for the reciprocal basis.