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>.

Solution

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

 Solution (source code)