Past exam of the mathematical methods I course of the University of Oxford / 2013 resit / 9 / b

The denominator for taking reciprocals a second time is $D^{\prime }=1/D$. The preceding cross-product result therefore givesCyclic permutation gives $\mathbf{b}$ and $\mathbf{c}$. Thus taking the reciprocal basis twice returns the original basis. The matrix statement is the same reciprocal basis involution: $(B^{-T}{)}^{-T}=B$. Noncoplanarity is essential, since otherwise the defining denominator and the matrix inverse do not exist.