= Solution
The denominator for taking reciprocals a second time is $D'=1/D$. The preceding cross-product result therefore gives
$$
\boxed{\frac{\mathbf b'\times\mathbf c'}{D'}=\frac{\mathbf a/D}{1/D}=\mathbf a.}
$$
Cyclic 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.
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