= Solution
Let $\Delta=v_1\cdot(v_2\times v_3)\ne0$, since the original vectors form a <basis>. The <reciprocal basis> is
$$
\boxed{u^1=\frac{v_2\times v_3}{\Delta},\qquad
u^2=\frac{v_3\times v_1}{\Delta},\qquad
u^3=\frac{v_1\times v_2}{\Delta}.}
$$
Taking the <dot product> with each $v_j$ gives $u^i\cdot v_j=\delta_{ij}$, using orthogonality of a <cross product> and cyclic symmetry of the <scalar triple product>. If $\sum_i a_i u^i=0$, taking its <dot product> with $v_j$ yields $a_j=0$. The three reciprocal vectors are therefore <linearly independent> and form a <basis> of $\mathbb R^3$.
Equivalently, if $V$ has columns $v_j$ and $U$ columns $u^i$, then $U^TV=I$, so $U=V^{-T}$. In particular any vector has coordinates $r=\sum_i(r\cdot u^i)v_i$, even when the original basis is not orthonormal.
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