 Solution

Let $\Delta =v_1\cdot (v_2\times v_3)\ne 0$, since the original vectors form a basis. The reciprocal basis isTaking 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 $U^{T}V=I$, 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.