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Write the row components as $E_{ai}=(e_a{)}_i$. Expanding the cross product in components,equals the alternating six-term expansion of the determinant of $E$. Therefore $(e_1\times e_2)\cdot e_3=\det E$.

Geometrically, $|e_1\times e_2|$ is the area of the parallelogram spanned by $e_1,e_2$, and its direction is perpendicular to that plane. A nonzero scalar triple product means both that this area is nonzero and that $e_3$ has a nonzero component normal to the plane. Thus the vectors are linearly independent and form a basis of ${\mathbb{R}}^3$, without needing to be orthogonal. Equivalently, their parallelepiped has nonzero volume.

The entries of $E{\hat{E}}^T$ are $e_a\cdot {\hat{e}}_b$, so the required duality is exactly $E{\hat{E}}^T=I$. Since $E$ is invertible, this has the unique solutionIts determinant is $1/\det E\ne 0$, so its rows are again a basis. Put $\Delta =(e_1\times e_2)\cdot e_3$. The cross product $e_2\times e_3$ is perpendicular to $e_2,e_3$, and its dot product with $e_1$ is $\Delta$. Consequently the reciprocal basis isThese vectors have precisely the required dot products, so uniqueness identifies them with the rows of $E^{-T}$.

For the final request, an orthonormal basis of rows gives $E{E}^T=I$. If instead the reciprocal rows are orthonormal, then $\hat{E}{\hat{E}}^T=I$ and substitution of $\hat{E}=E^{-T}$ gives $E^{-T}{E}^{-1}=I$, again implying $E{E}^T=I$. Thus in either case $E$ is an orthogonal matrix and $\hat{E}=E$.

The further printed claim that $E$ must be a rotation matrix is false without an orientation assumption. The matrix $E=\mathrm{diag}(1,1,-1)$ has orthonormal rows and is its own reciprocal basis matrix, but $\det E=-1$: it is a reflection. By the orientation of an orthonormal reciprocal basis, the corrected conclusion isA right-handed original ordered basis supplies exactly this missing condition.