Let $\{\mathbf b_1,\mathbf b_2,\mathbf b_3\}$ be a basis for $\mathbb R^3$ and consider three arbitrary vectors $\mathbf w_i=\sum_j\omega_{ij}\mathbf b_j$.
We define the following two $3\times 9$ matrices \begin{equation} B_\Lambda:= \begin{pmatrix} \mathbf b_1&\mathbf 0&\mathbf 0\\ \mathbf 0&\mathbf b_2&\mathbf 0\\ \mathbf 0&\mathbf 0&\mathbf b_3 \end{pmatrix}^T,\quad B_X:= \begin{pmatrix} \mathbf 0&\mathbf b_3&\mathbf b_2\\ \mathbf b_3&\mathbf 0&\mathbf b_1\\ \mathbf b_2&\mathbf b_1&\mathbf 0 \end{pmatrix}^T \end{equation} and define $W_\Lambda$ and $W_X$ in the same way (replacing $\mathbf b_i$ with $\mathbf w_i$ in the above definition).
Consider the following system of matrix equations: \begin{equation} B_\Lambda M=O_{3\times3},\quad B_X M=O_{3\times3},\quad W_\Lambda M=I_{3\times3}. \end{equation} These three equations uniquely determine a $9\times3$ matrix $M$ iff \begin{equation} \det\widehat\Omega\neq0\quad \text{where}\quad\widehat\Omega:= {\small\begin{pmatrix} 0&-\omega_{13}&\omega_{12}\\ \omega_{23}&0&-\omega_{21}\\ -\omega_{32}&\omega_{31}&0 \end{pmatrix}}. \end{equation} I would like to know if there is an elegant way to compute $W_XM$ under the given conditions.
One way to approach the problem is to let $\{\mathbf a_1,\mathbf a_2,\mathbf a_3\}$ be the reciprocal basis to $\{\mathbf b_1,\mathbf b_2,\mathbf b_3\}$, i.e. $\mathbf a_i\cdot\mathbf b_j=\delta_{ij}$, and to express $M$ as \begin{equation} M=\begin{pmatrix}\mathbf v_{11}&\mathbf v_{12}&\mathbf v_{13}\\ \mathbf v_{21}&\mathbf v_{22}&\mathbf v_{23}\\ \mathbf v_{31}&\mathbf v_{32}&\mathbf v_{33} \end{pmatrix},\quad \text{where }\mathbf v_{ij}=\sum_k\xi^i_{jk}\mathbf a_k. \end{equation} The three matrix equations then descent to equations for the tensor $\xi^i_{jk}$: the first two state that $\xi^i_{jk}$ is antisymmetric in its lower indices, while the third shows that the non-trivial components are \begin{equation} \begin{pmatrix}\xi^i_{23}\\\xi^i_{31}\\\xi^i_{12}\end{pmatrix} =\widehat\Omega^{-1}\mathbf e_i, \end{equation} where $\{\mathbf e_1,\mathbf e_2,\mathbf e_3\}$ is the standard basis of $\mathbb R^3$.
Having expressed $M$ in terms of the reciprocal basis, we can express $W_X$ in terms of the original basis through $\mathbf w_i=\sum_j\omega_{ij}\mathbf b_j$, and through tedious matrix multiplication obtain \begin{equation} W_XM= \frac{1}{\det\widehat\Omega} \begin{pmatrix} \omega_{21}c_{13}-\omega_{31}c_{12}& \omega_{12}c_{13}-\omega_{31}c_{21}& \omega_{21}c_{31}-\omega_{13}c_{12}\\ \omega_{32}c_{12}-\omega_{21}c_{23}& \omega_{32}c_{21}-\omega_{12}c_{23}& \omega_{23}c_{21}-\omega_{12}c_{32}\\ \omega_{31}c_{32}-\omega_{23}c_{13}& \omega_{13}c_{23}-\omega_{32}c_{31}& \omega_{13}c_{32}-\omega_{23}c_{31} \end{pmatrix}, \end{equation} where $c_{ij}$ is the $(i,j)$-cofactor of the matrix $\Omega:=(\omega_{ij})_{1\leq i,j\leq3}$.
While the found expression is certainly quite messy, the amount of regularity suggests that we might be able to rewrite it in a more insightful form. At first the appearance of the cofactors encouraged me to think of something of the form \begin{equation} W_XM\overset{?}{=}\frac{\det\Omega}{\det\widehat\Omega}\,\Omega^{-1}\widehat\Omega, \end{equation} but this doesn't seem to pan out.