Solving a system of linear equations in Clifford Algebra I’m interested in solving the following system of equations in $\mathbb{Cl}(2,0)$:
$$\boldsymbol{Z}_{11}\boldsymbol{x}_1+\boldsymbol{Z}_{12}\boldsymbol{x}_2+\ldots+\boldsymbol{Z}_{1n}\boldsymbol{x}_n=\boldsymbol{a}_1$$
$$\boldsymbol{Z}_{21}\boldsymbol{x}_1+\boldsymbol{Z}_{22}\boldsymbol{x}_2+\ldots+\boldsymbol{Z}_{2n}\boldsymbol{x}_n=\boldsymbol{a}_2$$
$$\boldsymbol{Z}_{m1}\boldsymbol{x}_1+\boldsymbol{Z}_{m2}\boldsymbol{x}_2+\ldots+\boldsymbol{Z}_{mn}\boldsymbol{x}_n=\boldsymbol{a}_m$$
where $\boldsymbol{Z}_{ij}$ are bivectors and $\boldsymbol{a}_i$ are vectors. The unknowns are $\boldsymbol{x}_i$ which are also vectors.
M. E. Horn used exterior algebra to solve a similar system where all the coefficients are scalars here (page 12).
Edited: How to solve if $\boldsymbol{Z}_{ij}$ are spinors? Then, multiplying both sides by $\boldsymbol{e}_{12}$ seems to be not enough.
 A: Since this is Cl(2,0), every bivector is a scalar multiple of the unit pseudoscalar (i = e_1 e_2).  Multiply both sides by i, and you are left with a set of vector equations, which you can solve using any method you desire.
For the special case of spinor $ Z_{ij} \in \mathbb{Cl}(2,0) $ (assuming spinor here means multivectors with only even grades),
we can setup this system of equations in matrix form, provided we are careful with the ordering of products.  For example:
$$\begin{bmatrix}   Z_{11} & Z_{12} \\    Z_{21} & Z_{22} \\ \end{bmatrix}\begin{bmatrix}   x_1 \\    x_2\end{bmatrix}=\begin{bmatrix}   a_1 \\    a_2\end{bmatrix}.$$
Here we must define matrix multiplication so that any products are taken in the order of the matrices.  That is
$$\begin{bmatrix}   Z_{11} & Z_{12} \\    Z_{21} & Z_{22} \\ \end{bmatrix}\begin{bmatrix}   x_1 \\    x_2\end{bmatrix}=\begin{bmatrix}   Z_{11} x_1 + Z_{12} x_2  \\    Z_{21} x_1 + Z_{22} x_2\end{bmatrix},$$
and not
$$\begin{bmatrix}   Z_{11} & Z_{12} \\    Z_{21} & Z_{22} \\ \end{bmatrix}\begin{bmatrix}   x_1 \\    x_2\end{bmatrix}=\begin{bmatrix}   x_1 Z_{11} + Z_{12} x_2  \\    Z_{21} x_1 + x_2 Z_{22}\end{bmatrix},$$
or any other permutation.  Because the even grade elements of $ Z_{ij} \in \mathbb{Cl}(2,0) $ all commute (this space is isomorphic to complex numbers),
we can just invert this leading matrix.
$$\begin{bmatrix}   x_1 \\    x_2\end{bmatrix}=\frac{1}{{ Z_{11} Z_{22} - Z_{12} Z_{21} }}\begin{bmatrix}   Z_{22} & -Z_{12} \\   -Z_{21} & Z_{11} \\ \end{bmatrix}\begin{bmatrix}   a_1 \\    a_2\end{bmatrix}.$$
We still must be careful with ordering when the $ 2x2 $ matrix inverse is multiplied with the matrix of vectors, but multiplying that out should solve the system.
This approach should also be possible for similar systems in this space with more variables.
