In Geometric Algebra, is there a geometric product between matrices? Thanks for your help in advance.
I literally just started to self-study about geometric algebra.
I have some coursework background in linear algebra and was trying to make an educational bridge between what I know and what I'm trying to learn.
My question:  Is there a geometric product for matrices in geometric algebra, like there is a geometric product for vectors?  If so, how would one compute the geometric product between matrices?
Thanks
 A: Let me address this more on the side of how linear algebra is presented in some GA material.
In traditional linear algebra, you use a lot of matrices and column/row vectors because this gives you an easy way to compute the action of a linear map or operator on a vector.  What I want to emphasize is that this is a representation.  It's a way of talking about linear maps, but it's not the only way.
In GA, there are reasons we don't often use matrices explicitly.  One reason is that we have a natural extension of a linear operator to all kinds of blades, not just vectors.  If you have a linear operator $\underline T$, and you want to compute its action on a bivector $a \wedge b$ with matrices, you would have to compute a totally different matrix from the one you would use just considering $\underline T$ acting on a vector (this matrix's components would describe its action on basis bivectors, not basis vectors).  This is one reason why matrices become rather useless.
Thus, since we tend to look at linear maps and operators merely as linear functions, we have to develop ways to talk about common linear algebra concepts without reference to matrices at all.  This is how we talk about a basis-independent of the determinant using the pseudoscalar $I$, saying $\underline T(I) = I \det \underline T$ for instance.  Texts on GA and GC also develop ways to talk about traces and other interesting linear algebra concepts without reference to matrices.
With all that in mind, since we don't talk about matrices when doing linear algebra in GA, we don't have to think about geometric products of matrices.  We just talk about compositions of maps (which would be represented through matrix multiplication) when applying several maps in succession.
A: I think you're giving undue distinction to matrices. 
Matrices are, after all, just fancily written vectors with $n^2$ entries. You can use the vector space $M_n(\Bbb R)$ and develop a geometric algebra containing it, but it would be the same as taking $\Bbb R^{n^2}$ with the same bilinear product and developing that geometric algebra.
The important thing about the geometric algebra is that you are taking the metric vector space $V$ that you're interested in and generating an algebra around it that has nice properties that we find useful. Nobody cares if the vectors are shaped like squares or hieroglyphs or ninja throwing stars, the only thing we care about is that it's a vector space with an inner product.

In case you are still looking for more materials on geometric algebra, you might find things with the Clifford-algebras tag useful, and solutions there, especially this one and also maybe this one. I found Alan Macdonald's online introductory stuff very helpful.
A: The only thing that is required to form matrices of multivectors is to take care to retain the ordering of any products, so if you have $ A = [a_{ij}] $ and $ B = [b_{ij}] $, where the matrix elements are multivector expressions, then your product is
$$A B = \begin{bmatrix}\sum_k a_{ik} b_{kj}\end{bmatrix},$$
and not 
$$A B = \begin{bmatrix}\sum_k b_{kj} a_{ik}\end{bmatrix}.$$
Such matrices can occur naturally when factoring certain multivector expressions.  See for example 
chapter: Spherical polar pendulum for one and multiple masses (Take II), where multivector matrix factors were used to express
the Lagrangian for a chain of N spherical-pendulums.
A: Linear algebra with its vectors and matrices is made entirely obsolete by Clifford algebra which provides a better way. Good riddance!
Gone is the awkward distinction between "row vectors" and "column vectors". In Clifford there is no distinction. And many weird abstract concepts become concrete spatial concepts that are easy to visualize. The determinant becomes a three-dimensional volume between three vectors, and the volume goes to zero as the vectors become parallel. 
In Clifford algebra a matrix is just an array of vectors that span a space. The geometric product has two effects: rotation and scaling. So a geometric product of a matrix will tend to rotate and scale the geometric shape.
I find that the most useful and interesting aspect of Clifford algebra is to try to picture all algebraic relationships as spatial structures, or operations by spatial structures on other structures.
