Why are algebra representations given with matrices? So I first saw representations with group theory.  However, I have now seen other algebras (lie, clifford, supersymmetric, quiver) also represented as matrices.  
Is there a fundamental reason for this or does it just so happen that matrices are useful?
 A: The Short answer is “Because we have worked with matrices for a very long time and know a lot about them”.
Instead of developing a totally new theory each time we have found a new interesting algebra, we can work our way to find a representation in matrices and then use our enormous baggage of knowledge about them to study this algebra. We can also imagine the elements in our brain, we can write them down on paper, we can use computer systems that can deal with matrices.
Finally, matrices are relatively easy math, so people get used to them a lot, whereas with some other representation you need to spend more time to teach people about your algebra.
A: Representation theory (of groups, algebras, lie algebras, quivers, etc) is concerned with mapping the object of interest to linear operators on some vector space over some field. In many cases the finite-dimensional vector spaces are much easier to deal with; a choice of a basis provides a correspondence between operators on such spaces and matrices.
So, in some sense, representation theory actually is about matrices satisfying some relations: compositional relations like $\pi(g_1)\pi(g_2)=\pi(g_1g_2)$ for groups, commutation relations $[\pi(g_1),\pi(g_2)]=\pi([g_1, g_2])$ for lie algebras, etc.
A: When an algebra $A$ has a "base field," $K,$ like $\mathbb Q,\mathbb R,\mathbb C,$ then $A$ forms a vector space over that field. 
In this case, for any $a\in A$ the map $\pi_a:A\to A$ sending $x\mapsto ax$ is a linear map on this vector space, since $$a(\lambda_1 x_1+\lambda_2 x_2)=\lambda_1(ax_1)+\lambda_2(ax_2)$$ for $\lambda_i\in K,x_i\in A.$ 
Automatically, you get $\pi_{a+b}=\pi_{a}+\pi_{b}$ by the distributive law. 
When multiplication is associative, you also get $\pi_{ab}=\pi_{a}\circ \pi_{b}.$
This means that $a\mapsto \pi_a$ gives an algebra homomorphism $A\to \operatorname{End}_K(A),$ the ring of endomorphism of $A$ as a vector space over $K.$
This means that you have a natural representation of $A$ as linear transformations. It is not necessarily "faithful." If $ab=0$ for all $a,b\in A$ then $\pi_a=0$ so the map $A\to \operatorname{End}_K(A)$ is just the zero transformation.
Matrices are just linear transformations - usually over finite-dimensional vector spaces.

Even if $A$ doesn't have a base field, like the group ring $\mathbb Z[G],$ this can be seen as a sub-ring of $\mathbb Q[G],$ which does have a base field. 
This is a special case. More generally, if $A$ is any algebra over a commutative ring $R$ and there is some homomorphism $f:R\to K,$ $K$ a field, we get an algebra $A\otimes_f K$ which has $K$ as a base field. 
When $A$ is nice, we might only need to study one such field, but in general, every algebra is an algebra over $\mathbb Z,$ and for any field $K$ there is a unique homomorphism $Z\to K.$
