Showing surjection and morphism Lets say I have 2 matrices,
matrix A$$\left[\begin{array}{ccc} a & b & c \\                                                  d & e & f \\                                                      g & h & i \end{array}\right]$$
and matrix B $$ \left[\begin{array}{ccc} j & k & l\\ m & n & o\\ p & q & r \end{array}\right]$$
How do I show that the mapping $$\theta: A \to B $$ is a morphism and surjective?
To show that the mapping $\theta$ is a morphism, do I just need to show that $$\theta(A+B)=....=\theta(A)+\theta(B)$$ and 
$$\theta(AB)=....=\theta(A)\theta(B)$$, is that it for a morphism?
How would I show surjection?
If I was trying to show that a quotient ring $$A/B is isomorphic to C, where c is some other matrix ring, could I just show that B is the kernel of A and then just show morphism and surjection?
 A: To be pedantic, you should say "elements of ring $A$ have the form" and not matrix $A$. $A$ is a collection of matrices; not a matrix in its own right. Same for $B$. 
Surjection is fairly easy. Given an element of $B$, what element of $A$ maps to i?. You're 90% of the way there considering how you phrased the question. 
Yeah, you would need to need to show $\theta(a+b)=\theta(a)+\theta(b)$, and$\theta(ab)=\theta(b)\theta(b)$. This is just some computation. Tedious but not difficult. 
You can't say $A/B$. It doesn't make sense because $B$ is not a sub-ring of $A$. You can use the first(?) isomorphism theorem and say $A/\text{ker}(\theta)$ is isomorphic to the image of $A$ under $\theta$,usually denoted  $\text{im}(A)$ or  $\theta(A)$. Now what is $\text{im}(A)$? If you have proven surjectivity, then $\text{im}(A)=B$.
Thus $A/\text{ker}(A) \simeq B$.
EDIT: The OP  originally had very specifically structured matrices. Some of what I posted is now wrong as the OP changed the $A$ and $B$ to be as general as possible. Before, A and B had a very certain formed that implied a unique $\theta$ and the problem was actually very concrete.  
