Isomorphism of a ring of matrices Is it possible for a ring of matrices to isomorphic a ring of numbers?
Suppose
$$R = \begin{pmatrix}
        a & b \\
        -3b & a \\
        \end{pmatrix} a,b \in \mathbb Z $$
Can $R$ isomorphic to $\mathbb C$?
 A: No, matrices in $\mathbb{Z}$ are countable whereas $\mathbb{C}$ is not. If you allow matrices with entries in $\mathbb{C}$ then the $1\times 1$-matrices do the job.
A: 
Is it possible for a ring of matrices to isomorphic a ring of numbers?

What hampers me the most here is that I have no idea what a "ring of numbers" is supposed to be. If it is supposed to be one of $\Bbb C$, $\Bbb R$, $\Bbb Q$ or $\Bbb Z$, or even $\Bbb H$, then yes, it is possible. For example, $\{\left[\begin{smallmatrix}a&b\\-b&a\end{smallmatrix}\right]\mid a,b \in \Bbb R\}\cong \Bbb C$.
For that matter, if $I_n$ is the square identity matrix of size $n$, then $\{rI_n\mid r\in R\}\cong R$ for any ring $R$. 
Your particular candidate does not work out, cardinality being the most obvious barrier, as others have pointed out.
Really the lesson to be learned here is not to make any artificial distinctions about what "numbers" are. Any element of a ring could broadly be described as a number, because rings are supposed to extend what was traditionally thought of as numbers. It is not common practice to call elements of rings "numbers," though.
A: As Mathematician 42 has already answered, the cardinality argument shows that $R$ cannot be isomorphic to $\mathbb{C}$. In fact, $R$ is isomorphic to $\mathbb{Z}[i\sqrt{3}]$ via the map $a + bi\sqrt{3} \mapsto \begin{pmatrix} a & b \\ -3b & a \end{pmatrix}$. It can be easily shown that the multiplication is just the same:
$$(a+bi\sqrt{3})(c+di\sqrt{3}) = (ac-3bd)+(ad+dc)i\sqrt{3}$$
$$\begin{pmatrix} a & b \\ -3b & a \end{pmatrix} \begin{pmatrix} c & d \\ -3d & c \end{pmatrix} = \begin{pmatrix} ac-3bd & ad+bc \\ -3(bc+ad) & -3bd+ac \end{pmatrix}$$
A: Yes, it can be isomorphic to a “ring of numbers”; more precisely, to a subring of the complex numbers.
It's easy to see that the ring $R$ is generated by
$$
J=\begin{bmatrix}0&1\\-3&0\end{bmatrix}
$$
in the sense that every element of $R$ can be written (in a unique way) as $aI+bJ$, for $a,b\in\mathbb{Z}$.
Notice that $R$ is indeed a ring, because $J^2=-3I$ and so
$$
(aI+bJ)(cI+dJ)=(ac-3bd)I+(ad+bc)J
$$
so $R$ is closed under multiplication; it's obviously an additive subgroup of the ring of all matrices.
We can define a unique homomorphism
$$
v\colon\mathbb{Z}[X]\to R
$$
by mandating $v(X)=J$. Then $v$ is surjective and its kernel is the ideal generated by $X^2+3$ (prove it).
Therefore $R$ is isomorphic to $\mathbb{Z}[X]/(X^2+3)$ which in turn is isomorphic to $\mathbb{Z}[\sqrt{-3}]$ (prove it).
