Why is the complex number $z=a+bi$ equivalent to the matrix form $\left(\begin{smallmatrix}a &-b\\b&a\end{smallmatrix}\right)$ 
Possible Duplicate:
Relation of this antisymmetric matrix $r = \!\left(\begin{smallmatrix}0 &1\\-1 & 0\end{smallmatrix}\right)$ to $i$

On Wikipedia, it says that:

Matrix representation of complex numbers
Complex numbers $z=a+ib$ can also be represented by $2\times2$ matrices that have the following form: $$\pmatrix{a&-b\\b&a}$$

I don't understand why they can be represented by these matrices or where these matrices come from.

Since you put the tag quaternions, let me say a bit more about performing identifications like that: 
Recall the quaternions $\mathcal{Q}$ is the group consisting of elements $\{\pm1, \pm \hat{i}, \pm \hat{j}, \pm \hat{k}\}$ equipped with multiplication that satisfies the rules according to the diagram
$$\hat{i} \rightarrow \hat{j} \rightarrow \hat{k}.$$
Now what is more interesting is that you can let $\mathcal{Q}$ become a four dimensional real vector space with basis $\{1,\hat{i},\hat{j},\hat{k}\}$ equipped with an $\Bbb{R}$ - bilinear multiplication map that satisfies the rules above. You can also define the norm of a quaternion $a + b\hat{i} + c\hat{j} + d\hat{k}$ as 
$$||a + b\hat{i} + c\hat{j} + d\hat{k}|| = a^2 + b^2 + c^2 + d^2.$$
Now if you consider $\mathcal{Q}^{\times}$, the set of all unit quaternions you can identify $\mathcal{Q}^{\times}$ with $\textrm{SU}(2)$ as a group and as a topological space. How do we do this identification? Well it's not very hard. Recall that 
$$\textrm{SU}(2) = \left\{ \left(\begin{array}{cc} a + bi & -c + di \\ c + di & a-bi \end{array}\right) |\hspace{3mm} a,b,c,d \in \Bbb{R}, \hspace{3mm} a^2 + b^2 + c^2 + d^2 = 1    \right\}.$$
So you now make an ansatz (german for educated guess) that the identification we are going to make is via the map $f$ that sends a quaternion $a + b\hat{i} + c\hat{j} + d\hat{k}$ to the matrix $$\left(\begin{array}{cc} a + bi & -c + di \\ c + di & a-bi \end{array}\right).$$
It is easy to see that $f$  is a well-defined group isomorphism by an algebra bash and it is also clear that $f$ is a homeomorphism. In summary, the point I wish to make is that these identifications give us a useful way to interpret things. For example, instead of interpreting $\textrm{SU}(2)$ as boring old matrices that you say "meh" to you now have a geometric understanding of what $\textrm{SU}(2)$. You can think about each matrix as being a point on the sphere $S^3$ in 4-space! How rad is that?
On the other hand when you say $\Bbb{R}^4$ has now basis elements consisting of $\{1,\hat{i},\hat{j},\hat{k}\}$, you have given $\Bbb{R}^4$ a multiplication structure and it becomes not just an $\Bbb{R}$ - module but a module over itself. 
