Numbers defined with matrices

This is to some extent a question of representation theory. So we do use these things.

For example, suppose you want to extend the field of rational numbers to include some weird number $\xi$ which satisfies $\xi^2 - N = 0$. You can represent multiplication by a number $a+b\xi$ with the 2x2 matrix $\pmatrix{a & Nb \\ b & a}$. Then yours is just a special case of $N = -1$, which (if the underlying elements $a$ and $b$ are in $\mathbb{R}$) is one way to represent complex numbers. If $N = 0$ then we have the dual numbers, and if $N=1$ then we have the split-complex numbers.

Note that extensions like this can cause problems and we may lose field properties, for instance there is no way to divide a real number by a pure dual number since eliminating the dual from the denominator constitutes division by zero. So always check that the basic rules of arithmetic are preserved, or if we need additional constraints. Just because we lose field properties doesn't mean the algebraic structure isn't interesting or useful.

This kind of extension can continue. If your underlying field is $\mathbb{Q}$ and you extend it to include $\xi_N$ (as above) and then want to extend it again to include another $\xi_M$ then you have a 4x4 matrix representation.

To put what you've done in algebraic context: you've identified a subset of the ring of 2x2 matrices (with real number entries I assume, since those are the type we use to represent the complex numbers) and showed that within this subset, multiplication is commutative, and the subset is closed under addition and multiplication, so it's a commutative subring. In quid's answer to this related question, quid shows that your subring is isomorphic to the direct product of two copies of $\mathbb{R}$. I refer you there to the algebraic proof: I'll show it for your example in a more concrete way.

Associate to each matrix $\begin{bmatrix} a & b \\ b & a \end{bmatrix}$ the pair $(a+b,a-b)$. It's clear that, conversely, given such a pair, you can recover the matrix. Let's check how matrix addition and multiplication affect the pair. $$\begin{bmatrix} a & b \\ b & a \end{bmatrix} + \begin{bmatrix} a' & b' \\ b' & a' \end{bmatrix} = \begin{bmatrix} a+a' & b+b' \\ b+b' & a+a' \end{bmatrix}$$, so $$(a+b,a-b)+(a'+b',a'-b') = (a+a'+b+b',a+a'-b-b')$$. And $$\begin{bmatrix} a & b \\ b & a \end{bmatrix} \cdot \begin{bmatrix} a' & b' \\ b' & a' \end{bmatrix} = \begin{bmatrix} aa'+bb' & ab'+ba' \\ ab'+ba' & aa'+bb'' \end{bmatrix}$$, so $$(a+b,a-b)\cdot(a'+b',a'-b') = (aa'+bb'+ab'+ba',aa'+bb'-ab'-ba')= ((a+b)(a'+b'),(a-b)(a'-b'))$$. Note that the matrix addition and multiplication on the matrices induce regular addition and multiplication on the pairs, performed independently on the first and second entries.

So it turns out these "numbers" are equivalent to pairs of numbers on which addition and multiplication are performed separately.

The point is that we want to have a field. At least, this is what is suggested here by speaking of "numbers". Now in general, subalgebras of $M_2(K)$ need not be a field. In the first case, however, we obtain a field, namey the field of complex numbers $\mathbb{C}$ . In the second example, we do not obtain a field. For a field $(K,+,\cdot)$, we need that $(K,+)$ and $(K^*,\cdot)$ are both abelian groups. This is not the case here.