# Why were the matrix multiplication rules chosen this way?

The currently standard matrix multiplication is isomorphic to split-quaternions (for 2x2 matrices, and similar for higher ranks, maybe this is called Clifford algebras). As such, the sets of square matrices is not commutative, has zero divisors, idempotents and nilpotents.

I wonder, why other matrix multiplication rules were not chosen. For instance, the tessarine-like that would be commutative?

Yes, usual quaternions or tessarines or split-numbers of any dimension can be represented by the matrices, but since the multiplication rules differ, one would need the matrices of squared rank to represent them.

• Because they reflect the composition of linear transformations (linear substitutions originally), see Why, historically, do we multiply matrices as we do? In applications people care about interpretation first, and about algebraic properties only if they help with that or with computations. There are other products defined on matrices, the entrywise one, Kronecker product, etc., but they are not as applicable. Mar 30, 2021 at 23:42
• Matrix operations were perhaps defined by Cayley. The definitions were made because of the useful applications. So in a sense, Cayley "discovered" matrices rather than "invented". Nowadays we see that, since composition of linear transformations is not commutative, neither is multiplication of matrices commutative. Mar 31, 2021 at 12:27

As with your previous question about why tessarines' commutativity doesn't win them more interest, commutativity is not mathematicians' goal; problem-solving is. We're interested in vector spaces, and in linear operations from such spaces to the same or other spaces (i.e. matrices), and in the composition of such linear operations (per @Conifold's comment) viz.$$(AB)v:=A(Bv)\implies(AB)_{ij}v_j=A_{ik}(Bv)_k=A_{ik}B_{kj}v_j\implies(AB)_{ij}=A_{ik}B_{kj}.$$You're welcome to consider other operations of the form $$(A\circ B)_{ij}:=A_{mn}B_{pq}C_{ijmnpq}$$; the example above is $$C_{ijmnpq}:=\delta_{im}\delta_{jq}\delta_{np}$$.$${}^\dagger$$ You're welcome to determine which $$C$$ obtain $$A\circ B=B\circ A$$. But the irony is this:
• Each $$\circ$$ comprises a two-vectors-to-one-vector bilinear operation, analogous to the cross product on $$3$$-dimensional space. (Does it bother you that that's anticommutative?)
• Such operations also have a natural characterization in terms of matrices as they're defined above. To wit:$$(v\circ w)_i:=K_{ijk}v_jw_k\implies(v\circ)_{ik}=K_{ijk}v_j,\,(\circ w)_{ij}=K_{ijk}w_k.$$In other words, matrices are still the natural software you're running on.
$${}^\dagger$$ No others will obey $$(AB)v=A(Bv)$$, so you'll just be treating "matrices" as "arrays of numbers I can do anything I want with" rather than quantities that represent an underlying something in a suitable basis, but I suppose mathematics doesn't force you to do something useful if you don't want to. On the other hand, it doesn't force people to be interested in what you find elsewhere.