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Thu Feb 28 12:36:19 CET 2013

## Rings and multiplicative inverses

```Turn the language into a ring?  I.e. use inverse instead of division
as primitive?  Maybe this isn't necessary: division for matrices is
actually more efficient to implement, as Gauss-Jordan elimination.

I.e. for transfer function:

o = (C (zI - A)^1 B + D) i

However.. multiplication is not commutative, so what would A/B mean?
A * B^1 or B^1 * A?  If it's the former, how to express the latter?
Explicit inverses are probably better from this respect.

In practice, for matrices the form will mostly be:

A^-1 * B

which corresponds to x as the solution of the linear system

A x = B

following this approach, a right inverse B * A^-1 corresponds to

B = x A

so it's the same thing (still a set of equations) but placed in
transposed form.  i.e. it is the same as

B^T = A^T x^T

Does it matter, the transfer function equation above, which RHS is
used to solve the system?  I.e. C or B ?

```
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