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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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