@Carl Love 

Yes. Actually division ring is enough (commutativity is automatic) by Wedderburn's theorem.

@Magma 

Thanks, everything is clear now. So, A is not an arbitrary binary matrix, it is probably derived from a matrix over GF.
I wonder whether this approach is faster than computing the determinants over GF(2^r).
I'll try to find an enhanced version soon.

@Magma

The MDS definition is not in terms of GF. Actually GF is not mentioned at all in the first paper.
That is why I have asked. So, do you know the answers (mainly the first one)?

@Magma 

I did not know about this technique. So, it seems that each r x r block of B corresponds to an element of GF(2^r).

Is it known an embedding of GF(2^r) into the M_r(Z2) ? Or, here it is used another definition for MDS?

@Jjjones98 

I don't see how such a solution could be useful. It is inherently huge (almost each coefficient being a symbolic expression).

Not to mention that for some values of the parameters the system will be inconsistent or undetermined, i.e. the obtained solution is generic only. A complete solution (with all possibilities) would be much much longer.
 

@CyberRob 

f is already in the form you describe. Or, use
collect(f, [k12, k21, Ve], distributed);
It would be better to present mathematically the entire problem (preferably with generic notations: x,y,...  the unknowns,  a,b,... the parameters).

@Carl Love 

OK, but we cannot assume that x-1 and x-3 are the only OP's polynomials.

@Carl Love 

Compare:

(x^2+12)/(x+1) mod 13;

Normal((x^2+12)/(x+1)) mod 13;

@Carl Love 

Probably OP wants

Normal( (x-1)/(x-3) ) mod 13; 
    (x+12)/(x+10)

(the field of fractions).

Or, maybe (in this case)

r := Rem(x-1,x-3,x,'q') mod 13:
q + r/(x-3 mod 13);
    1 + 2/(x+10)

@Stretto 

The control is less intuitive because all 3 angles theta, phi, psi  are changed using the mouse. In other CASs, psi=0; this is also true in the Classic Worksheet (exists only for 32 bit but it seems that it will be discontinued).
Probably psi=0 is enough and more intuitive: horizontal mouse move (anywhere) ==> theta, vertical ==> phi.

P.S. Instead of using psi, we could rotate the screen, at least for a laptop  :-)

@Christian Wolinski 

Nice, vote up.
It would have been better if the undefined entries were not treated as 0 by matrixplot.

@Christian Wolinski 

Both are linear, first order PDEs, with standard solutions. The option generalsolution is not needed in this case (if inserted, the result is the same).

@BeatrizSilva 

It works for me.

trabalho_final_2019-ok.mw

@Mariusz Iwaniuk 
It should be added that this is valid in general only for x>0 [not a bilateral expansion]

@Zeineb 

Try:

simplify(gg(x0) - alpha);