@MapleEnthusiast I mean a mathematical reason, not heuristics. Note also that you have simplified wrt size.

(1+e)^3 = 1 + 3e + 3e^2 + e^3: e appears once in LHS and 3 times in RHS. Is this "puzzling"? 

@MapleEnthusiast So, the solutions for A and B are different (and you also took only the numerators!). There is no reason to obtain similar results. If you construct a simpler system (for yourself!) you will be able to check by hand and understand & see the problem.

@MapleEnthusiast Why should they be the same? Note that the R[1,1] could be different in A and B.
If you want to understand exactly what happens, take a smaller system with 2 or 3 parameters and don't use (so many) indices.

@Carl Love Actually, for an expr having a type other than rtable,table, list, Grid:-Map returns 
op(0,expr)(result);
It could be indeed an oversight. Or maybe they want to avoid a (possibly huge) set in the answer.

@acer Yes, I know, but it is anyway unstable; mathematically Rank : R^(n^2) --> N is discontinuous.

So, you are asking for more than the mentioned NP-hard problem!

@MapleEnthusiast The matrix has many symbolic parameters, so the pseudo inverse will be huge. But what do you want to do with it? For a system Ax=b, the pseudo inverse A' is generally used to compute the least squares solution A'b (as Carl said). Your system is homogeneous (b=0), so, A'b=0.

@MapleEnthusiast The More-Penrose inverse exists for any matrix (even non square or 0). But I don't see how could it be useful here.
As in my example, the matrix of your system is singular if the relations are considered. So, it does not make sense to solve it before replacing the dependent parameters. 

@MapleEnthusiast 

@nm Now I can't find this info either. Probably I read about it elsewhere.

@nm PDEtools:-Solve says clearly that the multiplicity is ignored.

@MapleEnthusiast In E is computed the regular inverse (and the result is simplified). The matrix C-A is invertible, so the pseudo inverse equals the regular inverse and using method=pseudo just complicates the computations.

@MapleEnthusiast Your system is a simple polynomial system and homogeneous linear  in the three unknowns. The rank being 2 (in general) it has a unique solution (modulo a multiplicative constant). Obviously It cannot be solved in another way.

So, your new questions make no sense.

P.S. You have very long names in your system. Not knowing anyway their meaning, it would have been much easier for us to see x1,x2,x3 (unknowns), a1,a2...,an (parameters). 

@zphaze  It's about maths, not other CAS-es (check again with Mathematica and Maxima!).
The equation (in z)   arctan(z) = a   has a real solution iff -Pi/2 < a < Pi/2. What is a in your case?

@RohanKarthik I would  have formulated the question this way:
Among all the terms A*x^a*y^b  satisfying a - b \in {0,-1}  of a polynomial f(x,y) ,
find the one for which a+b is minimal.

Or, solve:

min { a+b  |  x^a*y^b is a monomial in f(x,y) and a - b \in {0,-1} }.