@dharr For the EqN for all parameters 1, the three equations have a common factor, that can be divided out so that you could work on solving a simpler system. If this still works with parameters in, that could afford a considerable simplification. 

@MaPal93 In general, I expect the solvability to depend on the parameters. This seems like a very complicated problem with too many parameters - how about solving a simpler version of the problem with a smaller set of parameters first? But if I were doing this problem I would step back and analyze the matrices. Are lambda's eigenvalues? if so a lot can be said about their possible values depending on the structure of the matrix of which they are eigenvalues, e.g., we know many classes of matrices that are guaranteed to have one or more real eigenvalues.

@dharr For solution 1 you can find real solutions as follows (but real and positive is harder). 

Download EqNsol.mw

@MaPal93 Some answers:

1) what is exactly Eqn? could you load my solution instead of hard coding expressions?

I definitely did not reenter this by hand! This was extracted from your worksheet by opening a new worksheet with the same Maple engine - try EqN; in your own worksheet to check.

2) is this system indeterminate (infinitely many solutions)?

Yes.

3) would it be preferred if I chose backsubstitute = true?

Probably the answer is too complicated to analyse, and maybe too slow to achieve - but you could try.

4) what happened to my parameters P := indets(MyEqs, name) minus MyVars ? I was hoping to obtain solutions as functions of those.

I just looked at what you were passing to the solver, i.e., EqN. Looks like in the line EqN := ((`~`[((numer@evala)@(:-Norm))@numer])@eval)(MyEqs, `=`~(P, 1)) you set all the parameters to 1. 

5) is this solving approach (a) reliable and (b) preferable to engine=groebner or standard solve()?

I don't see why the answer wouldn't be reliable. It worked for the case I tried at random. I don't know much about different engines; solve probably passes to one or other of these.

6) how to enforce RealDomain?

If you want only real solutions, try RealTriangularize in the RegularChains package, which was a recent suggestion somewhere here on Mapleprimes. But I really don't know much about solving polynomial equations beyond the bivariate case.

@MaPal93 Here's my understanding. You chose backsubstitute=false, so you can carry out the backsubstitution yourself, e.g.

Download EqNsol.mw

@occipitalbaker Thanks.

I did something similar before here but for the case where the walk had to start and end at a particular vertex. (Perhaps that is the same problem?) There is a significant difference between just finding the length and finding the walk itself. Are you OK with just the length?

@Axel Vogt Vote up. The second one can be done without a change of variable.

Download int.mw

@rockyicer You're welcome. If you want to convert the .lib file to .mla, you can use the LibraryTools:-ConvertVersion command.

Yes, development is always going on - new integration methods were added in the latest (2023) version, see

https://www.maplesoft.com/support/help/Maple/view.aspx?path=updates%2fMaple2023%2fAdvancedMath

@mmcdara If you try Eigenvalues(SQRT),Eigenvalues(Q_L); you will see that two of the eigenvalues are negative of what they should be, i.e., the method has effectively chosen the wrong sqrt. I'm guessing that can be solved by a different starting guess, but not sure how exactly that could be done. Note that for the Eigenvalue method, you can work completely in reals (and more efficiently) by telling it C is symmetric:
 

LEC := proc(C)
  local L, E;
  L, E := LinearAlgebra:-Eigenvectors(Matrix(evalf(C),shape=symmetric)):
  E . DiagonalMatrix(sqrt~(L)) . E^(-1)
end proc:

But if you use a numerical version of my routine above (with shape = symmetric) then you can avoid the numerical inverse (just a transpose now) and it might be more efficient (but the normalization will slow it down)

Msqrtf:=proc(A::Matrix(square,datatype=float))
  local evs,evals,evecs,U,i,j,k,g,Lambda;
  uses LinearAlgebra;
  evs:=Eigenvectors(Matrix(A,shape=symmetric),output='list');
  evals:=table();evecs:=table();
  j:=0;
  for i in evs do
    if i[2] = 1 then
      j:=j+1;
      evals[j]:=i[1];
      evecs[j]:=Normalize(i[3][],'Euclidean');
    else
      g:=GramSchmidt(i[3],conjugate=false,normalized=true);
      for k to i[2] do
        j:=j+1;
        evals[j]:=i[1];
        evecs[j]:=g[k];
      end do;
    end if;
  end do;
  Lambda:=DiagonalMatrix(map(sqrt,convert(evals,list)));
  U:=convert(convert(evecs,list),Matrix);
  U.Lambda.U^%T;
end proc:

@sand15 Thanks - I remembered that and had a quick look for it but didn't find it. The emphasis there was on the numerical case, but here on symbolic. For symbolic use, the method here requires that the exact eigenvalues can be found. Here even though the matrices are large, the eigenvalues and eigenvectors are easily found (for M1..M3, the eigenvalues are actually integers). The method here avoids finding a matrix inverse, or more accurately the inverse is just the hermitian transpose so is easy to compute.

My recollection is that Maple's method can suceed in more cases, at the expense of being slow. The solution is presumably for MatrixPower and/or MatrixFunction to do more preselection of different algorithms depending on the matrix type.

convert(expr,elementary) is intended to convert hypergeoms etc to elementary functions, but doesn't seem to work on these cases. 

@lcz Isomorphic triangulations don't necessarily correspond to isomorphic binary trees, so I think a different approach is needed.

@OtherAlloyMonkey I left all the equations in terms of beta, and made all the constants exact. Then it is easy to see that the 5th and 6th rows of the matrix are identical, and therefore the determinant must be zero.

Model_Derivation_3.mw