Yes! a and b are regarded as the coefficients!

I apologize for being so late.

The rank of the matrix is 5.

In[35]:= MatrixRank[{{1, 0, 0, -1, 0, 0}, {0, 1, 0, 0, -1, 0}, {0, 0, 
   1, 0, 0, -1}, {a1, a2, a3, 0, 0, 0}, {b1, b2, b3, 0, 0, 0}}]

Out[35]= 5

Excellent! This gives an explanation that the dimension of the set of solutions of eq2 is equal to 1 (6-5 : the number of variables minus the rank of the matrix).

So we can apply now eliminate:

In[36]:= Eliminate[eq2, {z1, z2, z3, u3}]

Out[36]= a3 b1 u1 + a3 b2 u2 - a2 b3 u2 == a1 b3 u1

More the more, we can apply Solve[]:

In[37]:= Solve[eq2, {u1, u2, u3, z1, z2, z3}]

During evaluation of In[36]:= Solve::svars: Equations may not give solutions for all "solve" variables.

Out[37]= {{u2 -> -(((a3 b1 - a1 b3) u1)/(a3 b2 - a2 b3)), 
  u3 -> -1 - ((-a2 b1 + a1 b2) u1)/(a3 b2 - a2 b3), z1 -> u1, 
  z2 -> -(((a3 b1 - a1 b3) u1)/(a3 b2 - a2 b3)), 
  z3 -> -(((-a2 b1 + a1 b2) u1)/(a3 b2 - a2 b3))}}

Both from Eliminate[] and Solve[] we obtain the same result relative to the dimension of the set of the solutions. It's equal to 1.

Additionally, we can apply Solve[] to the initial system:

In[38]:= Solve[{a1 u1 + a2 u2 + a3 (1 + u3) == 0, 
  b1 u1 + b2 u2 + b3 (1 + u3) == 0}, {u1, u2, u3}]

During evaluation of In[38]:= Solve::svars: Equations may not give solutions for all "solve" variables.

Out[38]= {{u2 -> -(((a3 b1 - a1 b3) u1)/(a3 b2 - a2 b3)), 
  u3 -> -1 - ((-a2 b1 + a1 b2) u1)/(a3 b2 - a2 b3)}}

Evidently, this system and eq2 has the same solution set. So, we must admit that the transformations are equivalent!

Hello Daniel, I am busily reading Cox, Little, and O'Shea. It is hard work, but well spent and not only "digestible" but even delicious. Here is what I found so far ( hopefully I understood everything o.k. and am not barking up the wrong tree):

Lets start with the equation from Eliminate (from above):

a3+a1 u1+a2 u2+a3 u3==0 &&
b3+b1 u1+b2 u2+b3 u3==0 &&
a1 (b3+b2 u2+b3 u3)==b1 (a3+a2 u2+a3 u3)

I claim that this set is linear dependent. As you pointed out, I wrongly claimed that the third equation is dependent on the first two. Because you said this, I stopped thinking (intimidation-by-expert-spell!) and it took me some time to get over it and realize that the set IS linear dependent. You can e.g. express the first one by the remaining ones:

eq= {a3+a1 u1+a2 u2+a3 u3,
b3+b1 u1+b2 u2+b3 u3,
a1 (b3+b2 u2+b3 u3)-(b1 (a3+a2 u2+a3 u3)) };

{a1/b1,-1/b1} . eq[[2;;3]] == eq[[1]] //Simplify
True

Well, why does "Eliminate " give linear dependent answers? Let's look what "Groebner Basis" has to tell. We start with eq2, rewritten from equation to polynomial form and calculate the Groebner basis (using lexicographical ordering. Remember z and u are variables, a and b parameters):

eq2a={
z1-u1,
z2-u2,
z3-1-u3,
a1 u1+a2 u2+a3 (1+u3),
b1 u1+b2 u2+b3 (1+u3)};

vars={z1,z2,z3,u1,u2,u3};
GroebnerBasis[eq2a,vars]

{-a3 b1+a1 b3-a2 b1 u2+a1 b2 u2-a3 b1 u3+a1 b3 u3,
b3+b1 u1+b2 u2+b3 u3,
a3+a1 u1+a2 u2+a3 u3,
-1-u3+z3,
-u2+z2,
-u1+z1}

I claim that this is not a reduced Groebner basis and therefor not minimal. To be a reduced basis, no monomial term of any basis polynomial can lie in the affine variety of the leading terms of the remaining polynomials. (Whow quite a mouth full!)

We need only look at the leading terms. The leading terms are:
{(-a2 b1+a1 b2) u2, b1 u1, a1 u1, z3, z2,z1}

it is quite obvious that the second and third term are dependent. I think this is the reason that "Eliminate" gives linear dependent solution sets.

Daniel, thanks a lot. The hint about "CoefficientDomain" really helps, I do not think I would have realized this myself.

Now the problem with "Eliminate". Is it too naive an approach to simply test the output of Eliminate for linear dependencies and eliminate if necessary. Or is the problem more subtle?

O.k. I take your advice. GroebnerBasis seems to be the way to go. And it also has a nice psychological aspect by using a new technique just learned. thank's for all, Daniel

You are right, I must have made a mistake when checking for dependency. However, now it is even worse,the third equation is additional information. Where does it coming from?