using "solve" and getting rid of unknown parameters

I am trying to solve a system of equations whose answer will depend on arbitrary values.

For example, consider the following system of 2 equations and 3 unknowns:

syms x1 x2 x3
E=[x1+2*x2+x3;x3-x1+x2];
a=solve(E==0,x1,x2,x3)
[a.x1 a.x2 a.x3]

this gives me an answer that depends on an arbitrary value z.

I would like MATLAB to directly give me an answer (with no arbitrary values). So for instance, I would like to get an answer where the x1 value is positive. Is there a way to do that?

I am using "solve" in a subroutine, and after that, I would like to test the answer to see if it is a non negative array. But whenever the code reaches the answer, it tells me there is an error (obviously) because of the unknown parameter.

Answer 1

[Azzi Abdelmalek]

You have 2 equations and 3 unknowns variables, which means there are infinite solutions. It's like your are solving

E=[x1+2*x2+z;z-x1+x2];
b=solve(E==0,x1,x2)

[Del]

Thanks for your answer, but I am aware there will be infinitely many solutions. However, what I would like MATLAB to give me among all those solutions, is one answer that satisfies specific conditions.

For instance, I would like one answer such that x1>=0. So, an example of an acceptable answer is:

   [1 -2 3]

[Azzi Abdelmalek]

Then just replace z by a positive number, for example

z=3
x1=eval(a.x1)
x2=eval(a.x2)
x3=eval(a.x3)

[Del]

I would like to do that, the only problem is, like I said, the equation that I am solving is in a subroutine, and I don't necessarily know how many arbitrary "parameters" it will have at each iteration or even, the names of the unknown parameters.

For instance, for the small example that I gave, I know that there is one arbitrary parameter, and the name is z, only because I typed the problem in the MATLAB command Window by itself. However, if it is a big problem, and I want to solve an equation as a part of the code, I will not know the names of the parameters involved.

[Azzi Abdelmalek]

As long as there are infinite solutions, Matlab will generate those solutions, you have to add, clearly, the conditions that allows to choose some of them.

[Del]

sure, my question, is what is the correct way in MATLAB " of adding , clearly, the conditions that allows to choose some of them."

[Azzi Abdelmalek]

There are your conditions, if for example you said that the first solution must be positive. The below code will choose one solution.

syms x1 x2 x3
E=[x1+2*x2+x3;x3-x1+x2];
a=solve(E==0,x1,x2,x3)
n= size(fieldnames(a),1)
z=solve(a.x1==1)
for k=1:n
    sol(k)=eval(a.(sprintf('x%d',k)));
end

Answer 2

[Walter Roberson]

You might be able to use numeric::solve

That routine is not exposed at the MATLAB level so you would have to feval() or evalin() the symengine.