Assuming you want to solve this problem:
2*A1 = A2
A1 + A2 = A3
then you could use the ndgrid solution, as shown by Walter. In fact, that is surely the solution for such a small problem, since there are only 1000 possible combinations.
Of course, on this specific problem, the answer is simple. Replace a2 in the second equation, and we see
A1 + 2*A1 = A3
So we now have two euations:
A2 = 2*A1
A3 = 3*A1
So we can choose any value for A1 that will not cause a problem with A3. Thus we trivially have
That is clearly the set of all possible solutions. Is this always a viable approach? If the system were far more complex, of course there could be problems. For example, suppose you have some more complex system? I chose a random set of numbers here.
Now suppose we want to find all solutions of this form
A*[x;y;z] == B
ans =
where (x,y,z) all come from that set? Clearly, I have no clue as to whether such a solution even exists. We can find if a particular solution exists. For example, intlinprog will do it.
xyz = intlinprog(ones(3,1),[1 2 3],[],[],A,B,ones(3,1),ones(3,1)*10)
LP: Optimal objective value is 12.600000.
No feasible solution found.
Intlinprog stopped because no integer points satisfy the constraints.
xyz =
[]
Now we know that no solution exists at all, for that problem. If I change B, though,
xyz = intlinprog(ones(3,1),[1 2 3],[],[],A,B2,ones(3,1),ones(3,1)*10)
LP: Optimal objective value is 10.000000.
Optimal solution found.
Intlinprog stopped at the root node because the objective value is within a gap tolerance of the optimal value, options.AbsoluteGapTolerance = 0 (the default value). The intcon variables are integer
within tolerance, options.IntegerTolerance = 1e-05 (the default value).
then we do find a solution. But not all possible solutions.
which tells us the problem reduces to
z == 5, and
x + y == 5
And now we can trivially find all solutions from the admissable set.
