> If I start by assigning values to the N unknowns and then recalculate > the values recursively:
Huh? If you assign values to the n unknowns, then the n - 1 equations are all true or one or more is false.
> (a) Will they always reach a single solution (assuming there is a > solution for the equations in the first place), and
> (b) If a single solution can be reached, is there a way to determine it > in terms of the initial values without having to go through the > recursion, and if so, what is it?
> Example: > 3a - b - c - d = 9 > -a + 3b - c - d = 3 > -a - b + 3c - d = -3 > -a - b - c + 3d = -9
That's not three equations in four unknowns; it's four equations in four unknowns.
> If you combine the first three equations, you get a + b + c - 3d = 9, > which is -1 times the fourth equation, so it is effectively 3 equations > in 4 unknowns.
What do you mean by combining? I get a + b + c - 3d = 9 by adding the first three equations.
> If the values are initially set to a = 3, b = 1, c = -1, and d = -3, and > the equations repeated recalculated recursively, they eventually become > a = 9/4, b = 3/4, c = -3/4, and d = -9/4.
You can't set d. What d is, is determined by setting a,b and c. What's a recalculation?
> Can those values be determined in a non-recursive method from the > equations and the initial values?
Given the three equations 3a - b - c - d = 9 -a + 3b - c - d = 3 -a - b + 3c - d = -3 that is 3a - b - c = 9 + d -a + 3b - c = 3 + d -a - b + 3c = -3 + d we have 2a + 2b - 2c = 12 + 2d; a + b - c = 12 + d -2a + 2b + 4c = 2d; -a + b + 2c = d 2b + 2c = 12 + 2d; b + c = 6 + d from which we should be able to calulate a,b,c in terms of d. So d is the only parameter that can be set and by setting d, a,b,c are sets.