# How to solve

1. Feb 7, 2004

### franz32

I hope some can help me here.

What is the best strategy in solving 4 linear equations with five unknown variables?

2. Feb 7, 2004

### HallsofIvy

Well, of course, you cannot "solve" four equations in five unknowns. What you can do is select any specific value for one of the five unknowns and then solve the four equations for the other four.

Equivalently, you could solve for four of the unknowns "in terms of the fifth". That is, you treat one of the unknowns as if it were a constant and solving for the other four so that you have something of the form:
a= a function of e, b= a function of e, c= a function of e, d= a function of e, where a, b, c, d, e are the five unknowns.

Now you could choose any value you like for e and calculate the other four. Whether it is best to solve the four linear equations by substitution or Gaussian elimination depends on the particular equations.

3. Feb 7, 2004

### himanshu121

General Rule

If above is not true u will end up in finding relation

4. Feb 9, 2004

### franz32

Can anyone help me again

Hello guys! I did understand what you are talking about. =)
But not that really understand fully. I tell you what I don't
understand.

Let say D = {v1, v2, v3, v4, v5} where v1 = (1, 1, 0, -1); v2 = (0, 1, 2, 1); v3 = (1, 0, 1, -1); v4 = (1, 1, -6, -3) and v5 = (-1, -5, 1, 0). In order for me to find a basis for the subspace W = Span D of R^4, I must show that D spans R^4 right? If this fails, then, can I still find a basis for it?

(a,b,c,d) = k1v1 + k2v2+ k3v3 + k4v4 + k5v5. When I made an
augmented matrix out of it, I reached ... " 0 0 0 0 | b + c -3d - 4a.
What does it mean?

When do I know if a subspace does not span?

5. Feb 9, 2004

### HallsofIvy

Re: Can anyone help me again

No, in order to find a basis for subspace W= Span D, you do NOT need to show that D spans R^4.
Suppose v1= (1, 1, 0, -1), v2= (2, 2, 0, -2), v3= (3, 3, 0, -3), v4= (4, 4, 0, -4) and v5= (5, 5, 0, -5). D certainly does NOT span R^4. W= Span D is the set of all vectors of the form (a, a, 0, -a) where a is any real number. That is one-dimensional and has any one of the vectors given as basis: {v1} will do nicely.

In the example you give, I would set this up as a matrix having each of those vectors as row and "row-reduce" I get 4 non-zero rows (the last row all zeroes) so, yes, the example you give does span all of R^4 and a perfectly good basis is {(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)}.

If I did that with the example I gave, I would see all rows except the first become all zeroes and would know that that first row constituted the basis.