Augmented matrices and solution sets - Please help - exam is tomorrow

[stukbv]

Basically there are 2 equations ;

x+2y+3z = 1 2x+4y+6z=2
I put them into a matrix and row reduce to get

1 2 3 | 1
0 0 0 | 0

so we can say x = 1 - 2y -3z and let y and z = 0 to get a solution is (1,0,0)
Now i need to find the nullspace to find the whole solution set;

so x + 2y + 3z = 0

Ive been told the full answer to the set of solutions is
(1,0,0)+ { a(2,-1,0) + b(-3,0,1) | a,b are in reals}

How do they get those solutions for the nullspace, i can see they have set y = 0 and z=0 to get the 2 vectors but how do you know which ones to set = 0, i.e. why couldn't i set x =0 to get a solution in the nullspace??

Thanks so much!

 

[micromass]

Basically there are 2 equations ;

x+2y+3z = 1 2x+4y+6z=2
I put them into a matrix and row reduce to get

1 2 3 | 1
0 0 0 | 0

so we can say x = 1 - 2y -3z and let y and z = 0 to get a solution is (1,0,0)

Hmm, I don't quite get why you're going to search the nullspace now. You already solved the system, that is, you know that the set of solutions is

$$\{(1-2a-3b,a,b)~\vert~a,b\in \mathbb{R}\}=\{(1,0,0)+a(-2,1,0)+b(-3,0,1)~\vert~a,b\in \mathbb{R}\}=(1,0,0)+\{a(-2,1,0)+b(-3,0,1)~\vert~a,b\in \mathbb{R}\}$$

It's not wrong to search for the nullspace, but it's a bit useless...

 

[stukbv]

ok thanks