# Am i setting up equations right? Matrices, finding nullspace

1. Dec 11, 2005

### mr_coffee

Hello everyone i'm confused if i'm setting these equations up right:
i have:
1 0 1
0 1 -2
0 0 0

so i said:
x + z = 0;
y - 2z = 0;
z = ? because its a whole row of 0's, so u have no info about what z could be
so i said let z = a;
x + a = 0;
y -2a = 0;
z = a;

x = a;
y = 2a;
z = a;

so let a = 1;
my nullspace is:
1
2
1

am i doing that right?

2. Dec 11, 2005

### shmoe

You lost a minus sign, it should be x=-a.

The nullspace is given by all solutions to this system, not a specific one. That is all vectors (-a,2a,a) make up the nullspace. You can look at a specific non-zero one if you're after a basis of the nullspace (since it has dimension one in this case).

3. Dec 11, 2005

### mr_coffee

i c, thank u!! but the synatx i used is correct? -a, 2a, a correct? so if i factorted out an a i would get
a[-1,21] because my professor got a different answer: he got:

from this matrix:
which is a not simplified version of the matrix i gave above:
2 1 0
0 -1 2
0 0 0

2x + y = 0;
-y + 2z = 0;
x = -a/2;
z = a/2
y = a;

null(A) = [-1/2,1/2]
and i thought he did it wrong.

4. Dec 11, 2005

### shmoe

Strictly as written, null(A)=[-1/2,1,1/2] is wrong, the nullspace is not one vector, but a set of vectors (even if it's just the zero vector).

null(A)=span{[-1/2,1,1/2]} would be a way to write it.

5. Dec 11, 2005

### mr_coffee

Yep that makes sense to me, thanks shmoe