Question about nullity and rank

[georgeh]

So i have the following matrix:
A= [2,0,-1; 4,0,-2;0,0,0]
I do r-r-e
I get
[1,0,-1/2;0,0,0;0,0,0]
So my rank for A is 1, because I only have 1 leading one.
Now for my nullity, i get the following
x_1 - 1/2 X_3 = 0
-->
x_1=1/2 x_3
therefore

[x_1,X_2,X_3] =[1/2;0;1]t
Which would imply that nullity = 1
which then wouldn't satisfy the Dimension theorem for matrices..
which states
rank(a)+nullity(a) = n, where n is the # of columns.
..
so my question is.. what am I doing wrong?

 

[nocturnal]

How did you come up with x_2 = 0?

 

[HallsofIvy]

So i have the following matrix:
A= [2,0,-1; 4,0,-2;0,0,0]
I do r-r-e
I get
[1,0,-1/2;0,0,0;0,0,0]
So my rank for A is 1, because I only have 1 leading one.
Now for my nullity, i get the following
x_1 - 1/2 X_3 = 0
-->
x_1=1/2 x_3
therefore

[x_1,X_2,X_3] =[1/2;0;1]t
Which would imply that nullity = 1
which then wouldn't satisfy the Dimension theorem for matrices..
which states
rank(a)+nullity(a) = n, where n is the # of columns.
..
so my question is.. what am I doing wrong?

The equation x₁- (1/2)x₃= 0 tells you that x₃= 2x₁ alright but tells you nothing about x₂. A basis for the null space is [1, 0, 2] (equivalent to your [1/2, 0, 1]) and [0, 1, 0]. The nullity is 2.