Question

can someone tell me whether my results are correct?

Answer 1

I get rank to 2

Answer 2

Nullity to 2

Answer 3

and null space to

Answer 4

can you show me how you get that answer?

Answer 5

what is your rref?

Answer 6

this is the rref of A, not null space, the question asks you about nullity, space of null space. have to translate that rref into null space

Answer 7

the definition of nullity is number of free variable, you have 1 free (either x or y) so, nullity is 1

Answer 8

Arrh...

Answer 9

the rank of null space is 2, you are correct then. I don't know how do you get it, but whatever, you are right.

Answer 10

:)

Answer 11

How can I find the rank and nullity for the null space?

Answer 12

base on your rref, I continue :
x+y =0 --> x =-y
z = 1/2
I choose y is free variable, therefore, the null space can be expressed as
\[\left(\begin{matrix}-y \\ y\\1/2\end{matrix}\right)= y \left(\begin{matrix}-1 \\ 1\\0\end{matrix}\right)+ 1/2\left(\begin{matrix}0 \\ 0\\1\end{matrix}\right)\]

Answer 13

that the way I express null space, with 1 free variable is y , 2 vector base --> rank is 2 , 1 free --> 1 nullity

Answer 14

I thinkt you are missing that the rref is without the right hand-side
\[x1+x2=0\]\[x3-(1/2)x4=0\]

so:
\[x1=-x2\]and
\[x3=(1/2)x4\]

Answer 15

oops, the way you post the matrix makes me misunderstanding. I thought that is not homogeneous system, that the last column is the RHS of the equations.
Sorry for that. Please, close the post and open another one with the original system to make the question clear.

Answer 16

you mean you have 3equations, 4 variables? and after getting rref, you have 2 independent equations left?

Answer 17

terribly sorry for my hastily given answer.