Question

Find a basis for the null space of A.

Answer 1

My book says that the null space is the solution space of Ax = b when b = 0. The example in the book shows this...

Answer 2

@TuringTest @asnaseer @CliffSedge

Answer 3

Do I just solve it and that's all...?

Answer 4

I'm also a bit confused how they got the column matrix's filled out... Maybe that's my issue...

Answer 5

pretty much, yeah
just solve it and rewrite each variable as a new vector, i.e. one for r, one for s, and one for t

Answer 6

hmm okay maybe this is easier than I'm thinking let me try it. My example is

A = (Matrix(3, 4, {(1, 1) = 1, (1, 2) = 4, (1, 3) = 5, (1, 4) = 2, (2, 1) = 2, (2, 2) = 1, (2, 3) = 3, (2, 4) = 0, (3, 1) = -1, (3, 2) = 3, (3, 3) = 2, (3, 4) = 2}));
print(`output redirected...`); # input placeholder

Answer 7

woah... LOL

Answer 8

1 4 5 2
2 1 3 0
-1 3 2 2

Answer 9

now do I have to augment it with the 0 column-vector?

Answer 10

go ahead and row reduce it as much as you can
brb, gotta go to get dinner

Answer 11

the book does 2 eq's which is really annoying....

Answer 12

how they would both be the same, so it seems like it's not an augmented matrix....

Answer 13

row reduce, what do you get?

Answer 14

OOOOOOOOOOOOOOOo There's a Null Space/Nullity thing on Maple hehehe. I guess it isn't augmented then :D

Answer 15

no, it's not

Answer 16

1 0 1 -2/7
0 1 1 4/7
0 0 0 0

Answer 17

now I actually found something that says the num space is the subspace of vectors x satisfying A. x = 0

Answer 18

it shows the null space as

2/7
-4/7
0
1

then another column matrix next to it.

-1
-1
1
0..

Answer 19

Is there a matrix thing in the equations..... This is stupid writing it out like this :P.

Answer 20

oh I found it.... :)

Answer 21

setting up matrices with latex is a pain, you are doing it the easy way

Answer 22

HJHAHAH

Answer 23

\[\left(\begin{matrix}2/7 \\-4/7\\ 0\\ 1\end{matrix}\right), \left(\begin{matrix}-1 \\-1\\ 1\\ 0\end{matrix}\right)\]

Answer 24

yes, you get that from
1 0 1 -2/7
x1+r-2/7s=0->x1=-r+2/7s
0 1 1 4/7
x2+r+4/7s=0->x2=-r-4/7s

collect the r's in one vector and the s's in another

Answer 25

my eyes they burn.... :)

Answer 26

you're so smart TT :Pe

Answer 27

first row of your matrix reduced:
1 0 1 -2/7
this is the same as the equation
x1+r-2/7s=0

which leads to
x1=-r+2/7s

and thanks, but I need to get better at computer stuff now, so maybe I'll need your help

sorry, back in 15

Answer 28

yeah yeah yeah!