Question

Given the Matrix A and it's reduced row echelon form find

Answer 1

Basically I just need help finding how to set up the variables to be able to do this EQ

@TuringTest @phi @satellite73 @farmdawgnation

Answer 2

This question is asking about the column space, row space, null space
you can find the null space by setting the bottom variable to 1 or 0

Answer 3

Yes, I know, but I thought we need to find the variables first to find the null space and such... I was looking back on my past homeworks so I could be wrong, because I forgot how to do this mostly :P.

Answer 4

We have to solve for the variables though right?

Answer 5

I use a shortcut so often that the actual details are fuzzy. I just take the last 2 columns (the ones without a pivot) and negate all the entries in a "pivot row"
then I put 1 and 0 in the remaining slots
See http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-7-solving-ax-0-pivot-variables-special-solutions/

Answer 6

hmmm........

Answer 7

basis for row space are the pivot rows
basis for the col space are the pivot cols of the original matrix

Answer 8

From looking at my homework again I guess I don't really need to set up anything. It just seems that the last 2 columns have to do with it, which might be what you're getting it?

Answer 9

Remark Observe that the basis vectors , , and in the last example are the vectors that result by successively
setting one of the parameters in the general solution equal to 1 and the others equal to 0.


hmm......

Answer 10

It also seems to me that they just take each row and make r1, r2, r3, etc and tha'ts the basis for the rows....

Answer 11

yes, that sounds right.
I know the first vector in N(A) is
[13/7 -1/2 -5/3 -2/7 1]

Answer 12

It also seems to me that they just take each row and make r1, r2, r3, etc and tha'ts the basis for the rows....

the rows with a pivot.

Answer 13

same idea for the col space, except you take the cols from A not rref(A)

Answer 14

got to go...

Answer 15

Oh noes :(.

Answer 16

wait crap.... :(

Answer 17

I know the first vector in N(A) is
[13/7 -1/2 -5/3 -2/7 1] isn't a row? :( a col maybe that's negated :P.

Answer 18

oic you meant null basis :P