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KonradZuse
 3 years ago
Given the Matrix A and it's reduced row echelon form find
KonradZuse
 3 years ago
Given the Matrix A and it's reduced row echelon form find

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KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0Basically I just need help finding how to set up the variables to be able to do this EQ @TuringTest @phi @satellite73 @farmdawgnation

phi
 3 years ago
Best ResponseYou've already chosen the best response.1This 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

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, 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.

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0We have to solve for the variables though right?

phi
 3 years ago
Best ResponseYou've already chosen the best response.1I 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/1806linearalgebraspring2010/videolectures/lecture7solvingax0pivotvariablesspecialsolutions/

phi
 3 years ago
Best ResponseYou've already chosen the best response.1basis for row space are the pivot rows basis for the col space are the pivot cols of the original matrix

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0From 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?

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0Remark 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......

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0It also seems to me that they just take each row and make r1, r2, r3, etc and tha'ts the basis for the rows....

phi
 3 years ago
Best ResponseYou've already chosen the best response.1yes, that sounds right. I know the first vector in N(A) is [13/7 1/2 5/3 2/7 1]

phi
 3 years ago
Best ResponseYou've already chosen the best response.1It 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.

phi
 3 years ago
Best ResponseYou've already chosen the best response.1same idea for the col space, except you take the cols from A not rref(A)

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

KonradZuse
 3 years ago
Best ResponseYou've already chosen the best response.0oic you meant null basis :P
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