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KonradZuse

  • 3 years ago

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

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  1. KonradZuse
    • 3 years ago
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  2. KonradZuse
    • 3 years ago
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    Basically I just need help finding how to set up the variables to be able to do this EQ @TuringTest @phi @satellite73 @farmdawgnation

  3. phi
    • 3 years ago
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    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

  4. KonradZuse
    • 3 years ago
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    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.

  5. KonradZuse
    • 3 years ago
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    We have to solve for the variables though right?

  6. phi
    • 3 years ago
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    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/

  7. KonradZuse
    • 3 years ago
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    hmmm........

  8. phi
    • 3 years ago
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    basis for row space are the pivot rows basis for the col space are the pivot cols of the original matrix

  9. KonradZuse
    • 3 years ago
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    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?

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  10. KonradZuse
    • 3 years ago
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    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......

  11. KonradZuse
    • 3 years ago
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    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....

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

  13. phi
    • 3 years ago
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    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.

  14. phi
    • 3 years ago
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    same idea for the col space, except you take the cols from A not rref(A)

  15. phi
    • 3 years ago
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    got to go...

  16. KonradZuse
    • 3 years ago
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    Oh noes :(.

  17. KonradZuse
    • 3 years ago
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    wait crap.... :(

  18. KonradZuse
    • 3 years ago
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    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.

  19. KonradZuse
    • 3 years ago
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    oic you meant null basis :P

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