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mazehaq

  • one year ago

Question involves spans & vectors, see attached picture.

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  1. mazehaq
    • one year ago
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  2. mazehaq
    • one year ago
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    I know I'm supposed to form a matrix, but I keep getting the answers wrong. Maybe I need to see one of these attempted by you.

  3. UnkleRhaukus
    • one year ago
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    What is the question exactly?

  4. mazehaq
    • one year ago
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    Let me get attach the question, I'm sorry I just noticed it was cut out from the pic I posted

  5. mazehaq
    • one year ago
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  6. mazehaq
    • one year ago
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    Just attached the question

  7. UnkleRhaukus
    • one year ago
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    And what was the modified version of the algorithm mentioned?

  8. mazehaq
    • one year ago
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    I have no idea haha, all I know is to create a matrix using what's given

  9. mazehaq
    • one year ago
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    I'll show #11

  10. mazehaq
    • one year ago
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    lol I'm having trouble making a matrix on here

  11. mazehaq
    • one year ago
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    3 1 1 ; 3 5 1 2 ; 4 -2 4 -3 ; 5

  12. UnkleRhaukus
    • one year ago
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    how are you gonna get rid of the 5 , and -2, in the first column ?

  13. mazehaq
    • one year ago
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    working on it right now

  14. mazehaq
    • one year ago
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    1 5 -2 ; 8 0 7 -3 ; 11 0 14 -7 ; 21

  15. mazehaq
    • one year ago
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    so far

  16. mazehaq
    • one year ago
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    am I even doing the right thing lol

  17. mazehaq
    • one year ago
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    :(

  18. amistre64
    • one year ago
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    v1 v2 v3 S = {(3,5,-2),(1,1,4),(1,2,-3)} b = (3,4,5) can b be formed by a linear combination of the vectors in S? b = c1v1 + c2v2 + c3v3 ?? can we find coefficients that create b, from the vectors in S? c1v1 = 3(c1), 5(c1), -2(c1) c2v2 = 1(c2), 1(c2), 4(c2) c3v3 = 1(c3), 2(c3), -3(c3) --------------------------- b = 3 , 4 , 5 in other words: 3(c1) + 1(c2) + 1(c3) = 3 5(c1) + 1(c2) + 2(c3) = 4 -2(c1) + 4(c2) - 3(c3) = 5 this conforms to the matrix that you wrote up to start your solution, so yes you are on the right track

  19. amistre64
    • one year ago
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    3 1 1 3 5 1 2 4 -2 4 -3 5

  20. mazehaq
    • one year ago
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    So do I go for reduced row echelon form?

  21. amistre64
    • one year ago
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    the long version is to use elimination to solve the system of equation 3x + y + z = 3 5x + y + 2z = 4 -2x +4y - 3z = 5

  22. mazehaq
    • one year ago
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    apparently the answer is 1/2 v1 + 3/2 v2

  23. mazehaq
    • one year ago
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    I don't get how that's the answer

  24. amistre64
    • one year ago
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    it is quite possible that the vectors in S are not independant ... that one or more can be written as a linear combination of the others, what is the determinant of S? 3 1 1 5 1 2 -2 4 -3 1 1/3 1/3 1 1/5 2/5 1 -2 3/2 1 1/3 1/3 0 -2/15 1/15 0 -7/3 7/6 1 1/3 1/3 0 1 -1/2 0 1 -1/2 1 1/3 1/3 0 1 -1/2 0 0 0 1 0 1/2 0 1 -1/2 0 0 0 ................... what this tells us is that v3 is redundant as far as an efficient span goes, v3 can be formed as a linear combination of v1 and v2, thereby making any combination with v3 simplifies to v1 and v2 alone. v3 = (v1-v2)/2 3/2 -1/2 = 1 5/2 -1/2 = 2 -2/2 -4/2 = -3

  25. amistre64
    • one year ago
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    is b, a linear combination of v1 and v2? seeing that v3 is redundant (v3 is in the same plane as v1 and v2) v1 v2 b 3 1 3 5 1 4 -2 4 5 rref{{3,1,3},{5,1,4},{-2,4,5}} http://www.wolframalpha.com/input/?i=rref {{3%2C1%2C3}%2C{5%2C1%2C4}%2C{-2%2C4%2C5}}&dataset=

  26. amistre64
    • one year ago
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    firefox doesnt like to copy paste links correctly .. but the wolf agrees with 1/2v1+ 3/2v2

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