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Set of linear equation question

  1. Dec 9, 2008 #1
  2. jcsd
  3. Dec 9, 2008 #2

    Defennder

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    You need to write out equations for x,y,z such that alpha, beta cancels out. Eg. Suppose x = a + 1 and y = a, where a is the free variable, then a possible eg of an equation would be x - y = 1.
     
  4. Dec 9, 2008 #3
    i wrote a matrix
    and inputed alpha=1 and beta=2
    and this is my answer
    is that ok?
     
  5. Dec 9, 2008 #4

    Defennder

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    What matrix did you write? And I don't think you can arbitrarily set values of alpha, beta. Remember that these are free variables, so if you set some particular values for them you are in effect saying they are not parameters which may be varied.
     
  6. Dec 9, 2008 #5
    i need to give an example so the set that i was given
    would be its general solution(i guess i need to present one case)
    i have constructed this general matrix:

    1 a b
    2 a b
    3 a 0


    and inputed some a=1 b=2

    but i am not sure if this the correct way of solving it??
     
  7. Dec 9, 2008 #6

    Mark44

    Staff: Mentor

    transgalactic,
    I find that it is frequently difficult for me to understand what you are trying to do, which makes it hard to help you.

    From what you posted on imageshack, it appears to me that you are supposed to find a 3 x 3 matrix A so that <what you posted> is the general solution of Ax = b.

    The vectors with alpha and beta multipliers, (1, 1, 1)^T and (1, 1, 0)^T appear to me to be all of the solutions of Ax = 0. That is, these vectors form a basis for the nullspace of A.

    The other vector, (1, 2, 3)^T seems to be the particular solution of Ax = b, for some unknown vector b.

    Possibly there is enough information here so that you can find the coefficients of matrix A.
     
  8. Dec 10, 2008 #7
  9. Dec 10, 2008 #8
    i got a way of solving this question:
    (x,y,z)=(1+a+b) + (2+a+b) +(3+a)

    i say that z=0 because (3+a) is not dependent on b.
    i say that a+b=t

    x=1+t y=2+t
    x-y=-1
    y-x=1

    is that a prove??
     
  10. Dec 10, 2008 #9

    Defennder

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    No, that doesn't satisfy the required general solution. z is not zero here, since in the general solution z=3+a where a is arbitrary.
     
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