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Math Help - Linear Combination

  1. #1
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    Linear Combination

    I need to find if
    3
    -1
    2

    is a linear combination of

    1 3 2
    -1 1 2
    2 -2 -4

    I set it up so i had 3 equations they were
    3 = x1 + 3x2 +2x3
    -1 = -x1 + x2 + 2x3
    2 = 2x1 - 2x2 - 4x3

    where x1, x2, and x3 are variables. I should be able to solve these and then find the values of the variables and those are the #'s which will make the linear combination possible. But I can't figure it out!!! Help!!!
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  2. #2
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    Heir it is solved.
    Throught the inverse matrix.
    Attached Thumbnails Attached Thumbnails Linear Combination-picture1.gif  
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Heir it is solved.
    Throught the inverse matrix.
    So you have changed the task that needs to be done from
    solving a set of three linear simultaneous equations to that
    of inverting a 3x3 matrix. Other than that you have given the
    inverse, so having effectively given the answer, you leave the
    OP with a task they are less likely to be able to do.

    RonL
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  4. #4
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    why did you take the inverse of the matrix? what does that accomplish? I have just started linear algebra and am not sure what you mean or why you would take the inverse. Could you please explain... Thanks
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  5. #5
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    Quote Originally Posted by TreeMoney View Post
    I set it up so i had 3 equations they were
    3 = x1 + 3x2 +2x3
    -1 = -x1 + x2 + 2x3
    2 = 2x1 - 2x2 - 4x3
    Replace the second equation by the second plus the first, give us:

    Code:
    3 = x1   + 3.x2 +2.x3
    2 =      + 4.x2 +4.x3
    2 = 2.x1 - 2.x2 - 4.x3
    Now replace the third row by the third row minus twice the first
    row:

    Code:
    3  = x1   + 3,x2 +2.x3
    2  =      + 4.x2 +4.x3
    -4 =      - 8.x2 - 8.x3
    Now we see that the last two rows are multiples of one another,
    which leaves us with two effective equations in three unknowns.

    This leaves us with multiple solutions so assume a value for x3
    and then solve for x1 and x2 in terms of this.

    For instance x1=3/2, x2=1/2, x3=0 looks like a solution to me
    (assuming no arithmetic errors).

    RonL
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by TreeMoney View Post
    why did you take the inverse of the matrix? what does that accomplish? I have just started linear algebra and am not sure what you mean or why you would take the inverse. Could you please explain... Thanks
    Ignore it, first he has inverted the wrong matrix, second the matrix
    he should have inverted is singular, third even if the matrix was not
    singular he is replacing a simple process related to Gaussian elimination
    by a more complex problem without showing how to solve the new
    problem.

    RonL
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  7. #7
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    Thanks CaptainBlack and everyone else. I see what I was missing now. Thanks again!!!
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