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Math Help - Prove there always be a solution

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    Prove there always be a solution

    The system
    3x+4y+z=a
    y+4z=b
    2x+3y+2z=c
    has an obvious solution if [a b c] = [3 0 2]or[4 1 3]or[1 4 2]

    find these solutions.
    Prove or disprove: for any numbers \alpha, \beta, \gammathere is always a solution if
    [a b c] = \alpha[3 0 2] + \beta[4 1 3] + \gamma[1 4 2]

    Under what gerenal conditions on a,b,c will there always be a solution?



    the matrix [a b c] [3 0 2] [4 1 3] [1 4 2]supposed to be in column. I am kind of confused with the question



    I have solved the a part the 3 solution is \left(\begin{array}{c}1 \\0 \\0\end{array}\right), \left(\begin{array}{c}0 \\1 \\0\end{array}\right), and \left(\begin{array}{c}-5 \\4 \\0\end{array}\right).


    now how do I prove there is always a solution under what condition
    Last edited by 450081592; November 15th 2009 at 05:41 PM.
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