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Math Help - Matrix problem - number of solutions to a linear system of equation.

  1. #1
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    Matrix problem - number of solutions to a linear system of equation.

    Determine the values of a for which the system has no solutions, exactly one solution, or infinitely many solutions.

    x + 2y + z = 2
    2x - 2y + 3z = 1
    x + 2y - (a^2 - 3)z = a

    I made an augmented matrix.

    I tried doing that elimination thing. I ended up with (-a^2 + 2) = a - 2 for the final line.
    In order for the system to be inconsistent, -(a^2 + 2) has to equate to 0, so I solved for a and got rad 2.
    So it's rad 2 for no solution.
    A =/= radical 2 for unique solution (I think).

    I'm not sure about infinite solutions. Any hints would be greatly appreciated.
    Last edited by mr fantastic; September 14th 2011 at 07:10 PM. Reason: Re-titled.
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  2. #2
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    Re: Matrix problem. Help please.

    I think you'll find that you get no solutions when a=\pm\sqrt{2}. For any other value of a, the determinant is nonzero, and hence you get one solution. Thus, for no values of a can you get infinite solutions.
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  3. #3
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    Re: Matrix problem. Help please.

    Quote Originally Posted by Ackbeet View Post
    I think you'll find that you get no solutions when a=\pm\sqrt{2}. For any other value of a, the determinant is nonzero, and hence you get one solution. Thus, for no values of a can you get infinite solutions.
    Thanks. I forgot about - rad 2.
    I'm still not convinced about replacing a with 0. I'll plug it in and see what happens to the matrix.
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