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Math Help - Finding the values of k for which a matrix has no solutions

  1. #1
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    Finding the values of k for which a matrix has no solutions

    Problem:
    Find the values of k for which the system of equations

    has no solutions.


    I instinctively want to solve a question like this by adding/subtracting rows to turn the system into row echelon form, but the k:s in there make it awfully difficult. I've tried changing the order of the equations, but this hasn't helped me either. It seems like such an easy question, but I'm still stumped.
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  2. #2
    Super Member TheChaz's Avatar
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    Have you tried taking the determinant of the matrix?
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  3. #3
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    That will only help me determine if there is a unique solution or not, though, won't it? If the determinant is 0 there can be either an infinite amount of solutions or no solutions, but I can't distinguish between these two cases.
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  4. #4
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    Note that there is no value of k that will make the columns of your matrix linearly dependent. To see this, first solve the homogeneous system ( = 0 vector). The first equation you have implies that:

    <br />
kx + z = 0 \\<br />
    <br />
k = -z/x\\<br />

    But the second equation forces

    <br />
3x - y -\frac{z^2}{x} = 0\\<br />
\Rightarrow y = 3x + \frac{z^2}{x}<br />

    Using the third equation,

    <br />
5x + 3(3x + \frac{z^2}{x}) + z = 0<br />

    <br />
14x + 3\frac{z^2}{x} + z = 0 \\<br />

    <br />
14x^2 + 3z^2 + xz = 0 \\<br /> <br />

    which has no solution except x, z = 0, which will then force y = 0 to solve the homogeneous equation. Hence, your columns are linearly independent independent of your choice of k (bad pun there). You can safely take the determinant and solve.

    Incidentally, how do I insert line breaks (carriage returns)? The standard Latex " \\ " doesn't seem to be working.
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  5. #5
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    The first equation implies that kx+z=1, not 0. And still, how does taking the determinant help me find k?
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  6. #6
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    take the determinant. you will get a quadratic equation in k, which will factor over the integers, giving 2 rational values for k. use these in your matrix, and row-reduce each of the two. you will find both values of k lead to an inconsistent system, that (1,1,1) is not in the column space of either matrix.
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