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Math Help - Linear Algebra help

  1. #1
    Junior Member
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    Linear Algebra help

    hey, i need some hints on how do this problem. i really have no idea how to do it... i don't need an answer, i just need to understand how to do the problem...

    it is:

    x + y + kx = 1
    x + ky + z = 1
    kx + y + z = -2

    no solution, unique solution, or infinitely many solutions
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  2. #2
    MHF Contributor

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    Paris, France
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    The symmetry of the equations suggests to look at the particular case k=1 first. What happens in this case?

    Then, suppose k\neq 1. You can try to solve the system just as you would have with known numbers: for instance, using the second and the last equation, you can get an equation without z. Repeating this with another subset of two equations, you find a system of two equations in x and y (where the coefficients depend on k). Do these equations have one solution? none? infinitely many?

    I hope this helps. If you know matrices and determinants, there's a quicker solution by looking at the system only for the values of k which make the determinant of the system equal to 0 (you know 1 is one of them), since you know that if it is non-zero then there's a unique solution.
    Laurent.
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  3. #3
    Senior Member
    Joined
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    Quote Originally Posted by aeubz View Post
    hey, i need some hints on how do this problem. i really have no idea how to do it... i don't need an answer, i just need to understand how to do the problem...

    it is:

    x + y + kx = 1
    x + ky + z = 1
    kx + y + z = -2

    no solution, unique solution, or infinitely many solutions
    First of all, I'm assuming you made a misprint, and that by:

    \left\{\begin{array}{c}x +  y +  kx =  1 \\ x +  ky +  z =  1 \\ kx +  y +  z =  -2 \end{array}\right\}

    ...you actually meant:

    \left\{\begin{array}{c}x +  y +  kz =  1 \\ x +  ky +  z =  1 \\ kx +  y +  z =  -2 \end{array}\right\}

    I actually don't know how to do this using matrices, but you can do it using high school algebra. According to the first equation:

    x=1-y-kz

    Plug that into the second equation, then find y in terms of z:

    1-y-kz+ky+z=1

    y=z

    Now plug your x and y equations into the third equation to find z in terms of k:

    k(1-z-kz)+z+z=-2

    z=\frac{k-2}{(k+2)(k-1)}

    Which of course means:

    y=\frac{k-2}{(k+2)(k-1)}

    Which in turn means:

    x=1-\frac{k-2}{(k+2)(k-1)}-k\frac{k-2}{(k+2)(k-1)}

    or, simplified:

    x=\frac{2k}{(k+2)(k-1)}
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