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Thread: Find x1,x2,x3,x4,x5

  1. #1
    Super Member dhiab's Avatar
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    Find x1,x2,x3,x4,x5

    Find the real numbers $\displaystyle x_1 ,x_2 ,x_3 ,x_4 ,x_5 $$\displaystyle
    $ solutions this system, when m is the real parametr :

    $\displaystyle \left\{ \begin{array}{l}
    x_5 + x_2 = mx_1 \\
    x_1 + x_3 = mx_2 \\
    x_2 + x_4 = mx_3 \\
    x_3 + x_5 = mx_4 \\
    x_4 + x_1 = mx_5 \\
    \end{array} \right.$
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by dhiab View Post
    Find the real numbers $\displaystyle x_1 ,x_2 ,x_3 ,x_4 ,x_5 $$\displaystyle
    $ solutions this system, when m is the real parametr :

    $\displaystyle \left\{ \begin{array}{l}
    x_5 + x_2 = mx_1 \\
    x_1 + x_3 = mx_2 \\
    x_2 + x_4 = mx_3 \\
    x_3 + x_5 = mx_4 \\
    x_4 + x_1 = mx_5 \\
    \end{array} \right.$
    There is an obvious solution $\displaystyle (x_1,x_2,x_3,x_4,x_5) = (0,0,0,0,0)$. For all but three values of m, that will be the only solution. The exceptional values of m are $\displaystyle m=2$ and $\displaystyle m= \tfrac12(-1\pm\sqrt5)$.

    If $\displaystyle m=2$, the solution is $\displaystyle (x_1,x_2,x_3,x_4,x_5) = s(1,1,1,1,1)$ (for any real number s).

    If $\displaystyle m= \tfrac12(-1\pm\sqrt5)$ then m satisfies the equation $\displaystyle m^2+m-1=0$ and the solution is $\displaystyle (x_1,x_2,x_3,x_4,x_5) = s(-1,-m,m,1,0) + t(m,-m,-1,0,1)$ (for any real numbers s and t).
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  3. #3
    Senior Member pacman's Avatar
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    dhiab, your problem post is getting harder . . . .
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  4. #4
    Super Member dhiab's Avatar
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    Quote Originally Posted by pacman View Post
    dhiab, your problem post is getting harder . . . .


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