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Math Help - System of linear algebraic equations

  1. #1
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    System of linear algebraic equations

    Hi, i've been trying to solve following system for last two hours and cant find a solution. Can anyone help?

    Problem: Detremnie the values of k, for which the following system has a nontrivial solution:

    9x-3y=kx
    -3x+12y-3z=ky
    -3y+9z=kz


    Find a nontrivial solution in each case.

    Thanks in advance
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  2. #2
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    Re: System of linear algebraic equations

    Solve the first and last equation for y.

    From the first equation: y = \dfrac{9-k}{3}x

    From the last equation: y = \dfrac{9-k}{3}z

    Either y=0,k=9 or x=z. If y=0,k=9 then -3x-3z=0 implies x=-z, so the solution is nontrivial (although there are an infinite number of solutions where k=9).

    Next, suppose k\neq 9 and x=z. Then plug in what you know y=\dfrac{9-k}{3}x, z=x into the second equation:
    -3x + (12-k)\dfrac{9-k}{3}x-3x = 0

    (12-k)\dfrac{9-k}{3}x = 6x

    If x=z=0, then y=0, and this solution is trivial. So, suppose x\neq 0. Then multiply both sides by \dfrac{3}{x}. Hence (12-k)(9-k) = 18. Solve this quadratic: 108 - 21k + k^2 = 18. So k^2-21k + 90 = (k-6)(k-15)=0. So, you get k=6 or k=15.

    Hence, the three values for k that give nontrivial solutions for the system are k=6, k=9, k=15. Note that for any of those three values for k, there are an infinite number of solutions for x,y,z.
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