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Math Help - Matrices Again

  1. #1
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    Matrices Again

    For the following system of linear equations determine the values of k for which it has (i) no solutions, (ii) infinitely many solutions, and (iii) a unique solution. For those values of k where solutions exist, find what those solutions are.

    I am trying to do this with Gaussian or Gauss-Jordan elimination

    2x1 + 6x2 - 4x3 = 0
    3x1 + 9x2 - 6x3 = 0
    -4x1 - 12x2 + kx3 = 0

    I'm familiar with how to do Gaussian elimination, but I don't know how to go about solving each of the situations they ask for.

    Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Adrian View Post
    For the following system of linear equations determine the values of k for which it has (i) no solutions, (ii) infinitely many solutions, and (iii) a unique solution. For those values of k where solutions exist, find what those solutions are.

    I am trying to do this with Gaussian or Gauss-Jordan elimination

    2x1 + 6x2 - 4x3 = 0
    3x1 + 9x2 - 6x3 = 0
    -4x1 - 12x2 + kx3 = 0

    I'm familiar with how to do Gaussian elimination, but I don't know how to go about solving each of the situations they ask for.

    Thanks
    create the augmented matrix (from the coefficients):

    \left( \begin{array}{ccc|c} 2 & 6 & 4 & 0\\ 3 & 9 & -6 & 0\\ -4 & -12 & k  & 0\end{array} \right)

    since you are familar with Gaus-Jordan elimination, perform it on the system above, then read of your solutions at the end. do you know how to do that?
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  3. #3
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    Hello, Adrian!

    For the following system of linear equations, determine the values of k
    for which it has (i) no solutions, (ii) infinitely many solutions, and (iii) a unique solution.
    For those values of k where solutions exist, find what those solutions are.

    . . \begin{array}{ccc}2x +  6y - 4z& = &0\\<br />
3x + 9y - 6z & = & 0\\ \text{-}4x - 12y + kz & = & 0\end{array}

    We have: . \begin{vmatrix}\:2 & 6 & \text{-}4 & | & 0\: \\ \:3 & 9 & \text{-}6 & | & 0\: \\ \:\text{-}4 & \text{-}12 & k & | & 0\:\end{vmatrix}

    . . . . \begin{array}{c}\frac{1}{2}R_1 \\ \frac{1}{3}R_2 \\ \text{-}\frac{1}{4}R_3\end{array} \;\begin{vmatrix}\:1 & 3 & \text{-}2 &|& 0\: \\ \:1 & 3 & \text{-}2 &|& 0\: \\ \:1 & 3 & \text{-}\frac{k}{4} & | & 0\: \end{vmatrix}

    \begin{array}{c} \\ R_2-R_1\\R_3-R_1\end{array}\;<br />
\begin{vmatrix}\;1 & 3 & \text{-}2 & | & 0\: \\ \:0 & 0 & 0 & | & 0\: \\ \:0 & 0 & 2\,\text{-}\,\frac{k}{4} & | & 0\; \end{vmatrix}


    It turns out that the three equations are almost identical.

    If k = 8, the three equation are identical.
    We have one equation: . x + 3y - 2z\:=\:0
    . . which has infinitely many solutions.

    If k \neq 8, we have: . x + 3y \:=\:0,\;\;z \:=\:0
    . . which also has infinitely many solutions.


    I see no way that the system can have a unique solution or no solutions.

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  4. #4
    MHF Contributor red_dog's Avatar
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    The system is homogeneous and it has always at least a solution: (0,0,0).

    An homogeneous system has an unique solution ( (0,0,0) ) if the determinant of the matrix of system is not equal to 0.

    In this problem, the determinant is 0 for all real k. So, the system has more than one solution.
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  5. #5
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    Thanks a lot, that must've been why I was having trouble trying to find unique and no solutions.

    And sorry for the double post, I had no idea I did that.
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