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Math Help - Gauss-Jordan Elimination

  1. #1
    Senior Member Pinkk's Avatar
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    Gauss-Jordan Elimination

    Solve the following using Gauss-Jordan elimination:

    2x+2y+2z=0
    -2x+5y+2z=1
    8x+y+4z=-1

    I keep setting up augmented matrices but I do not know how to get it to reduced-eschelon form at all.

    Edit: I worked this out:

    x=-\frac{1}{7}-\frac{3}{7}r

    y=-\frac{3}{7}

    z=r

    Is this correct?
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  2. #2
    MHF Contributor

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    Yes, that is correct! Well, done!
    (It took me a moment to figure out where that "r" came from!)

    You don't say if you did that by row-reducing the augmented matrix but here's how I did it.

    The augmented matrix is
    \begin{bmatrix}2 & 2 & 2 & 0 \\-2 & 5 & -2 & 1\\ 8& 1 & 4 & -1\end{bmatrix}

    Divide the first row by 2, add the original first row to the second row, and subtract the original first row times 4 from the third row to get
    \begin{bmatrix}1 & 1 & 1 & 0 \\ 0 & 7 & 4 & -1\\0 & -7 & -4 & -1\end{bmatrix}

    Divide the second row by 7, add the original second row to the third row, and subtract the new second row from the first row to get
    \begin{bmatrix}1 & 0 & \frac{3}{7} & -\frac{1}{7} \\ 0 & 1 & \frac{4}{7} & \frac{1}{7} \\ 0 & 0 & 0 & 0\end{bmatrix}.

    The fact that the last row consists of all "0"s tells us that we will have an infinite number of solutions. From the first equation, x+ \frac{3}{7}z= -\frac{1}{7} so x= -\frac{1}{7}- \frac{3}{7}z and y+ \frac{4}{7}z= \frac{1}{7} so y= \frac{1}{7}- \frac{4}{7}z. Now let r= z to get your soltutions.
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  3. #3
    Senior Member Pinkk's Avatar
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    Bah, forgot to write the z for the y part on my paper. And yeah, I meant row-reduced since the following question asks to do it using Gaussian-elimination

    Thank you.

    And to just check, for Gaussian-elimination it would be:

    \begin{bmatrix}1 & 1 & 1 & 0 \\ 0 & 1 & \frac{4}{7} & \frac{1}{7} \\ 0 & 0 & 0 & 0\end{bmatrix}
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