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Math Help - How is this reduced ech form?

  1. #1
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    How is this reduced ech form?

    1 1 0 l 9
    0 1 1 l 7
    0 0 0 l 0

    Basically this is in a problem that says use Gauss jordan elination to solve the following system:

    x1 + x2 = 9
    x2 + x3 = 7
    2x1 + x2 -x3 = 11

    That is the end of elementary operations. To solve using gauss jordan elimination don't you have to work it all the way to reduced echelon form? The second row's leading coef. has a 1 above it so how is it reduced ech form? Doesn't reduced ech form have to have zeroes above all leading coefficents?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: How is this reduced ech form?

    It isn't the reduced echelon form of the matrix, if it was then 1 was the only non-zero element in each column. It's not always possible to bring the matrix in the reduced echelon form.

    Can you find the solution(s)?
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  3. #3
    MHF Contributor

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    Re: How is this reduced ech form?

    \begin{bmatrix}1 & 1 & 0 & 9 \\ 0 & 1 & 1 & 7 \\ 0 & 0 & 0  & 0\end{bmatrix}

    Technically, no, that is not in "reduced" row echelon form. To "reduce" it further you would subtract the second row from the first:
    \begin{bmatrix}1 & 0 & -1 & 2 \\ 0 & 1 & 1 & 7 \\ 0 & 0 & 0  & 0\end{bmatrix}
    but, since that introduces a -1 into the third position of the first row, it is not really any simpler. You can, in any case Find the general solution (since that last row is all 0's there exist an infinite number of solutions) equally easily from either form.
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