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Math Help - Row Echelon Form

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    Row Echelon Form

    Solve the given systems by reducing the corresponding augmented matrix to row-echelon form. Find the rank of the matrix of coefficients.

    A.) 2x + 3y + z =1
    x + y + z = 3
    3x + 4y + 2z = 4

    I'm having a LOT of trouble reducing this one to row echelon form. Can anyone help me out a bit? The third column is being a real pain to try and reduce...

    The best I've been able to get is

    | 1 1 1 | 3 |
    | 0 1 -1| -5|
    | 0 0 0 | 0 |

    I pretty much am given two unique equations with a derived one... Is there any way to actually get a more accurate answer than this? And would the rank be 2?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by VenomHowell View Post
    Solve the given systems by reducing the corresponding augmented matrix to row-echelon form. Find the rank of the matrix of coefficients.

    A.) 2x + 3y + z =1
    x + y + z = 3
    3x + 4y + 2z = 4

    I'm having a LOT of trouble reducing this one to row echelon form. Can anyone help me out a bit? The third column is being a real pain to try and reduce...

    The best I've been able to get is

    | 1 1 1 | 3 |
    | 0 1 -1| -5|
    | 0 0 0 | 0 |

    I pretty much am given two unique equations with a derived one... Is there any way to actually get a more accurate answer than this? And would the rank be 2?
    umm, it is in row-echelon form. do you mean you want to get it to reduced row-echelon form? if so:

    (assuming you were correct up to this point)

    \left| \begin {array}{ccc|c} 1 & 1 & 1 & 3 \\ 0 & 1 & {-1} & 5 \\ 0 & 0 & 0 & 0 \end {array} \right|

    subtract the second row from the first, and rewrite it as the first row, we get:

    \left| \begin {array}{ccc|c} 1 & 0 & 2 & 8 \\ 0 & 1 & {-1} & 5 \\ 0 & 0 & 0 & 0 \end {array} \right|

    and yes, rank(A) = 2 = \mbox { \# of leading 1's}

    where A, is of course, the matrix of coefficients
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