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Math Help - Quick Lin Alg question

  1. #1
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    Quick Lin Alg question

    If I have the following matrix (3x3):

    1 0 1 5
    0 1 0 -7/5
    0 0 5 7

    Is it legal to just take 1/5 of row 3 and say my new matrix is then:

    1 0 1 5
    0 1 0 -7/5
    0 0 1 7/5

    ? I'm assuming not bc ive checked my work a hundred times and this could be the only possible mistake (Note I am trying to get it in RREF)


    EDIT:

    Meh, actually I think you CAN... so who wants to help me find out where I went wrong:

    ORIG MATRIX:

    1 -1 0 5
    -1 1 5 2
    0 1 1 0

    Then switch R_2 and R_3

    1 -1 0 5
    0 1 1 0
    -1 1 5 2

    Then R_3 + R_1

    1 -1 0 5
    0 1 1 0
    0 0 5 7

    Then R_1 + R_2

    1 0 1 5
    0 1 1 0
    0 0 5 7

    Then R_2 - 1/5R_3

    1 0 1 5
    0 1 0 -7/5
    0 0 5 7

    Then 1/5R_3

    1 0 1 5
    0 1 0 -7/5
    0 0 1 7/5

    Then R_1 - R_3


    1 0 0 18/5
    0 1 0 -7/5
    0 0 1 7/5

    No idea
    Last edited by fifthrapiers; September 4th 2008 at 08:42 PM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by fifthrapiers View Post
    If I have the following matrix (3x3):

    1 0 1 5
    0 1 0 -7/5
    0 0 5 7

    Is it legal to just take 1/5 of row 3 and say my new matrix is then:

    1 0 1 5
    0 1 0 -7/5
    0 0 1 7/5

    ? I'm assuming not bc ive checked my work a hundred times and this could be the only possible mistake (Note I am trying to get it in RREF)
    If you're doing row reduction, that would be legal.

    however, you don't have the 3x3 identity matrix appearing in that 3x4 matrix!

    to fix that, do this:

    \left[\begin{array}{lllr} 1&0&1&5\\ 0&1&0&-\frac{7}{5}\\ 0&0&1&\frac{7}{5}\end{array}\right]


    -R_3+R_1\rightarrow R_1

    \color{red}\boxed{\left[\begin{array}{lllr} 1&0&0&\frac{18}{5}\\ 0&1&0&-\frac{7}{5}\\ 0&0&1&\frac{7}{5}\end{array}\right]}

    I hope this helps!

    --Chris
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  3. #3
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    Sorry I meant I had a 3x3 matrix with an augmented column. I edited my post. Could you look it over?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by fifthrapiers View Post
    Sorry I meant I had a 3x3 matrix with an augmented column. I edited my post. Could you look it over?
    Your work looks correct.

    The solution set would be \left(\tfrac{18}{5},-\tfrac{7}{5},\tfrac{7}{5}\right)

    --Chris
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  5. #5
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    It's not though. The solution is 2, -1, 1.

    EDIT: Turns out I copied down the wrong original matrix. No wonder I was spending like an hour trying to figure out what I did wrong. Thanks !
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    When you apply the Gauss-Jordan Method to the augmented matrix, your solution should have the form:

    if its a 2x3:

    \left[\begin{array}{llr}1&0&x_0\\ 0&1&y_0\end{array}\right]

    where \left(x_0,y_0\right) is the solution set.

    it its a 3x4:

    \left[\begin{array}{lllr}1&0&0&x_0\\ 0&1&0&y_0\\0&0&1&z_0\end{array}\right]

    where \left(x_0,y_0,z_0\right) is the solution set.

    I hope this helps you in solving future problems!

    --Chris
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