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Math Help - Linearly Independent (Matrix), help!

  1. #1
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    Linearly Independent (Matrix), help!

    Linearly independent (MAtrix)

    --------------------------------------------------------------------------------

    The rows of the matrix:
    Show it is linearly independent

    1 4 -2 3
    2 -2 1 1
    3 1 2 11

    I have the solution, but I don't understand.

    Solution:

    a(1 4 -2 3) + b(2 -2 1 1 ) + c(3 1 2 11) = (0 0 0 0)

    /\
    |
    \/
    why the above equation can change to the below equations??????


    a + 2b + 3c = 0
    4a - 2b + c = 0
    -2a + b + 2 c = 0
    3a + b + 11c = 0

    a=b=c=0, so it is linearly independent.


    Confused
    Last edited by orange890; May 19th 2006 at 05:27 AM. Reason: typing error
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by orange890
    Linearly independent (MAtrix)

    --------------------------------------------------------------------------------

    The rows of the matrix:

    1 4 -2 3
    2 -2 1 1
    3 1 2 11

    I have the solution, but I don't understand.
    You want to show that the rows of this matrix are linearly independent?

    If they are linearly independent the only solution of:


    a(1 4 -2 3) + b(2 -2 1 1 ) + c(3 1 2 11) = (0 0 0 0)
    is the trivial solution a=b=c=0.

    a + 2b + 3c = 0
    4a - 2b + c = 0
    -2a + b + 2 c = 0
    3a + b + 11c = 0
    This is just your previous vector equation written as four scalar equations,
    which you should proceed to solve to show that you must have: a=b=c=0,
    and so you can conclude that the rows are linearly independent.

    RonL
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  3. #3
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    a(1 4 -2 3) + b(2 -2 1 1 ) + c(3 1 2 11) = (0 0 0 0)
    [COLOR=DarkOrange]
    /\
    |
    \/
    How does it form??????

    a + 2b + 3c = 0
    4a - 2b + c = 0
    -2a + b + 2 c = 0
    3a + b + 11c = 0
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by orange890
    a(1 4 -2 3) + b(2 -2 1 1 ) + c(3 1 2 11) = (0 0 0 0)
    [COLOR=DarkOrange]
    Look at the first component of the expressions on the Left and Right hand sides:

    a + 2 b +3 c = 0

    Now the seecond component:

    4a-2b+c=0

    Third component:

    -2a+b+2c=0

    Fourth component:

    3a+b+11c=0

    RonL
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  5. #5
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    hi RonL,

    yeah really thanks for your helping.
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