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Math Help - Basic Linear Algebra question

  1. #1
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    Basic Linear Algebra question

    find a solution to the system:

    5a1 + 2a2 - 3a3 = 4
    a1 - a2 + 2a3 = -1

    Solution:

    we can eliminate a1 from the second row by using R1 - 5R2 -> R2, which
    gives:

    5a1 - 2a2 - 3a3 = 4
    7a2 - 13a3 = 9

    from the second equation we obtain:

    a2 = 1/7 (9+13a3)

    We substitute this expression in the first equation and solve for a1 in

    terms of a3. We find that

    a1 = 2/7 - 1/7 a3
    ....................

    I can understand everything up to this point. However I cannot see how this has been substituted or how one can do this. Can anyone illustrate precisely what has happened in the last step?

    Many thanks

    Biter
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  2. #2
    Senior Member
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    Jul 2006
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    Given a_2=\frac{1}{7}(9+13a_3),

    We substitute it into R1:

    5a_1 + 2a_2 - 3a_3 = 4 (R1)
    \implies 5a_1 + 2({\color{red}\frac{1}{7}(9+13a_3)}) - 3a_3 = 4

    EDIT: You made a typo the second time you wrote R1 and reversed a sign.
    Last edited by scorpion007; May 22nd 2009 at 06:23 AM.
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  3. #3
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    Hi thanks for helping me with this. The following step (or outcome) of the process is:

    a1 = 2/7 - 1/7 a3

    *************

    I cannot follow the solution resulting from the substitution. Does the answer need to be factorised?

    Thanks
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  4. #4
    Senior Member
    Joined
    Jul 2006
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    It's just basic algebraic manipulation from there:

    5a_1 + 2(\frac{1}{7}(9+13a_3)) - 3a_3 = 4

    \implies 5a_1 + \frac{18}{7}+\frac{26}{7}a_3 - 3a_3 = 4

    \implies 5a_1 +\frac{5}{7}a_3 = 4-\frac{18}{7}

    \implies 5a_1 = \frac{10}{7}-\frac{5}{7}a_3

    \implies a_1 = \frac{2}{7}-\frac{1}{7}a_3

    Hope that is clear.
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