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Math Help - Linear eqn's in 3 variables

  1. #1
    Junior Member cupid's Avatar
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    Linear eqn's in 3 variables

    I was solving a question when in the end the author wrote something like this:


    5x_1 - 2x_2 + x_3 = 0
    2x_1 + 2x_3 = 0
    -x_1 + 2x_2 + 3x_3 = 0

    Solving the first two, we have

    \frac{x_1}{-4} \ = \ \frac{x_2}{2-10} \ = \ \frac{x_3}{4}
    What just happened in the end?
    Please can someone explain me this ??
    Thanks a lot !!!!
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Re: Linear eqn's in 3 variables

    He found the cross product of the first two equations.

    \displaystyle \begin{array}{|ccc|}x_1 & x_2 & x_3 \\5 & -2 & 1 \\2 & 0 & 2\end{array} = (-4-0,\ 2-10,\ 0--4) = (-4,\ 2-10,\ 4)

    You can google cross product and see how it is worked out. Or just ask it here
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  3. #3
    MHF Contributor

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    Re: Linear eqn's in 3 variables

    Of course, a less "sophisticated" way to solve the equations would be:
    Add the first and third equations to get 4x_1+ 4x_3= 0.

    That last equation is obviously equivalent to x_1= -x_3. Replacing x_1 with -x_3 in the second equation gives 2x_1+ 2x_3= -2x_3+ 2x_3= 0 for all values of x_3. Putting -x_3 for x_1 in the first equation gives -5x_3- 2x_2+ x_3= -2x_2-4x_3= 0 so that x_2= -2x_3.
    That is (x_1, x_2, x_3)= (-x_3, -2x_3, x_3)= (-1, -2, 1)x_3 , for any value of x_3.

    Since x_1 and x_2 depend linearly on the single parameter x_3, that is a line in three space. Of course, if we multiply all of x_1, x_2, and x_3 by 4 (chosen for the "4" under the x_3 in the original given solution) we have (4x_1, -8x_2, 4x_3)= (-4, -8, 4)x_3 is also on that line. And, taking \frac{x_3}{4}= t as parameter, so that x_3= 4t, we have x_1= -x_3= -4t, x_2= -2x_3= -8t.

    Then t= \frac{x_1}{-4}= \frac{x_2}{-8}= \frac{x_3}{4}
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