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Thread: 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:


    $\displaystyle 5x_1 - 2x_2 + x_3 = 0$
    $\displaystyle 2x_1 + 2x_3 = 0$
    $\displaystyle -x_1 + 2x_2 + 3x_3 = 0$

    Solving the first two, we have

    $\displaystyle \frac{x_1}{-4} \ = \ \frac{x_2}{2-10} \ = \ \frac{x_3}{4}$
    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.

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

    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 $\displaystyle 4x_1+ 4x_3= 0$.

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

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

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