# Thread: Linear eqn's in 3 variables

1. ## 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 !!!!

2. ## 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

3. ## 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 $t= \frac{x_1}{-4}= \frac{x_2}{-8}= \frac{x_3}{4}$