gaussian elimination to solve a question (using a paramter) I want to solve :
    x2+x3=0
-x1   -x3=0
 x1-x2   =0

I got the $x_1 = -t,       x_2=-t,       x_3=t$.    But the book has  $x_1 = t,       x2=t,       x3=-t$.
Why don't we have the same answer? When I do Gaussian elimination I got this:
    x2+x3=0
 x1-x2   =0

Please help!
 A: You do have the same answer.
$t$ is a parameter; it simply states that given any $t$ in $\mathbb R$, if you give those values to $x_1, x_2, x_3$ you will find solution to your problem.
Since $t$ can be anything, you may as well change sign to $t$ and find the same solution of your book.
But $t$ in itself is not important, what is important is the relationship that $t$ introduces on the solutions (that is, $x_1, x_2, x_3$). And the relationship that both you and your book find is that $x_1 = x_2 = -x_3$. 
It is more convenient though to introduce an independent variable $t$. Look up parametrization
A: I tried again and i got an answer same as the book,
-x2   -x3=0
 x1-x2   =0

We let x3=-t  then we get x1=x2=t
A: The following is your system of linear equations:
$$
\begin{array}{lcl} x_1-x_2 &= 0\\ -x_1-x_3&=0\\ x_2+x_3&=0\end{array}
$$
Represent this system as an augmented matrix and perform the Guassian elimination:
$$\small
\begin{bmatrix}1&-1&0&0\\-1&0&-1&0\\0&1&1&0\end{bmatrix}\sim\begin{bmatrix}1&-1&0&0\\0&-1&-1&0\\0&1&1&0\end{bmatrix}\sim\begin{bmatrix}1&-1&0&0\\-1&0&-1&0\\0&0&0&0\end{bmatrix}\sim\begin{bmatrix}1&0&1&0\\0&1&1&0\\0&0&0&0\end{bmatrix}.
$$
Hence, we may represent the general solution as
$$
\begin{cases}
x_1 &= -x_3\\
x_2 &= -x_3\\
x_3 & \text{is free.}
\end{cases}\tag{1}
$$
If we let $x_3=t$, then $x_1=-t$ and $x_2=-t$, confirming your solution. However, if we let $x_3=-t$, then $x_1=t$ and $x_2 = t$, confirming the book's solution. You could come up with any sort of parameter you want to express $x_3$, but you will need to solve for $x_1$ and $x_2$ in $(1)$ accordingly. 
