gaussian elimination to solve a question (using a paramter)

Question

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!

Answer 1

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.

Answer 2

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

Answer 3

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