Determine a basis for the solution set of the homogeneous system 
Determine a basis for the solution set of the homogeneous system:
$$\begin{align*}
x_1 +x_2 +x_3 &=0\\
3x_1+3x_2+x_3 &=0\\
4x_1+4x_2+2x_3&=0
\end{align*}$$
Then the augmented matrix is:
$$
        \left[\begin{array}{ccc|c}
        1 & 1 & 1 &0\\
        3 & 3 & 1 &0\\
        4 & 4 & 2 &0\\
        \end{array}\right]
$$
Reduced Row Echelon Form $\to$
$$
        \left[\begin{array}{ccc|c}
        1 & 1 & 0 &0\\
        0 & 0 & 1 &0\\
        0 & 0 & 0 &0\\
        \end{array}\right]
$$

I already looked at this example but it didn't help much. I am wondering can someone help to find basis (choosing some parameter for variables $x_1,x_2,x_3$) from RREF.
 A: The first equation in the Reduced Row Echelon Form tells you that we need $x_1+x_2=0$ and the second equation says $x_3=0$. So If we take $x_2=t$ for $t\in\mathbb{R}$ then we must have $x_1=-x_2=-t$ and $x_3=0$.
Thus we have a 1-dimensional solution space determined by the vector 
$(x_1,x_2,x_3)=(-1,1,0)$.
This follows from the above discussion since taking $x_2=t$ corresponds to scalar multiplication by t on $(-1,1,0)$.
A: The equations to be satisfied are 
$x_1+ x_2+ x_3= 0$
$3x_1+ 3x_2+ x_3= 0$
$4x_1+ 4x_2+ x_3= 0$
The first thing I notice is that if we multiply the first equation by 3 and subtract the second equation we have $2x_3= 0$ so any solution to this system of equations must have $x_3= 0$.  Setting $x_3= 0$ in these equations, we have
$x_1+ x_2= 0$, $3x_1+ 3x_2= 0$, $4x_1+ 4x_2= 0$
all of which are equivalent to $x_1+ x_2= 0$ or $x_2= -x_1$.  Any solution to this system of equations is for the form $(x_1, x_2, x_3)= (x_1, -x_1, 0)= x_1(1, -1, 0)$.  
That is a one dimensional subspace of $R^3$ with basis $\{(1, -1, 0)\}$.
(Of course, $\{(-1, 1, 0)\}$ is equivalent.)
