Find base vectors and dim Find base vectors and dim of a space described by the following system of equation:
$$2x_1-x_2+x_3-x_4=0 \\ x_1+2x_2+x_3+2x_4=0 \\ 3x_1+x_2+2x_3+x_4=0$$
I did rref of the matrix and as a result i get:
$$\begin{pmatrix}
 1 & 2 & 1 & 2 \\ 
 0 & -5 & -1 & -5 \\ 
 0 & 0 & 0 & 0
\end{pmatrix} $$ Thus i think that the independent variables will be $x_1,x_2$ and base vectors are the solution of this sytem when $x_1=0,x_2=1 $ and $x_1=1,x_2=0$, but in the answer to the question is that independent variables are $x_2,x_4$. Am i doing something wrong?

My definition of RREF is different from yours it seems, and I calculated RREF form of augmented matrix to be:
$$ \left(
\begin{array}{cccc|c}
1 & 0 & 3/5 & 0 & 0\\
0 & 1 & 1/5 & 1 & 0\\
0 & 0 & 0 & 0 & 0
\end{array}
\right)
$$
From this you set $x_3=t, x_4=s$ where $t,s$ are (scalar) parameters, and you can describe your solution space with these two independent variables.
Then $x_1=-\frac{3}{5} t$, $x_2= -\frac{1}{5}t-s$. So
$$
(x_1,x_2,x_3,x_4)=\left(-\frac{3}{5} t,\; -\frac{1}{5}t-s,\; t,\; s\right)=\left(-\frac{3}{5},\; -\frac{1}{5},\; 1,\; 0\right)t+\left(0,\; -1,\; 0,\; 1\right)s
$$
and you easily see that dimension is $2$.
