How to find a set of vectors spanning the solution space of $Ax=0$, where How to find a set of vectors spanning the solution space of $Ax=0$, where

Basically I have tried many times to solve it and my answer consistently comes in the following form:
$\pmatrix{1 \\ -1 \\-1\\0}$
While my book gives an answer of:
$\pmatrix{-1 \\-1\\1\\0}$
 A: Do the following row-operations: 
$$R_2:-R_4, \; R_3:-2R_1, \; R_4: -R_1, \; R_4:-R_3, \; R_3: -R_2$$
which gives you 
$$\pmatrix{1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0}$$
which means $x_1+x_3=0, x_2+x_3=0, x_4=0$. Letting $x_1=x_2=t$, we have $$x=\pmatrix{t \\ t \\ -t \\ 0} \rightarrow \pmatrix{1 \\ 1 \\ -1 \\ 0}$$
A: Reducing:
$$\begin{pmatrix}1&0&1&0\\1&2&3&1\\2&1&3&1\\1&1&2&1\end{pmatrix}\longrightarrow\begin{pmatrix}1&0&1&0\\0&2&2&1\\0&1&1&1\\0&1&1&1\end{pmatrix}\longrightarrow\begin{pmatrix}1&0&1&0\\0&1&1&1\\0&0&0&\!\!-1\\0&0&0&0\end{pmatrix}$$
The general solution to the homogeneous system $\;A\vec x=\vec0\;$ is
$$x_4=0\;,\;\;x_1=x_2=-x_3\;\implies\;\left\{\;\begin{pmatrix}t\\t\\\!\!-t\\0\end{pmatrix}\;\;;\;\;\;t\in\Bbb F\right\}$$
whatever the field $\;\Bbb F\;$ is. Thus, a particular non-zero solution, and also a basis for the solution space, is for example
$$\begin{pmatrix}1\\1\\\!\!-1\\0\end{pmatrix}$$
A: Hint:
solve the system:
$$
\begin{cases}
x+z=0\\
x+2y+3z+t=0\\
2x+y+3z+t=0\\
x+y+2z+t=0
\end{cases}
$$
note that you can find $z=-x$ from the first equation and solve for the first three only since the fourth equation is a linear combination.
You find that the solution is the space of vectors of the form $(y,y,-y,0)^T$
so the solution of your book is correct ($y=-1$).
