Finding two linearly independent solutions for a homogeneous linear system I'm having difficulty getting the same answer as a textbook solution to a problem.
The basis of the problem is a finding two linearly independent solutions to a homogeneous linear system of the form:
$$u - y = 0\\v +2y - 3z = 0\\ w - z =0\\ x+z=0 $$
MY SOLUTION: Putting this into an augmented matrix it appears to already be in row echelon form...
$$ \left[
    \begin{array}{cccccc|c}
      1&0&0&0&-1&0&0\\
      0&1&0&0&2&-3&0\\
    0&0&1&0&0&-1&0\\
    0&0&0&1&0&1&0
    \end{array}
\right] $$
with free variables y and z. Setting $$y = s$$ and $$z=t$$ the general solution is given as
$$x_H =  \begin{pmatrix}
        u \\
        v \\
        w \\
        x \\
    y \\
    z \\
        \end{pmatrix} 
= s\begin{pmatrix}
        1 \\
        -2 \\
        0 \\
        0 \\
    1 \\
    0 \\
        \end{pmatrix} + t\begin{pmatrix}
        0 \\
        3 \\
        1 \\
        -1 \\
    0 \\
    1 \\
        \end{pmatrix}  $$ 
However the solution in the book gives $$u = -1/2,\\ v= 1,\\ w=x=0,\\ y =-1/2,\\ z=0
$$
and 
$$u = 3/2,\\ v = 0,\\ w = 1,\\ x=-1,\\ y=3/2,\\ z= 1$$
Any help with where I've gone wrong/ misunderstood the method would be much appreciated. I am currently away from my old notes on the topic and haven't quite found the solution online.
Cheers
 A: Hint;
The reduced form should be ( if we order columns left to right  x , y , z, v , u , w )
$$\begin{bmatrix} 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & 1 \\ 0 & 2 & -3 & 1 & 0 & 0 \\ 0 & -1 & 0 & 0 & 1 & 0 \\ \end{bmatrix}$$
Can you see why? Try reducing this.
A: The first answer they give is $-1/2$ times your first answer.  Yours and theirs are both perfectly fine. I prefer yours, no fractions, and fewer minus signs.  
The second given answer is $3/2$ times your first vector plus your second vector. Perfectly correct, as is yours. 
A: For the first case, consider $s = -\dfrac 12$ and $t = 0$. Thus we have that the first solution is:  $$x_1 = -\dfrac 12 \cdot \vec e_1 + 0 \cdot  \vec e_2 .$$
For the second case, consider $s = \dfrac 32$ and $t = 1$. Thus, we have that second solution is of the form:
$$x_2 = \dfrac 32 \cdot \vec e_1 + 1 \cdot \vec e_2.$$ 
These solutions are linearly independent due to the fact that:
$$\begin{vmatrix} -\dfrac 12 & 0 \\ \dfrac 32 & 1 \end{vmatrix} \neq 0$$
Basically, in order to have two linearly independent solutions, you can choose any $2-$ tuples of the form $(s_i,t_i),\ i = 1,2$ such that the determinant
$$\begin{vmatrix} s_1 & t_1 \\ s_2 & t_2  \end{vmatrix} \neq 0$$
