How to solve system of linear equations with negative scalars over finite field? I need to solve the homogeneous system of linear equations over $Z_7$ via matrix:
\begin{cases}
    \frac{2}{3}x-\frac{1}{2}y+4z -t = 0 \\
    4x - z +3t = 0
  \end{cases}
So this is the matrix:
\begin{bmatrix}
    \frac{2}{3} & -\frac{1}{2} & 4 & -1 \\
    4 & 0 & -1 & 3
  \end{bmatrix}
I can't arrive to the correct solution which I checked in Wolfram Alpha http://www.wolframalpha.com/input/?i=((2%2F3)x+-+0.5y%2B4z+-t)+mod+7%3D0,+(4x+-+z+%2B3t)+mod+7%3D0. Can you please point the incorrect transitions I'm making? (I have to use Gaussian elimination for this)
First of all, my assumption is that if we have negative numbers I need to translate them immediately into  $Z_7$ so the matrix becomes after switching first and second row:
1)
\begin{bmatrix}
     4 & 0 & 6 & 3 \\
    \frac{2}{3} & 6.5 & 4 & 6
  \end{bmatrix} 
2) $R_1 \to 4^{-1}R_1$
 \begin{bmatrix}
     1 & 0 & 5 & 3 \\
    \frac{2}{3} & 6.5 & 4 & 6
  \end{bmatrix} 
3) $R_2 \to R_2 - \frac{2}{3}R_1$
 \begin{bmatrix}
     1 & 0 & 5 & 3 \\
    0 & 6.5 & \frac{2}{3} & 4
  \end{bmatrix} 
4) $R_2 \to \frac{10}{65}R_2$ because $6.5^{-1} = \frac{10}{65}$ over $Z_7$
 \begin{bmatrix}
     1 & 0 & 5 & 3 \\
    0 & 1 & \frac{4}{39} & \frac{8}{13}
  \end{bmatrix} 
Now the matrix looks canonical but I can clearly see that it's not the solution from Worlfram. 

you have 2 equations and 4 unknowns there will be a 2 dimensional solution
$z = z\\
t = t\\
4x = z-3t\\ 
x = \frac 14 z - \frac 34 t\\
\frac 12 y = \frac 23 x + 4z-t\\
y = 2(\frac 23)(\frac 14 z - \frac 34t) + 8z - 2t\\
y = \frac 13 z - t + 8z - 2t\\
y = \frac {25}{3} z - 3t$
or:
$\begin{bmatrix}x\\y\\z\\t\end{bmatrix} = \begin{bmatrix}\frac 14\\\frac {25}3\\1\\0\end{bmatrix}r + \begin{bmatrix}-\frac 34\\-3\\0\\1\end{bmatrix}s$
Now lets get that mod 7
$\begin{bmatrix}x\\y\\z\\t\end{bmatrix} = \begin{bmatrix}3\\ 100\\12\\0\end{bmatrix}r + \begin{bmatrix}-3\\-12\\0\\4\end{bmatrix}s$
$\begin{bmatrix}x\\y\\z\\t\end{bmatrix} = \begin{bmatrix}3\\ 2\\5\\0\end{bmatrix}r + \begin{bmatrix}4\\2\\0\\4\end{bmatrix}s$
