<s>[INST] For Which $a$ And $b$ Above $\mathbb{Z}_5$ There Are Solutions? 
For which values of $a$ and $b$ above $\mathbb{Z_5}$ the following equations have no solution/one solution, infinite solutions  
\begin{cases}
ax+4y+3z=0   \\
2y+3z=1\\
3x-bz=3
\end{cases}

So the matrix is 
$$
  \left[\begin{array}{rrr|r}
    a & 4 & 3 & 0 \\
    0 & 2 & 3 & 1 \\
    3 & 0 & -b & 3
  \end{array}\right]
$$
After $-\frac{3}{a}R_1+R_3\rightarrow R_3$ and $\frac{6}{a}R_2+R_3\rightarrow R_3$ and assuming $a\neq 0$
$$
  \left[\begin{array}{rrr|r}
    a & 4 & 3 & 0 \\
    0 & 2 & 3 & 1 \\
    0 & 0 & -b+\frac{9}{a} & 3+\frac{6}{a}
  \end{array}\right]
$$
So I have no solution if $-b+\frac{9}{a}=0$ and $3+\frac{6}{a}\neq 0$
Infinite solution if $-b+\frac{9}{a}=0$ and $3+\frac{6}{a}=0$
And one solution in all the other cases? 
 [/INST] from the equation (III) we get
$$x=1+\frac{b}{3}z$$
plugg in (I)
$$a+z\left(\frac{ab}{3}+3\right)+4y=0$$
from (II) we have
$$4y=2-6z$$  and we obtain
$$z\left(\frac{ab}{3}-3\right)=-(2+a)$$
</s>