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?
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 $$z\left(\frac{ab}{3}-3\right)=-(2+a)$$