# 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?

• I do not see how you come up with the last columnn $(0, 0, -b+\frac{9}{a}, 3+\frac{6}{a})$ – Cornman Nov 11 '17 at 20:11
• @Cornman edtied – gbox Nov 11 '17 at 20:13
• @Cornman sorry typo it is $\frac{6}{a}$ not $-\frac{6}{a}$ – gbox Nov 11 '17 at 20:24
• I deleted my comment, since I responded before I saw that you make 2 calculations to come up with your 3rd row. – Cornman Nov 11 '17 at 20:25

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 $$z\left(\frac{ab}{3}-3\right)=-(2+a)$$
• If look for not solution I get $ab=9$ and $a\neq -2$ or $ab=4$ and $a\neq 3$ (mod 5) – gbox Nov 11 '17 at 20:28