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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?

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

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  • $\begingroup$ If look for not solution I get $ab=9$ and $a\neq -2$ or $ab=4$ and $a\neq 3$ (mod 5) $\endgroup$ – gbox Nov 11 '17 at 20:28

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