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I'm having difficulties proceeding with this problem:

We have the following linear equations:

$$\begin{array} 1x + 2y + 2z = 0 \\ 3x – 2y + 2z =1\\ 2x + y + z =3\end{array}$$ $$A=\left(\begin{array}{@{}ccc@{}} 1 & 2 & 2 \\ 3 & -2 & 2 \\ 2 & 1 & 1 \end{array}\right) \\$$ Solve this over $\mathbb{F_2}, \mathbb{F_3}, \mathbb{F_5}$

$det{A}=12\equiv 0 \mod{2}$
$det{A}=12\equiv 0 \mod{3}$

So for $\mathbb{F_2}$ I found after one gaussian elimination step that the solution set $E=\emptyset$. However, for $\mathbb{F_3}$ the last row becomes $0$, so it is an underdeterminate system, in "normal" numbers that would yield a parameter solution, i.e a line equation.

If I did the modulo arithmetic correctly I have sth like: $$\begin{array} 1x + 2y + 2z = 0 \\ y + 2z =1\end{array}$$

How to proceed from here. Is it the same as for non modular equations? Could I have solutions in $\mathbb{Q}$? I'm a bit confused what to do next, and I'd appreciate any input.


PS: for $\mathbb{F_5}$ and all $p>5$ we would have only one solution since $det(A)=12 \not \equiv 0 \mod{5,7,11,...}$

How many solutions $x,y,z \in \mathbb{Z}$ are there?

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What does $L$ stand for? – Gerry Myerson Nov 13 '12 at 5:25
Thanks for this hint. I realized in different languages it's a different letter. I meant to say the solution set L or E or any name. – user41725 Nov 13 '12 at 9:05
up vote 2 down vote accepted

For ${\mathbb F}_3$ your modulo arithmetic is not quite right: since $-2 \equiv 1 \mod 3$ you should have $y + 2 z = 1$, not $-y + 2 z = 1$. Now, just as you are used to over $\mathbb R$, you can express the solutions in terms of the arbitrary variable $z$: $y = 1 - 2 z = 1 + z$, $x = -2y - 2z = y + z = 1 + 2 z$. Taking each possible value for $z \in {\mathbb F}_3$, you should get three solutions.

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I changed my mistake. Thanks for notifying, and for your reply! – user41725 Nov 13 '12 at 3:57

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