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.
Thanks!
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?
Best Answer
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.