Calculate the solution set of this system of equations in $\mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3}$
$$I: x_{1}+2x_{2}+x_{4}=2$$
$$II: 2x_{1}+x_{2}+x_{3}+2x_{4}=0$$
$$III: 2x_{2}+x_{3}+x_{4}=0$$
$$IV:x_{1}=1$$
It is important for me to understand the meaning of $Z_{3}$. I think it means that we could add a $\text{ mod }3$ in every equation which has been ignored in this example, right?
So actually it should be saying $\text{ I: }(x_{1}+2x_{2}+x_{4}) \text{ mod } 3=2$ instead, right?
If so, how would you solve this because this modulus will make things completely different :o I could solve it if there wasn't this modulus.. or maybe we can leave it away?
You solve it in exactly the same way as you would if you worked over one of the usual fields, such as $\Bbb R$. Namely, you could write it in matrix form
$$\begin{pmatrix} \bar 1 & \bar 2 & \bar 0 & \bar 1 \\ \bar 2 & \bar 1 & \bar 1 & \bar 2 \\ \bar 0 & \bar 2 & \bar 1 & \bar 1 \\ \bar 1 & \bar 0 & \bar 0 & \bar 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} \bar 2 \\ \bar 0 \\ \bar 0 \\ \bar 1 \end{pmatrix} ,$$
notice that the determinant of the system matrix is $\bar 0$ (meaning that the system has multiple solutions), then noticing that the lower left $3 \times 3$ determinant is non-zero, choosing $x_1, x_2, x_3$ as main unknowns and $x_4$ as a parameter and solve the new $3 \times 3$ system using Cramer's method. The last step would be to give $x_4$ the values $\bar 0, \bar 1, \bar2$ and write down the corresponding values for $(x_1, x_2, x_3)$. This would be an algorithmic approach.
It is more convenient though, in this very specific problem, to use a shortcut: since we are already given that $x_1 = \bar 1$, replace it in all the other equations to get a new system
$$\left\{ \begin{eqnarray} \bar 2 x_2 + x_4 &=& \bar 1 \\ x_2 + x_3 + \bar 2 x_4 &=& - \bar 2 = \bar 1 \\ \bar 2 x_2 + x_3 + x_4 &=& \bar 0 . \end{eqnarray} \right.$$
Subtracting the first equation from the third gives us $x_3 = - \bar 1 = \bar 2$. Replacing this in the new system, the first and third equation will now coincide, so we write only one of them, in order to get yet another system
$$\left\{ \begin{eqnarray} \bar 2 x_2 + x_4 &=& \bar 1 \\ x_2 + \bar 2 x_4 &=& \bar 2 . \end{eqnarray} \right.$$
Finally, this means that from the first equation of this last system we may obtain $x_4 = \bar 1 - \bar 2 x_2 = \bar 1 + x_2$ (we cannot do better because this small system has determinant $\bar 0$, meaning that its solution is not unique), which means that the general solution of the original system is $(\bar 1, x_2, \bar 2, \bar 1 + x_2)$. Since $\Bbb Z _3 = \{ \bar 0, \bar 1, \bar 2 \}$, for each value of $x_2$ in this list of elements we shall obtain a corresponding solution of the original system; namely, these will be
$$(\bar 1, \bar 0, \bar 2, \bar 1), \quad (\bar 1, \bar 1, \bar 2, \bar 2), \quad (\bar 1, \bar 2, \bar 2, \bar 0) .$$