Decide if these 3 vectors are linear independent 
Let $n,p \in \mathbb{N}$ whereby $p$ is a prime number. We have the
  $n$-th cartesian product $\mathbb{Z}_{p}^{n}=\mathbb{Z}_{p} \times ... \times \mathbb{Z}_{p}$ and
  an addition / multiplication is defined by:


We have the vectors


Decide if the vectors $v_{1},v_{2}$ and $v_{3}$ as vectors in
  $\mathbb{Z}_{3}^{3}$ are linear independent.

I wasn't sure if I did it right so that's why I ask. I did this:
$$\begin{pmatrix}
 & (1+1+0) \text{ mod }3 & \\ 
 & (1+0+1) \text{ mod }3 & \\ 
 & (1+0-1) \text{ mod }3 & 
\end{pmatrix}= \begin{pmatrix}
 &2& \\ 
 &2&\\ 
 &0& 
\end{pmatrix}$$
What confuses me is $Z_{3}^{3}$ else I would just have taken $v_{1},v_{2}$ and $v_{3}$, get a linear system and find out if it's linear independent. But what now..?
 A: So, we want to know whether or not we are capable of expressing the zero vector as a non-trivial linear combination of the vector $v_1,v_2,v_3$, so all we have to do is verify if the only possible solution for the system: 
$$\bar{a} \begin{pmatrix}
\bar{1} \\
\bar{1}\\
\bar{1}\\
\end{pmatrix} + \bar{b} \begin{pmatrix} \bar{1}\\ \bar{0} \\ \bar{0} \end{pmatrix}+\bar{c} \begin{pmatrix} \bar{0} \\ \bar{1} \\ \bar{2}\end{pmatrix} = \begin{pmatrix} \bar{0} \\ \bar{0} \\ \bar{0}\end{pmatrix}$$
(remember that, in $\mathbb{Z}_3$, $\ \bar{-1}=\bar{2}$), so the system is :
$$\begin{cases} \bar{a}+\bar{b} = \bar{0} \\ \bar{a}+\bar{c} = \bar{0}\\ \bar{a}+\bar{2c}=\bar{0}
\end{cases}$$
So, subtracting the second equation from the third, we get $\bar{c}=\bar{0}$, necessarily, which implies that the only solution ofr the system is indeed the trivial.
Now, as you said, what confused you is the fact that we are dealing with $\mathbb{Z}_3$, otherwise, you would just solve the linear system, probably using Gauss elimination. However, I would like to point out that in any field, Gauss elimination algorithm is well defined, since in a field, all the non-zero elements have inverses. So in this case too, all you have to do is, as you said yourself, "get a linear system".
