Could somebody please have a look at my attempt to solve this LES? It's the first time I'm solving a LES over a residue class so I'm not quite sure if it's right.
Solve the given linear equation system over $\mathbb{Z}/3\mathbb{Z}$.
For simplicity, let $[\lambda] \hat{=} \lambda$ for all $\lambda \in \mathbb{Z}/3\mathbb{Z}$.
What I did:
$\left(\begin{array}{@{}cccc|c@{}} 1 & 2 & 2 & 2 & 1 \\ 1 & 0 & 2 & 1 & 1 \\ 2 & 2 & 1 & 0 & 2 \\ 1 & 1 & 0 & 2 & 0 \end{array}\right)$ $\text{II-I;} \space \text{III-2} \cdot \text{I;} \space \text{IV-I} \leadsto$ $\left(\begin{array}{@{}cccc|c@{}} 1 & 2 & 2 & 2 & 1 \\ 0 & 1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 2 & 0 \\ 0 & 2 & 1 & 0 & 2 \end{array}\right) \\ $
II and III are identical, thus $x_4$ can be choosen arbitrarily.
Let $x_4 = \lambda$ for $\lambda \in \mathbb{Z}/3\mathbb{Z}$.
II (or III) yields: $x_2 = [0] - [2\lambda] = [-2] \cdot \lambda = [1] \cdot \lambda = \lambda$,
IV yields: $x_3=[2]-[2\lambda] = [2]+[-2]\cdot[\lambda] = [2]+[1]\cdot\lambda = [2+\lambda]$ and
I yields: $x_1 = [1] - [2\lambda] - [4+2\lambda] - [2\lambda] = [1]+[\lambda]-[1+2\lambda]+[\lambda] = [1+\lambda-1-2\lambda+\lambda] = [0]$.
The solution set $\mathbb{L}$ therefore is $$\mathbb{L} = \left\{\begin{pmatrix} [0]\\ [\lambda] \\ [2+\lambda] \\ [\lambda] \end{pmatrix}\right\}$$ for any $\lambda \in \mathbb{Z}/3\mathbb{Z}$.
Also: Would the solution set be any different if the field was $\mathbb{Z}/5\mathbb{Z}$, as no coefficient is bigger than $[3]$?
Thanks in advance!
The answer is right.
Regarding the original LES, it holds that:
I: $[2\lambda]+[4]+[2\lambda] + [2\lambda] = [6\lambda] + [4] = [1]$,
II: $[4]+[2\lambda]+[\lambda]=[4]+[3\lambda]=[1]$,
III: $[2\lambda]+[2]+[\lambda]=[3\lambda]+[2]=[2]$ and
IV: $[\lambda]+[2\lambda]=[3\lambda]=[0]$.