Basis of Image in a GF(5) matrix with variables consider $A_\alpha := \begin{pmatrix} \alpha & 3 & 4\\ 3 & 1 & 3 \\ 4 & 2 & \alpha \end{pmatrix} \in \mathbb{Z_5}^{3\times3}$. Determine, dependant on $\alpha$, a Basis and the dimension of the Image of the standardinterpretation $F_{A_\alpha}$. What is the dimension of the kernel of $F_{A_\alpha}$? The standardinterpretation is defined as $F_A : K^{n\times1} \rightarrow K^{m \times 1}, x \mapsto F_A(x):=Ax $.
Okay. To determine the Image of a Matrix we just row reduce it and then we will see which columns are in the Image of that matrix.
Let's row reduce the matrix
$\begin{pmatrix}
\alpha &  3 & 4\\ 
 3&  1& 3\\ 
 4&  2& a
\end{pmatrix} \underset{\overset{R1 \leftarrow R_1+2\cdot R_2}{\longrightarrow}}{\overset{R_3 \leftarrow R_3 + 3 \cdot R_2}{\longrightarrow}} 
\begin{pmatrix}
a+1 &  0& 0\\ 
 3&  1& 3\\ 
 0&  0& a+4
\end{pmatrix}$.
This doesn't look too bad, let's just look at the cases we get:
Let $\alpha = 0$ then we get $\begin{pmatrix}
1 &  0& 0\\ 
 3&  1& 3\\ 
 0&  0& 0
\end{pmatrix} \overset{R_2 \leftarrow R_2+2\cdot R_1}{\longrightarrow}
\begin{pmatrix}
1 &  0& 0\\ 
 0&  1& 3\\ 
 0&  0& 0
\end{pmatrix}
$ then we can see that a basis for $Image(A_\alpha)$ is  $\{\begin{pmatrix} a \\ 3 \\ 4 \end{pmatrix}, \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix}\}$. Since we have 2 elements in our basis for the Image, we know that $dim(Image)=2$.
We also remember that $dim(V) = dim(kernel) + dim(Image)$ therefore $3 = dim(kernel) + 2$ so we get $dim(kernel)=1$.
Do I have to repeat this process for all other 4 cases ($\alpha = 1,2,3,4$) or am I completly wrong about my approach? Thanks for any hints on how to solve this problem.

Your row reduction is indeed correct. Your steps from there are problematic.  We have the reduced matrix
$$
\pmatrix{
a+1 &  0& 0\\ 
 3&  1& 3\\ 
 0&  0& a+4}.
$$
It is easy to see that for $\alpha \notin \{1,4\}$, further row-reduction brings us to the identity so that $A$ is invertible. In the case that $\alpha = 1$, we get
$$
\pmatrix{
2 &  0& 0\\ 
 3&  1& 3\\ 
 0&  0& 0} \leadsto
\pmatrix{
2 &  0& 0\\ 
 0&  1& 3\\ 
 0&  0& 0}.
$$
We deduce that the first two columns of $A$ form a basis of the image. That is, the image has basis
$$
\{(\alpha,3,4),(3,1,2)\} = \{(1,3,4),(3,1,2)\}.
$$
So, the dimension of the image is $2$ and the dimension of the kernel is $3 - 2 = 1$.
Similarly, for $\alpha = 4$, we get
$$
\pmatrix{
0 &  0& 0\\ 
 3&  1& 3\\ 
 0&  0& 3} \leadsto
\pmatrix{ 
 3&  1& 0\\ 
 0&  0& 3\\
0&0&0}.
$$
We deduce that the first and third columns of $A$ form a basis of the image. So, the dimension of the image is $2$ and the dimension of the kernel is $3 - 2 = 1$.
