Notation: https://en.wikipedia.org/wiki/Elementary_matrix#Elementary_row_operations
$ a\in \mathbb{F}_{5}\\ A\in\mathbb{F}_{5}^{3x3}\\ A=\begin{pmatrix} a & a+1 & a+2 \\ a+1 & a+2 & a+3 \\ a+2 & a+3 & a+4\end{pmatrix}\\ L_{1,2}(-a)\cdot T_{1,2}\cdot L_{2,3}(-2)\cdot L_{1,3}(-1)\cdot L_{1,2}(-1) \cdot A= \begin{pmatrix} 1&1&1 \\ 0&1&2 \\ 0&0&0 \end{pmatrix} \\ $
My problem is that I don't know how to continue. In the cases I studied before, I could always transform a matrix B with elementary row operations to this form: $E_{k}\cdot ... \cdot E_{1}\cdot B=I_{n}$ , in which $E_{i}$ is an random elementary matrix and $I_{n}$ the identity matrix. The next step I always took was $B=(E_{1})^{-1}\cdot ... \cdot(E_{k})^{-1} $ So I multiplied the inverse of every used elementary matrix from the left and then I only needed to compute the matrix multiplication.
But how can I write A as product of elementary matrices if rk(A)<3?
PS: I'm not used to write about math in english, please ask if something doesn't makes sense to you.
You will end up with the identity matrix if and only if the original matrix was invertible. Yours was not :-)
To get to the reduced row echelon form, you would also need to multiply by $L_{2,1}(-1)$ in your notation, i.e. subtract row 2 from row 1, transforming row 1 to $(1,0,-1)$.
The end product would be simpler, but not an identity matrix still.