Problem with Smith normal form over a PID that is not an Euclidean domain This is an homework exercise of the Algebra lecture.  

I need to evaluate the Smith normal form of the following matrix 
  $$A:=\begin{pmatrix}1 & -\xi & \xi-1\\2 \xi&8&8\xi+7\\\xi& 4 & 3\xi +2 \end{pmatrix} \in M(3\times 3;\Bbb{Z}[\xi]),$$
  where $\xi := \frac{1+\sqrt{-19}}{2}$. 

We have seen in the lecture, that similarly as for rings, there exists the Smith Normal Form also for PID's. To solve the problem (we don't need to find $S, T \in M(3 \times 3; \Bbb{Z}[\xi])$, such that $SAT$ is in Smith Normal Form) we applied row, and column operations, paying particular attention to not multiply rows and columns by nonunits, since this is not allowed. (Also according to this answer). The steps that we performed are:
\begin{align*}\begin{pmatrix}1 & -\xi & \xi-1\\2 \xi&8&8\xi+7\\\xi& 4 & 3\xi +2 \end{pmatrix} &\overset{2\text{column}+\xi \text{ first column}}{\leadsto}
\begin{pmatrix}1 & 0 & \xi-1\\2 \xi&8+2\xi^2&8\xi+7\\\xi& 4+\xi^2 & 3\xi +2 \end{pmatrix}\\ \overset{3\text{column}-(\xi-1) 
\text{ first column}}{\leadsto} \begin{pmatrix}1 & 0 & 0\\2 \xi&8+2\xi^2&-2\xi^2+10\xi+7\\\xi& 4+\xi^2 & -\xi^2+4\xi+2 \end{pmatrix}
&\overset{2\text{row}-(2\xi) 
\text{ first row}}{\leadsto}
\begin{pmatrix}1 & 0 & 0\\0&8+2\xi^2&-2\xi^2+10\xi+7\\\xi& 4+\xi^2 & -\xi^2+4\xi+2 \end{pmatrix}\\
\overset{3\text{row}-(\xi) 
\text{ first row}}{\leadsto}
\begin{pmatrix}1 & 0 & 0\\0&8+2\xi^2&-2\xi^2+10\xi+7\\0& 4+\xi^2 & -\xi^2+4\xi+2 \end{pmatrix}& 
\overset{2\text{row}-2 
\text{ third row}}{\leadsto}
\begin{pmatrix}1 & 0 & 0\\0&0&2\xi+3\\0& 4+\xi^2 & -\xi^2+4\xi+2 \end{pmatrix}\\
\overset{\text{swap columns}}{\leadsto}
\begin{pmatrix}1 & 0 & 0\\0&2\xi +3&0\\0& -\xi^2+4\xi+2 & 4+\xi^2 \end{pmatrix} &
\\
\overset{\text{second column $+$ third column}}{\leadsto}
\begin{pmatrix}1 & 0 & 0\\0&2\xi +3&0\\0& 4 \xi +6 & 4+\xi^2 \end{pmatrix}
& \overset{\text{third row } - 2 \text{ second row}}{\leadsto} \begin{pmatrix}1 & 0 & 0\\0&2\xi +3&0\\0& 0 & 4+\xi^2 \end{pmatrix}=: B
\end{align*} 
Unfortunately the term $b_{33}$ is not divisible by the term $b_{22}$ as it should be. We have done the steps many many times, but we can't find the error (I'm doing this homework exercise with my classmates). Do you maybe find it? Thank you in advance for any help.

$\xi^2=\xi-5$, and therefore $4+\xi^2=\xi-1$ (which is a prime element). On the other side, $2\xi+3=2(\xi-1)+5=2(\xi-1)+\xi-\xi^2=-(\xi-1)(\xi-2)$, so $b_{33}\mid b_{22}$. 
The SNF of your matrix is $$\begin{pmatrix}1 & 0 & 0\\0&\xi-1&0\\0& 0 & (\xi-1)(\xi-2) \end{pmatrix}.$$
