Primary Decomposition of group $G=\mathbb{Z}^{3}/N$ where $N\leq G$

Question

I'm still trying to get the hang of this whole primary decomposition thing. In our notes, we were given the following worked out example:

Let $G=\mathbb{Z}^{3}/N$ where $N$ is a subgroup of the free abelian group $\mathbb{Z}^{3}$ generated by two elements $(3, 3, 3)$ and $(4,5,6)$ expressed in some basis $B$ of $\mathbb{Z}^{3}$. We start with writing the coordinate matrix expressing the generating set of $N$ in the basis of $B$:

$M=\begin{pmatrix} 3&3&3\\ 4&5&6\end{pmatrix}$

Then we perform a sequence of transformations of $M$ from among the following:

1. Permutations of rows and columns of $M$

2. Changing signs of rows and columns of $M$

3. Adding multiples of rows and columns of $M$

The goal of the first step is to make the entry $r_{11}$ of $M=(r_{in})$ the greatest common divisor of all entries of $M$ and gen make zero all other entries in the first row and column of $M$. Then we perform recursively a similar transformation with the submatrix of $M$ obtained by removing the first row and first column.

First Step: $\begin{pmatrix} 3&3&3 \\ 4&5&6\end{pmatrix}\, \implies \begin{pmatrix} 3&3&3\\ 1&2&3\end{pmatrix} \, \implies \, \begin{pmatrix} 0& -3& -6\\ 1&2&3 \end{pmatrix} \, \implies \, \begin{pmatrix} 1&2&3 \\ 0&-3&-6\end{pmatrix} \, \implies \, \begin{pmatrix} 1&0&3\\ 0&-3&-6\end{pmatrix}\,\implies \, \begin{pmatrix} 1&0&0 \\ 0&-3&-6\end{pmatrix}$

Second Step:$\begin{pmatrix} 1&0&0 \\ 0&-3&-6\end{pmatrix}\, \implies \, \begin{pmatrix} 1&0&0\\ 0& 3&-6\end{pmatrix} \, \implies \, \begin{pmatrix} 1&0&0\\ 0&3&0\end{pmatrix}$

From the final matrix, we see that $N$ is generated by elements $(1,0,0)$ and $(0,3,0)$ written in some basis for $\mathbb{Z}^{3}$. Hence, $G\simeq \mathbb{Z}^{3}/N\simeq (\mathbb{Z}/\mathbb{Z})\oplus (\mathbb{Z}/3\mathbb{Z}) \oplus \mathbb{Z}\simeq \mathbb{Z}_{3}\oplus \mathbb{Z}$ is the primary composition of $G$.

I tried to work out two exercises based on this example but because the numbers seemed a little weird, I wanted to post this here and ask 1) if I did it correctly, and 2) if I didn't do it correctly, based on the example I included, how would I fix what I did?

1. $N$ is generated by the set $X=\{(5,8,5), (8,5,8)\}$

First Step:

$M= \begin{pmatrix} 5&8&5\\8&5&8\end{pmatrix}\, \implies \, \begin{pmatrix} 5&8&5\\ 3&-3&3\end{pmatrix}\, \implies \, \begin{pmatrix} 3&-3&3 \\5&8&5\end{pmatrix}\, \implies \, \begin{pmatrix} 3&-3&3\\ 2&11&2\end{pmatrix}\, \implies\, \begin{pmatrix} 1 &-14&1\\ 2&11&2\end{pmatrix}\, \implies \, \begin{pmatrix} 1&-14&1\\ 0&39&0\end{pmatrix}$

Second Step: $\begin{pmatrix} 1&-14&1\\0&39&0\end{pmatrix}\,\implies\, \begin{pmatrix}1&0&1\\ 0&39&0\end{pmatrix}\, \implies \, \begin{pmatrix} 1&0&0\\ 0&39&0\end{pmatrix}$

This final matrix shows that $N$ is generated by elements $(1,0,0)$ and $(0,39,0)$. Written in some basis for $\mathbb{Z}^{3}$. Hence, $G\simeq \mathbb{Z}^{3}/N\simeq (\mathbb{Z}/\mathbb{Z})\oplus (\mathbb{Z}/39\mathbb{Z})\oplus \mathbb{Z}\simeq \mathbb{Z}_{39}\oplus \mathbb{Z}$ is the primary decomposition of $G$. (The 39 seems weird to me).

2. $N$ is generated by the set $X=\{(4,4,2), (16,20,0)\}$

Here, the first step and second steps kind of melded together; if there's a better way to do this where the steps are separated out, please let me know.

$M=\begin{pmatrix}4&4&2\\16&20&0\end{pmatrix}\, \implies \, \begin{pmatrix} 2&4&4\\0&20&16\end{pmatrix}\, \implies \, \begin{pmatrix}2&0&4\\0&4&16\end{pmatrix}\, \implies \, \begin{pmatrix}2&0&0\\0&4&16\end{pmatrix}\, \implies \, \begin{pmatrix} 2&0&0\\0&4&0\end{pmatrix}$

The final matrix shows that $N$ is generated by elements $(2,0,0)$ and $(0,4,0)$ written in some basis for $\mathbb{Z}^{3}$. Hence, $G\simeq \mathbb{Z}^{3}/N\simeq (\mathbb{Z}/2\mathbb{Z})\oplus (\mathbb{Z}/4\mathbb{Z})\oplus \mathbb{Z}\simeq \mathbb{Z}_{2}\oplus \mathbb{Z}_{4}\oplus \mathbb{Z}$ is the primary decomposition of $G$.

Thank you ahead of time for your help! :)