Compute the rank and nullity of the $\,3\times 3$ matrix over $\Bbb Z^3$ 
Compute the rank and nullity of the following matrix over $\mathbb{Z}^3$
$$ \begin{pmatrix} 1&1&2 \\ 2&1&2 \\ 2&0&0 \end{pmatrix} $$

Okay, by reducing the matrix, I get:
$$ \begin{pmatrix} 1&0&0 \\ 0&1&2 \\ 0&0&0 \end{pmatrix} $$
Great, so easily it is rank $=2$ and nullity $=1$. 
I feel that my answer is wrong, that question seemed to fast to complete and the fact that it was in $\mathbb{Z}^3$ was irrelevant (when I solved it) makes me worried that I did this problem wrong. So, I am hoping just for some advice, did I do the above problem incorrectly?
Thanks

I'm assuming that the question refers to a matrix over ${\Bbb Z}_3$, the integers modulo $3$, not over $Z^3$, which makes no sense as far as I know.
To reduce row $2$ you could for example add row $1$. giving
$$2+1=3=0\ ,\quad 1+1=2\ ,\quad 2+2=4=1\ .$$
Continuing in the same way gives your result.
It is true that in this case if you reduce using normal arithmetic (that is,over ${\Bbb R}$ or ${\Bbb Q}$) you get the same result.  For a more interesting example you could try
$$\pmatrix{2&1&2\cr 1&1&2\cr 0&2&1\cr}\ .$$
This has rank $2$ over ${\Bbb Z}_3$ but rank $3$ over ${\Bbb R}$.
