Consider the quotient $\mathbb{Z}-\text{module}$
$$M=\frac{\mathbb{Z}^3}{\langle (3,3,1),(2,2,2)\rangle}$$
Prove that $M$ is isomorphic to $\mathbb{Z} \oplus \mathbb{Z}_4$.
I think the first step is to find a basis of $M$. Clearly $(1,0,0)$ and $(0,0,1)$ are in a basis of $M$. I don't know how to find the third element in that basis.
Also the basis in $\mathbb{Z} \oplus \mathbb{Z}_4$ are $(1,0)$ and $(0,1)$.
On
The standard technique is to apply ($\mathbb{Z}$-invertible) elementary row and column operations to the matrix $$\left[\begin{array}{ccc}3&3&1\\2&2&2\end{array}\right].$$ You will get $$\left[\begin{array}{ccc}1&0&0\\0&4&0\end{array}\right],$$ which means there is a basis of $\mathbb{Z}^3$ such that the module you are factoring out is generated by $(1,0,0)$ and $(0,4,0)$; hence the quotient is $$\{0\}\oplus\mathbb{Z}_4\oplus\mathbb{Z}.$$
$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$$\newcommand{\Z}{\mathbb{Z}}$You should apply the standard algorithm to the matrix $$ \begin{bmatrix} 3 & 3 & 1\\ 2 & 2 & 2\\ 0 & 0 & 0\\ \end{bmatrix}. $$ First exchange column 1 and 2 to get $$ \begin{bmatrix} 1 & 3 & 3\\ 2 & 2 & 2\\ 0 & 0 & 0\\ \end{bmatrix} $$ then subtract from row 2 twice row 1 to get $$ \begin{bmatrix} 1 & 3 & 3\\ 0 & -4 & -4\\ 0 & 0 & 0\\ \end{bmatrix} $$ then subtract from column 2 thrice column 1 and from column 3 thrice column 1 to get $$ \begin{bmatrix} 1 & 0 & 0\\ 0 & -4 & -4\\ 0 & 0 & 0\\ \end{bmatrix} $$ and finally subtract from column 3 column 2, and multiply row 2 by $-1$ to get $$ \begin{bmatrix} 1 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 0\\ \end{bmatrix} $$ So the quotient is isomorphic to $$ \Z / \Span{1} \oplus \Z / \Span{4} \oplus \Z / \Span{0} \cong \Z / \Span{4} \oplus \Z . $$