Show that $M=\frac{\mathbb{Z}^3}{\langle (3,3,1),(2,2,2)\rangle}$ is isomorphic to $\mathbb{Z} \oplus \mathbb{Z}_4$ as $\mathbb{Z}$-modules 
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 $(0,1)$.
 A: 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
$$\{0\}\oplus\mathbb{Z}_4\oplus\mathbb{Z}.$$ 
A: $\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 .
$$
