Confusion about the quotient $\mathbb Z^4/H$ Let $f:\mathbb Z^3\to\mathbb Z^4$ be the group homomorphism given by $$f(a,b,c)=(a+b+c,a+3b+c,a+b+5c,4a+8b).$$ Let $H$ be the image of $f$. Find an element of infinite order in $\mathbb Z^4 /H$ and find the order of the torsion subgroup of that quotient.
This problem certainly has to do with presentation matrices, but I'm too confused. Let $A$ denote the matrix with rows $(1,1,1),(1,3,1),(1,1,5),(4,8,0)$. Then $A\mathbb Z^3=H$. So $A$ must be a presentation matrix for $\mathbb Z^4/H$ according to the definition from here (or from here, both of them agree with Artin's definition). So in the quotient module the relations $$v_1+v_2+v_3=0,\\v_1+3v_2+v_3=0,\\v_1+v_2+5v_3=0,\\ 4v_1+8v_2=0$$ hold. But then according to Artin's text, the matrix should be $3\times 4$, not $4\times 3$ (see Example 14.5.2 in this question). So what is the size of a presentation matrix in this case? My size doesn't agree with Artin's notation, even though I use his definitions and conventions.

Example 14.5.2 The $\mathbb{Z}-$module or an abelian group $V$ that is generated by three elements $v_1, v_2, v_3$ with the compete
  set of relations 
$$ 3v_1+2v_2+v_3=0\\ 8v_1+4v_2+2v_3=0\\ 7v_1+6v_2+2v_3=0\\
 9v_1+6v_2+v_3=0 $$
is presented by the matrix
$$
         A=\begin{bmatrix}
         3 & 8 & 7 & 9 \\
         2 & 4 & 6 & 6 \\
         1 & 2 & 2 & 1 \\
          \end{bmatrix}. $$
Its columns are the coefficients of the (above) relations: $(v_1, v_2,
 v_3)A=(0, 0, 0, 0)$.

 A: As I mentioned in the comments, conventionally, the shape of the matrix ought to be $4\times 3$ as you've worked out. So we have
$$A=\newcommand\bmat{\begin{pmatrix}}\newcommand\emat{\end{pmatrix}}\bmat 1 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 5 \\ 4 & 8 & 0 \emat.$$
Now row operations are isomorphisms of $\Bbb{Z}^3$, so we can freely row reduce this matrix and preserve the image of $A\Bbb{Z}^3$ as a subset of $\newcommand\ZZ{\Bbb{Z}}\ZZ^4$. I.e. applying row operations to a matrix $A$ that presents $\ZZ^4/A\ZZ^3$ to get a matrix $A'$ means that $A'$ also presents $\ZZ^4/A\ZZ^3$.
Subtracting the first row from the others to make the other entries of the first column zero, we have
$$A'=\bmat 1 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \\ 0 & 4 & -4 \emat.$$
Then subtracting twice the second row and adding the third row to the fourth row, we get 
$$A''=\bmat 1 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \\ 0 & 0 & 0 \emat.$$
Thus $$v_4=\bmat 0 \\ 0 \\ 0 \\ 1 \emat$$ is never in the image of $A$, even over $\Bbb{Q}$, hence $v_4$ has infinite order in the cokernel.
Finally, if we don't care about the coordinates on $\ZZ^4$, column operations on $A''$ correspond to applying automorphisms of $\ZZ^4$, and hence preserve the isomorphism class of $\ZZ^4/A''\ZZ^3$. However, if we subtract the first column from the second and third, we get
$$A''' = \bmat 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \\ 0 & 0 & 0 \emat.$$
Then since $A'''$ is diagonal, $\ZZ^4/A'''\ZZ^3$ splits as a direct sum, $$\ZZ/\ZZ \oplus \ZZ/2\ZZ \oplus \ZZ/4\ZZ\oplus \ZZ/(0),$$
so the torsion subgroup of $\ZZ^4/A\ZZ^3$ has size 8.
