The matrix corresponds to a homomorphism $\mathbb{Z^2} → \mathbb{Z^3}$ In Aluffi's Algebra: Chapter $0$ in 4.3. Reading a presentation it says:

For example, take $M=\begin{pmatrix}1&3\\2&3\\5&9\end{pmatrix}$; this matrix corresponds to a homomorphism $\mathbb{Z^2} → \mathbb{Z^3}$, hence to a $\mathbb Z$-module.



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*What does it mean it corresponds to a homomorphism $\mathbb{Z^2} → \mathbb{Z^3}$? For example, does it mean  $\begin{pmatrix}1&3\\2&3\\5&9\end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}a'\\b'\\c'\end{pmatrix}$ holds for any $a,b,a',b',c'$?

*What does it mean it corresponds to  a $\mathbb Z$-module? For example, in an $R$-module $M$ we have a map $R \times M \to M$ so how $\mathbb{Z^2} → \mathbb{Z^3}$ instead of $\mathbb{Z^2} → \mathbb{Z^2}$ if it is a $\mathbb Z$-module?     
 A: 
  
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*What does it mean it corresponds to a homomorphism $\mathbb{Z^2} → \mathbb{Z^3}$? For example, does it mean  $\begin{pmatrix}1&3\\2&3\\5&9\end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}a'\\b'\\c'\end{pmatrix}$ holds for any $a,b,a',b',c'$?
  

That equation as written doesn't make any sense (it certainly won't be true for arbitrary $a,b,a',b',c'$).  But it's the right idea: what it means is the homomorphism which sends $(a,b)\in\mathbb{Z}^2$ to the $(a',b',c')\in\mathbb{Z}$ such that your equation is true.


  
*What does it mean it corresponds to  a $\mathbb Z$-module? For example, in an R-module M we have a map $R \times M \to M$ so how $\mathbb{Z^2} → \mathbb{Z^3}$ instead of $\mathbb{Z^2} → \mathbb{Z^2}$ if it is a $\mathbb Z$-module? 
  

The corresponding $\mathbb{Z}$-module is the $\mathbb{Z}$-module presented by this map, which is the quotient of $\mathbb{Z}^3$ by the image of the given homomorphism.  The idea is that you constructing a module by "generators and relations": the generators are the three standard basis vectors $e_1,e_2,e_3$ of $\mathbb{Z}^3$ (which freely generate $\mathbb{Z}^3$ as a $\mathbb{Z}$-module), and then impose relations which say $e_1+2e_2+5e_3=0$ and $3e_1+3e_2+9e_3=0$.  Imposing these relations corresponds exactly to modding out the submodule that $e_1+2e_2+5e_3$ and $3e_1+3e_2+9e_3$ generate, which is the image of the homomorphism $\mathbb{Z}^2\to\mathbb{Z}^3$ given by the matrix $\begin{pmatrix}1&3\\2&3\\5&9\end{pmatrix}$ (since $e_1+2e_2+5e_3$ and $3e_1+3e_2+9e_3$ correspond to the columns of the matrix).
A: In linear algebra, our $3 \times 2$ matrix corresponds to the map
$$
\pmatrix{a\\b} \mapsto \pmatrix{1&3\\2&3\\5&9} \pmatrix{a\\b} = \pmatrix{a + 3b\\2a + 3b\\5a+9b}
$$
This map is linear when $a,b$ are taken to be field elements, which is to say that a matrix of this shape traditionally represents a module homomorphism from $\Bbb F^2$ to $\Bbb F^3$.  Similarly, when we take $a,b$ to be elements of $\Bbb Z$, we end up with a module homomorphism from $\Bbb Z^2$ to $\Bbb Z^3$.
I'm not sure about the answer to your second question, but I suspect that somebody else might.  The full excerpt is as follows:

this matrix corresponds to a homomorphism $\Bbb Z^2 → \Bbb Z^3$, hence to a $\Bbb Z$-module; that is, a finitely generated abelian group $G$. The reader should figure out what $G$ is more explicitly (in terms of the classification of §IV.6, cf. Exercise 2.19) before reading on. (Aluffi, p. 345)

The referenced exercise:

Re-prove Corollary IV.6.5 as a corollary of Proposition 2.11 [viz. classification of finite abelian groups]. In fact, prove the more general fact that every finitely generated abelian group is a direct sum of cyclic groups. (Aluffi, p. 327)

