A question on finitely generated modules over Z in matrix form in my class on module theory I have been given this problem on finitely generated $ \mathbb{Z} $ modules (Abelian groups) stating the following:

We define the vectors $ v_1 = (1,0,-1) $ $ v_2 = (2,-3,1) $ $ v_3 = (0,2,1) $ 
  $ v_3 = (3,1,5) $. We are to show that these vectors span the whole of $ \mathbb{Z}^3 $ and also that any subset of them consisting of three vectors does not span $ \mathbb{Z}^3 $

Now I know that there is the brute force solution with systems of linear equations and taking all 4 possible subsets consisting of three elements but I figured it has something to do with matrix form, but unfortunately I cannot do this, so I am kindly asking for help in a solution to this with some explanation, I thank all helpers
 A: I miscopied, I put a $3$ where it should be $-3.$ I will leave this here.
There is an error. Take the matrix $R$ with the first three vectors as columns. This has determinant $1.$ If we have a column vector $A$ of three integers and want to represent it as $RX$ with $X$ an unknown column of integers, we just need $X = R^{-1}A.$
If $X$ is the column vector with entries $x,y,z,$ then the vector (as a column) corresponding to $x v_1 + y v_2 + z v_3$ is the $RX$ I mentioned. You should check that by hand.
$$
R =
\left(
\begin{array}{ccc}
1 & 2 & 0 \\
0 & 3 & 2 \\
-1 &   -1 & 1
\end{array}
\right)
$$
$$
R^{-1} =
\left(
\begin{array}{ccc}
5 & -2 & 4 \\
-2 & 1 & -2 \\
3 &   -1 & 3
\end{array}
\right)
$$
This topic is in "integer lattices," which are positive definite quadratic forms with all integer coefficients. 
A: Corrected, I hope.
$$
R =
\left(
\begin{array}{ccc}
1 & 2 & 0 \\
0 & -3 & 2 \\
-1 &   -1 & 1
\end{array}
\right)
$$
All we really need to know is that the determinant is not $\pm 1,$ it is actually $-5.$
$$
R^{-1} =
\frac{1}{5}
\left(
\begin{array}{ccc}
1 & 2 & -4 \\
2 & -1 & 2 \\
3 &   1 & 3
\end{array}
\right)
$$
Is it possible to represent $(1,0,0)$ as $x v_1 + y v_2 + z v_3?$ No, because we see from $R^{-1}$ that $x = 1/5, y = 2/5, z= 3/5.$ That is, $(1,0,0)$ is not in the span of the first three given vectors. 
Do the same for the other three triples. Then, for the four triples, you get a rectangular matrix, use Gaussian elimination to show that there is an integer quadruple to represent any target vector $(a,b,c).$ ADDENDUM: no, only represent $(a,b,c)$ with $a+b+c \equiv 0 \pmod 3.$ Some error, not mine this time.
