Question concerning the group over GL$(n,\mathbb{Z})$ 
Is every vector $[a_1,a_2,\dots, a_n]$ with $\gcd(a_1,a_2,\dots,a_n)=1$ a column in some matrix $A\in  GL(n,\mathbb{Z})$?

I don't think this is a duplicate:
Let me rephrase this questions using the notation and solution to the thread Does there exist such an invertible matrix? 

Given a PID $A$ and $a_1\in A^n$, when does there exists $a_2,\dots, a_n\in A^n$ such that $$A a_1\oplus \cdots \oplus A a_n= A^n\ ?$$ 

By the solution in the thread  "Does there exist such an invertible matrix?", given the existence of $a_2,\cdots,a_n$, this implies the existence of a matrix $B\in  GL(n,\mathbb{Z})$ with $a_1$ being the first column.  But given $a_1$ there won't always exists such $a_2,\cdots,a_n$.  For instance, if $A=\mathbb{Z}$, then if there is such $a_2,\cdots,a_n$, it is clear that $a_1$ would have to satisfy $a_1= a_{1j}e_1 +\cdots + a_{1n} e_n$ with $\gcd(a_{11},\cdots, a_{1n})=1$. I am trying to prove the converse.  Perhaps this is more trivial, but this is what I am not sure how to prove.
 A: By replicating the Euclidean algorithm, we can find a matrix $P\in GL(n,\mathbb Z)$, which is the product of a permutation matrix and of transvection (shear) matrices corresponding to elementary row operations, such that
$$
P\begin{bmatrix}a_1\\a_2\\\vdots \\a_n\end{bmatrix} =
\begin{bmatrix}1\\0\\\vdots \\0\end{bmatrix}
$$
Now, convince yourself that the matrix $P^{-1} \in GL(n,\mathbb Z)$ answers the problem.

Example: the operations
$$
\begin{bmatrix}15\\6\\10\end{bmatrix} \xrightarrow{\substack{L_1 := L_1 - 2L_2\\L_3 := L_3 - L_2}}
\begin{bmatrix}3\\6\\4\end{bmatrix} \xrightarrow{\substack{L_2 := L_2 - 2L_1\\L_3 := L_3 - L_1}}
\begin{bmatrix}3\\0\\1\end{bmatrix} \xrightarrow{\substack{L_1 := L_1 - 3L_3}}
\begin{bmatrix}0\\0\\1\end{bmatrix} 
$$
lead to
$$
P = \begin{bmatrix}0 & 0& 1\\0 & 1 & 0\\1 & 0 & 0\end{bmatrix} \begin{bmatrix}1 & 0& -3\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}
\begin{bmatrix}1 & 0& 0\\-2 & 1 & 0\\-1 & 0 & 1\end{bmatrix}
\begin{bmatrix}1 & -2 & 0\\0 & 1 & 0\\0 & -1 & 1\end{bmatrix}
$$
and finally
$$
P^{-1} = \begin{bmatrix}15 & 2 & 5\\6 & 1 & 2\\10 & 1 & 3\end{bmatrix}
$$
