integers with greatest common divisor $1$ fit in a row of invertible matrix

Question

If $a,b$ are integers with greatest common divisor $1$, then $[a,b]$ sit in a row of an invertible matrix over $\mathbb{Z}$. This is easy to see: we can write $ma+nb=1$ for some integers $m,n$. Consider the matrix with one row $[a,b]$ and other row $[-n,m]$.

Q. If $a_1,a_2,\ldots,a_n$ are integers with greatest common divisor $1$, then do they sit in a row of an $n\times n$ invertible matrix over $\mathbb{Z}$? Is this also valid if $\mathbb{Z}$ is replaced by principal ideal domain?

Answer 1

Yes, this is true. Keith Conrad has a nice blur about primitive vectors. In general, the statement is true for any Dedekind domain, which is a more general class of ring than PID.

In fact, a more general statement is true: If you have $r$ vectors of length $n$ with $r \leq n$, then they form the first $r$ rows of an invertible $n \times n$ matrix if and only if the greatest common divisor of all of the $r \times r$ minors is $1$.