Problem on the determinant of a Matrix Suppose we have $n-1$ linearly independent vectors  $a_1, \ldots,a_{n-1} \in \mathbb{Z^n}$. Is it possible to find another vector $a_n\in \mathbb{Z^n}$ such
that the determinant of the matrix $M$ whose row vectors are $a_1,\ldots,a_n$ is
1? 
 A: Let $\rm M_k$ be the matrix formed by cutting out the $\rm k$th column of the $\rm a_i$ row vectors and putting them together. Then the condition that $\rm \det M=1$ imposed on $\rm \vec{v}=a_n$ is equivalent to
$$\rm (\det M_1)\,v_1-(\det M_2)\,v_2+\cdots\pm (\det M_n)\,v_n=1.$$
By Bezout's lemma, this has a solution in $\rm \vec{v}$ if and only if $\rm \gcd(\det M_1,\cdots,\det M_n)=1$.
This is not always the case. Take e.g. in $\rm n=3$, with $\rm v,w\ne0$ and $\rm u\ne0,\pm1$,
$$\rm a_1^T=\begin{pmatrix}u \\ 0 \\ \rm v\end{pmatrix} \quad and \quad a_2^T=\begin{pmatrix} 0 \\ \rm u \\ \rm w \end{pmatrix}$$
where we have $\rm \gcd(\det M_1, \det M_2,\det M_3)=\gcd(-uv,uw,u^2)\ne 1$. More generally, for arbitrary $\rm n$ following you's suggestion, let $\rm a_i=x_i\,\vec{e}_i$ with $\rm \gcd(x_1,\cdots,x_n)\ne1$. The resulting matrix will be diagonal, so its determinant is the product of the diagonal entries, which must be $>1$ in magnitude if any of the $\rm x_i$ are.
A: Hint $a_1=(2,2,2,..., 2)$.
