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Given $a_1,a_2,\dots,a_n \in \mathbb{Z}$ with $\gcd(a_1,a_2,\dots,a_n)=1$, is it possible to find $M \in GL_n(\mathbb{Z})$ with first row $(a_1,a_2,\dots,a_n)$?

Clearly for $n=2$ the result is true. For $a,b \in \mathbb{Z}$ with $\gcd(a,b)=1$, there exists $x,y \in \mathbb{Z}$ such that $ax+by=1$. Then the required matrix is $ M= \left[ {\begin{array}{cc} a & b \\ -y & x \\ \end{array} } \right] $. But I am not being able to extend it to general case. Even the converse of this result is true. For $M \in GL_n(\mathbb{Z})$ we can prove that every row and column has gcd 1.

user26857
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snehashis mukherjee
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  • See also https://math.stackexchange.com/questions/76645/can-any-set-of-n-relatively-prime-elements-be-extended-to-an-invertible-matrix where it is noted that there's a relation to the Quillen-Suslin Theorem, aka Serre's Conjecture. – Gerry Myerson Apr 29 '19 at 03:41

1 Answers1

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yes. By a finite sequence of elementary, integer matrices (determinant one), we can eventually write the first row as $(1,0,0,0,...,0).$ Call the original row $R,$ we now have $RG = (1,0,0,...,0).$ It follows that the top row of $G^{-1}$ is the original row, while both $G$ and $G^{-1}$ have integer entries

Will Jagy
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