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Let $n >1$ be an integer , then is it true that for any integer $k$ , there exist a matrix $A \in M(n,\mathbb Z)$ with first row of $A$ as $(1,2,...,n)$ such that $\det A=k$ ?

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  • $\begingroup$ Yes. That's basically because the gcd of $1,2,\ldots,n$ is $1$. $\endgroup$ – Ewan Delanoy May 16 '16 at 7:59
  • $\begingroup$ @EwanDelanoy : could you please elaborate ... $\endgroup$ – user228169 May 16 '16 at 8:01
  • $\begingroup$ @user228169 See e.g. here: math.stackexchange.com/questions/1411324 $\endgroup$ – punctured dusk May 25 '16 at 8:47
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Take the transpose of $$ \left(\begin{array}{ccccccc} 1 & -1 & 0 & 0 & \ldots & 0 & 0 \\ 2 & -1 & 0 & 0 & \ldots & 0 & 0\\ 3 & 0 & 1 & 0 & \ldots & 0 & 0 \\ 4 & 0 & 0 & 1 & \ldots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ n-1 & 0 & 0 & 0 & \ldots & 1 & 0\\ n & 0 & 0 & 0 & \ldots & 0 & k\\ \end{array}\right) $$

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    $\begingroup$ Now I see your answer ... how about taking an upper triangular matrix with first row as desired and the diagonal entries all $1$ except the $n$ th diagonal entry being $k$ ? $\endgroup$ – user228169 May 16 '16 at 8:08
  • $\begingroup$ @user228169 this is exactly what I'm saying, I just initially wrote my matrix with first column as desired instead of first row, so I siad "take the transpose". Note that the matrix is not completely triangular $\endgroup$ – Ewan Delanoy May 16 '16 at 8:11
  • $\begingroup$ yeah , your matrix is not triangular . But the way I am saying , we can make an upper triangular matrix right ? And the determinant is just the product of diagonal entries ... $\endgroup$ – user228169 May 16 '16 at 8:14
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    $\begingroup$ @user228169 Indeed. it is a little simpler $\endgroup$ – Ewan Delanoy May 16 '16 at 8:14
  • $\begingroup$ Thank you :) all those $0$ s in your construction triggered the idea .. $\endgroup$ – user228169 May 16 '16 at 8:17

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