# For integers $n>1$ , $k$ , does there exist matrix $A$ with integer entries and first row $(1,2,…,n)$ such that $\det A=k$?

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 $\det A=k$ ?

• Yes. That's basically because the gcd of $1,2,\ldots,n$ is $1$. – Ewan Delanoy May 16 '16 at 7:59
• @EwanDelanoy : could you please elaborate ... – user228169 May 16 '16 at 8:01
• @user228169 See e.g. here: math.stackexchange.com/questions/1411324 – punctured dusk May 25 '16 at 8:47

Take the transpose of $$\left(\begin{array}{ccccccc} 1 & -1 & 0 & 0 & \ldots & 0 & 0 \\ 2 & -1 & 0 & 0 & \ldots & 0 & 0\\ 3 & 0 & 1 & 0 & \ldots & 0 & 0 \\ 4 & 0 & 0 & 1 & \ldots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ n-1 & 0 & 0 & 0 & \ldots & 1 & 0\\ n & 0 & 0 & 0 & \ldots & 0 & k\\ \end{array}\right)$$
• 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 $k$ ? – user228169 May 16 '16 at 8:08
• Thank you :) all those $0$ s in your construction triggered the idea .. – user228169 May 16 '16 at 8:17