Can the determinant of an integer matrix with a given row be any multiple of the gcd of that row?

Solution 1:

The conjecture appears true:

For any $a_1,\ldots,a_n\in\mathbb Z$, the determinant of a matrix with them in its first row can be any multiple of their $\gcd$.

It clearly suffices to handle the case in which their $\gcd$ is $1$. I found a recursive procedure that seems to work. For $n=2$ we take $$\begin{pmatrix}a_1&a_2\\-b_2&b_1\end{pmatrix}$$ where $b_1a_1+b_2a_2=1$. Now let $n\geq3$ and suppose we've constructed $M$ such that $$\left|\begin{array}{c}a_2&\cdots&a_n\\\hline\\ &\large M\end{array}\right|=\gcd(a_2,\ldots,a_n)=a.$$

Imagine expanding this determinant along the first row. What we get is a linear combination of $a_2,\ldots,a_n$ with the $n-2\times n-2$ minors of $M$, say $m_2,\ldots,m_n$, as coefficients. Note that this implies $\gcd(m_2,\ldots,m_n)=1$, so there exist $c_2,\ldots,c_n\in\mathbb Z$ such that $$\left|\begin{array}{c}c_2&\cdots&c_n\\\hline\\ &\large M\end{array}\right|=1.$$ Because $\gcd(a_1,a)=1$ we have $xa_1+ya=1$ for some $x,y\in\mathbb Z$. Now observe that $$\left|\begin{array}{c|ccc}a_1&a_2&\cdots&a_n\\ -y&xc_1&\cdots&xc_n\\\hline 0\\ \vdots&&M\\ 0\end{array}\right|=a_1\cdot\left|\begin{array}{c}xc_2&\cdots&xc_n\\\hline\\ &\large M\end{array}\right|+y\cdot\left|\begin{array}{c}a_2&\cdots&a_n\\\hline\\ &\large M\end{array}\right|=1.$$

Note: the obtained matrix will look like $$\begin{pmatrix}a_1&a_2&a_3&\cdots&a_{n-1}&a_n\\ *&*&*&\cdots&*&*\\ 0&*&*&\cdots&*&*\\ 0&0&*&\cdots&*&*\\ \vdots&\ddots&\ddots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&*&*\end{pmatrix},$$ which is called an upper Hessenberg matrix.