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Let $n,\ k$ be two positive integers such that $n \ge 2$ and $1 \le k \le n-1$. If the matrix $A\in \mathcal{M}_n(\mathbb{C})$ has exactly $k$ null minors of order $n-1$, then $\det A \neq 0$.

source: Romanian Mathematical Olympiad, final round , 2012

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    What do you mean by minor?2012-06-24
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    And what is a "null minor"? Minor I know, and even principal or chief minor, but null minor I never heard of.2012-06-24
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    @DonAntonio I understand the assumption as: among all $(n-1)\times (n-1)$ submatrices, exactly $k$ have zero determinant.2012-06-24
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    @Don Antonio: Leonid Kovalev is right. I`m sorry for my enghlish. I`ve searched in the dictionary and "null" means equal to 0.2012-06-24

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