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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? –  Davide Giraudo Jun 24 '12 at 15:12
And what is a "null minor"? Minor I know, and even principal or chief minor, but null minor I never heard of. –  DonAntonio Jun 24 '12 at 15:15
@DonAntonio I understand the assumption as: among all $(n-1)\times (n-1)$ submatrices, exactly $k$ have zero determinant. –  user31373 Jun 24 '12 at 15:30
@Don Antonio: Leonid Kovalev is right. I`m sorry for my enghlish. I`ve searched in the dictionary and "null" means equal to 0. –  Claudiu Mindrila Jun 24 '12 at 15:32
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1 Answer

I assume the olympiad is over and it's legitimate for us to discuss this problem. Anyway, I hide the answer as to not spoil the fun.

Let's use the adjugate matrix. We know that exactly $k$ entries of $\mathrm{adj}\,A$ are zero. Suppose that $\det A=0$; then the product of $A$ and $\mathrm{adj}\,A$ must be the zero matrix. Now, the rank of $A$ is exactly $n-1$, which implies that the kernel of $\mathrm{adj}\,A$ is $(n-1)$-dimensional. In other words, $\mathrm{adj}\,A$ has rank $1$. But any rank 1 matrix is of the form $u\otimes v$, making it impossible for it to have exactly $k$ zero entries.

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Yes, the olympiad is over (in fact I`ve recieved this problem in the contest). –  Claudiu Mindrila Jun 24 '12 at 19:50
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