A condition for a matrix to be invertible 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

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.

