Making non-singular matrices singular What is the minimum value of $k$ such that every non-singular $n\times n$ real matrices can be made singular by switching EXACTLY $k$ entries  with ZERO ? 
 A: You can always make an $n \times n$ matrix singular by making $n$ entries ( all in one row or one column) zero.  For a matrix $(a_{ij})$ whose entries are 
algebraically independent,  it is impossible to do with fewer than $n$.  You can prove (e.g. using Hall's marriage theorem) that if you make fewer than $n$ 
elements zero there will still be a permutation $\pi$ of $[1,\ldots,n]$
such that all $a_{i,\pi(i)}$ are unaffected, so the corresponding term in the
determinant will still be there and can't be cancelled by the other terms.
EDIT: This is, of course, over a field that has $n^2$ algebraically independent elements.  You can do it over any infinite field, with a bit more work, by constructing inductively a sequence $c_j$ of nonzero elements such that for each $n$, all terms of
the following form are distinct and nonzero: $\sum_{i=1}^k \prod_{j \in S_i} c_j$ where $S_1, \ldots, S_k$ are distinct subsets of $\{1, \ldots, n\}$, $1 \le k \le 2^n$.  On the other hand, it doesn't work over finite fields.  Thus over $GF(2)$ you can always
make a nonsingular matrix singular by changing one element to $0$.
