TITLE: Can every nonsingular $n\times n$ matrix with real entries be made singular by changing exactly one entry?
QUESTION [29 upvotes]: I was just thinking about this problem:

Can every nonsingular $n\times n$ matrix with real entries be made singular by changing exactly one entry?

Thanks for helping me.

REPLY [13 votes]: Here's a fast solution. The adjugate matrix $\def\adj{\operatorname{adj}}\adj A$ satifies (more or less by definition)
$$
  \adj A\cdot A=\det(A)I_n,
$$
and the right hand side is nonzero by the nonsingular hypothesis. Now looking at what contributes the diagonal entry $\det(A)$ at position $k,k$ on the right, one sees that of row $k$ of $\adj A$, which I will denote $(c_1~\ldots~c_n)$ and column $k$ of $A$, which I will denote $(x_1~\ldots~x_n)^t$, produce a matrix product $c_1x_1+\cdots+c_nx_n=\det(A)$. Clearly $c_ix_i\neq0$ for at least one $i$. Choose such $i$, and form a matrix $A'$ by replacing the entry $x_i$ of $A$ (which is at position $(i,k)$) by $x_i-\frac{\det(A)}{c_i}$. Row $k$ of $\adj A'$ is still $(c_1~\ldots~c_n)$ (since it does not depend on column $k$ of $A$), and the $(k,k)$ entry of $\adj A'\cdot A'$ equals $c_1x_1+\cdots+c_nx_n-c_i\frac{\det(A)}{c_i}=0$. But it is also equal to $\det(A')$, so $A'$ is singular.
This shows that not only the answer is "yes, you can make any real invertible matrix singular by changing a single entry", you can even specify in advance the column in which an entry must change. Or the row, by transposition of the argument (but not both). And it works over any field.