When every term of the determinant is zero The determinant of $A$ is defined as 
$$ \det(A) = \sum \pm a_{1,i_1} a_{2,i_2} \cdots a_{n,i_n}$$
Suppose $A$ is a real matrix such that every term in the above sum
is zero. Is it true that $A$ has a zero row or zero column?
 A: Counterexamples:
$A=\begin{pmatrix}1&0&1\\0&1&0\\0&1&0\end{pmatrix}$ or 
$B=\begin{pmatrix}1&1&1\\0&1&0\\0&1&0\end{pmatrix}$.
If you look at matrix $B$:


*

*every product of the above form contains an element from the first column

*if it contains $0$, we are fine

*if it contains $1$ from the position $(1,1)$ then the rest of the product must be some of the summands in the subdeterminant $\begin{vmatrix}1&0\\1&0\end{vmatrix}$.

A: Building upon the answer of @MartinSleziak, I saw a more general pattern emerge. I'm not sure yet if my observed method will yield a necessary and sufficient condition, but at least it is sufficient.
So let $A$ be an $n \times n$ matrix, such that there are $k$ columns $a_i, i \in I$, with the property that:
$\quad \forall j \in J: a_{i,j} = 0$
(with $I,J \subseteq \{1, \ldots, n\}, \#(I) = k, \#(J) = n-k+1$).
Now let $\sigma \in S_n$, so that $\displaystyle \prod_{l = 1}^n a_{l, \sigma(l)}$ is a term of $\det A$ (disregarding sign as it will become zero anyway). 
Consider the set $\{\sigma(i): i \in I\} \subseteq \{1, \ldots, n\}$. It has $k$ elements, so since $J$ has $n-k+1$ elements, there is necessarily an $i \in I$ such that $\sigma(i) \in J$.
By the above assumption on $J$, it follows that $a_{i, \sigma(i)} = 0$, and we conclude:
$\quad \displaystyle \prod_{l = 1}^n a_{l, \sigma(l)} = 0$
Since $\sigma$ was arbitrary, all terms of $\det A$ are zero. 
Swapping $k$ for $n-k+1$ proves the same argument for rows. Probably, there is a connection with non-vanishing minors, as was already pointed out.
