I am considering an interesting question, given an $\{0,1\}$-matrix $M \in \{0,1\}^{n\times p}$, we define that an k-operation to the matrix is changing value of $M$ to be $\tilde{M}$ such that $$ \tilde{M}_{ij}=\begin{cases} 0, & i \in\{1,\ldots,n\}\text{,and } j \in\{l\in\{1,\ldots,n\}|M_{kl}=1\} \\ M_{ij} ,& \text{otherwise.} \\ \end{cases} $$ We renew $M$ with $\tilde{M}$, operation finished.
A k-operation is essentially putting all columns that has the $k$th row element 1 to be $\boldsymbol{0}$.
The question is how to pick minimum number of the k s to do operations such that $M$ becomes the zero matrix (which is the matrix with all entries $0$)?
My idea is each step picking the $k$ which flips most number of $1$s to become $0$ in a greedy way. However, I could not prove this, could someone please help me with it?
What you describe is the set cover problem, and greedy is not optimal.