In a LP problem Ax = b, how to solve for A instead of x?
Let $a_{ij}$ represent member of matrix $A$. This can either be the constant $0$ if you want $a_{ij}$ to be $0$ in the solution matrix, or a variable for a linear programming problem. Then you want to add linear constraints:
\begin{alignat}{2}
&amp;\!\min_{a}        &amp;\qquad&amp; \sum_{i} s_i\\
&amp;\text{subject to} &amp;      &amp; \forall j \left(\sum_{i}a_{ij} \leq 1 + \epsilon \right),\\
&amp;                  &amp;      &amp; \forall j \left(\sum_{i}a_{ij} \geq 1 - \epsilon \right),\\
&amp;                  &amp;      &amp; \forall_i \left(\sum_{j}a_{ij}x_j \leq s_i\right),\\
&amp;                  &amp;      &amp; \forall_i \left(\sum_{j}a_{ij}x_j \geq -s_i\right).
\end{alignat}
This finds the optimal matrix $A$, with the desired elements set to $0$, that has columns adding up to $1$ (within a certain precision) and minimizes $\left\lVert Ax\right\rVert_1$.