Let $n\in \Bbb N$ and $\mathcal{M}_n(\Bbb R)$ be the set of the square real-valued matrices.
Find all $A \in \mathcal{M}_n(\Bbb R)$ such as $|\det(A)| = \prod_{i=1}^n\left( \sum_{j=1}^n |a_{i,j}| \right)$.
What I have tried:
By definition:
$\begin{align*} \lvert \det A \rvert = \lvert \sum_{\sigma \in S_n} \varepsilon(\sigma) \prod_{i=1}^n a_{i,\sigma(i)} \rvert & \leq \sum_{\sigma \in S_n} \prod_{i=1}^n \lvert a_{i, \sigma(i)} \rvert = \prod_{i=1}^n \sum_{i=1}^n \lvert a_{ij} \rvert \end{align*}$
We have equality in the previous inequality if, for all $\sigma \in S_n$ such that $\varepsilon(\sigma) = -1$, we have, for all $i \in [[1, n]], a_{i, \sigma(i)} = 0$.
But I cannot find an explicit form of such matrices, I guess it have to do with Hadamard inequality, but unsure.
Your intuition is right. Clearly, if $A$ is singular, your inequality holds iff $A$ has a zero row. Suppose $A$ is nonsingular. Hadamard's inequality for complex matrices says that $|\det A|\le\prod_i\|a_{i\ast}\|_2$. So, if your inequality holds for some nonsingular $A$, we will have $\prod_i\|a_{i\ast}\|_1\le\prod_i\|a_{i\ast}\|_2$. However, as the $2$-norm is dominated by the $1$-norm, we must have $\|a_{i\ast}\|_1=\|a_{i\ast}\|_2$ for every $i$ and in turn $a_{i\ast}$ has at most one nonzero entry. As $A$ is nonsingular, every row of $A$ must have exactly one nonzero entry, and in turn $A$ has exactly one entry on every column too. This means $A$ is the product of a nonsingular diagonal matrix and a permutation matrix.