The maximal number of $1$s in a $n$ by $n$ matix 
In a matix $A=a_{ij_{1\le i,j\le n}}$ (where $n$ is a postive integer), we put ones and zeroes. If there is a unique permutation $\sigma$ of $\{1,2,...,n\}$ such that: $a_{1\sigma(1)}=\dots =a_{n\sigma(n)}=1$, then what is the maximal number of $1$s in $A$.

This is fairly easy, of course permuting the rows, does nothing to our problem. So, WLOG, that permutation is the identity. Now, let us consider the pairs $(a_{ij},a_{ji})_{1\le i \neq j \le n}$ , each pair contains at most one $1$, otherwise $a_{kk}=a_{ij}=a_{ji}=1$ $\forall k \in \{1,\dots,n\}-\{i,j\}$, and we have found another permutation. Thus the number of $1$s cannot exceed $n+\frac{n(n-1)}2=\frac{n(n+1)}2$, and one can easily construct a matrix with that maximal number.
My question is: the condition implies clearly $|\det A|=1$. But if we were just given that $|\det A|=1$, would we be able to find such maximal number?
 A: Set $E=(1)_{1\leq i,j \leq n}$, $D=\text{diag}((1)_{1\leq i\leq n-1},0)$ and $M=E-D$.
$M$ has $n^2-n+1$ entries $1$. If it had one entry $1$ more, by the pigeon hole principle we had two identical rows (having all entries $1$) and such a matrix would have determinant $0$. Thus the number of $1$s in $M$ is maximal.
Since elementary row operations (those used in Gaußian elimination) do not change the determinant, we can substract the last row (which is all $1$s) from each of the other rows in turn. We get a matrix $N$ which is almost a diagonal matrix featuring only $-1$s, exept for the last row which is all $1$s. From this we get $\text{det}(M)=(-1)^{n-1}$.
Short answer: Yes.
Long answer: Yes, we can find such a maximal number and it is $n^2-n+1$.
Remark: Trivially the answer is: yes. Fixing $n$, there are only a finite number of $0$-$1$-matrices, among them the identity matrix $I_n$ with $\text{det}(I_n)=1$. The set of matrices of given property is finite (a subset of the finite set of all $0$-$1$-matrices) and non-empty (contains $I_n$), thus the maximum is attained. The interesting question is: What is that number? Which is actually answered above.
