I have the following problem of linear algebra:
Let $A=(a_{ij})_{i,j=\overline{1,n}}$ be a $r\times r$ matrix with non-negative integer entries with $\det A = \pm 1$. Suppose that for any $j\neq k$ there exists $i$ such that $a_{ij}>0$ and $a_{ik}=0$. Prove that $A$ is a permutation matrix.
I was able to prove the cases $n=2$ and $n=3$.
In the case $n=2$: Since $1\neq 2$, there exists $i$ such that $a_{i1}>0$ and $a_{i2}=0$. Also, since $2\neq 1$, there exists $i$ such that $a_{i2}>0$ and $a_{i1}=0$. We have two possible cases:
(a) $a_{11}>0, a_{12}=0, a_{21}=0, a_{22}>0$ or (b) $a_{21}>0,a_{22}=0, a_{12}>0,a_{11}=0$.
Since $\det(A)=\pm 1$, it follows that (a) $a_{11}=a_{22}=1$ or (b) $a_{12}=a_{21}=1$. Hence $A$ is a permutation matrix.
In the case $n=3$, I used a similar argument, but more elaborated. However, I have no idea how to generalize it. Can anyone give me a clu or a reference? Thank you in advance!
I have a relatively neat method for proving the result for 3 x 3 matrices if you are interested. Applied to 4 x 4 matrices I think it gives a counterexample:-
$\pmatrix{2&0&1&0\\ 0&1&0&1\\ 1&1&0&0\\0&0&1&1}$