Determinant of matrix defined by set partition Let $n$ be a positive integer and $|S| = n$. Let $S_1,S_2,\cdots, S_{2^n-1}$ be all nonempty subsets of $S$, numbered arbitrarily. 
Let $A$ be a $2^n-1 \times 2^n-1$ matrix given by $a_{ij}= 1$ if $S_i \cap S_j  \neq \varnothing$ and $a_{ji}=0$ if $S_i\cap S_j = \varnothing$.
Calculate $\det A$.
I think that the determinant is not depend on the way we number the subsets, but I haven't done it yet.
Thanks for your help.
 A: I believe the determinant is always $-1$. Here's an argument showing that the determinant is either $\pm1$. We repeatedly use the fact that adding a multiple of one row of a matrix to another row doesn't change the matrix's determinant.
The row corresponding to $\{1,\dots,n\}$ has all $1$s. There are $n$ rows corresponding to the $(n-1)$-element subsets, each of which has a single $0$; subtract the all-$1$ row from each of these, resulting in $n$ rows with a single $-1$.
Now find the $\binom n2$ rows corresponding to the $(n-2)$-element subsets. Each of these has three $0$s; subtract the all-$1$ row from each of these, resulting in rows with three $-1$s. Two of these $-1$s are in the columns corresponding to the single-element subsets, and each of these has a corresponding row with a $-1$ in that column and $0$s elsewhere. By adding these rows appropriately, we obtain $\binom n2$ more rows with a single $-1$ and $0$s elsewhere.
Now do the same with the $\binom n3$ rows corresponding to the $(n-2)$-element subsets, each of which has seven $0$s that turn into seven $-1$s after subtracting the all-$1$ row; six of these are in columns that can be killed by the single-$-1$ rows already constructed. And keep going in this way.
At the end, there are $2^n-2$ rows with a single $-1$, and the original all-$1$ row; add all the $2^n-2$ rows to the final row, and the result is a permutation matrix up to signs.
