determinant of specific circulant matrices I got problem in determining the determinant of specific circulant matrix $C$ formed by shifting the vector $1\cdots101\cdots10\cdots0$. The number of $1$'s in the first sequence of $1$'s is $k$ and the one of the second sequence of $1$'s is $k+1$. These two sequences of $1$'s are separated by exactly one $0$. The number of $0$'s at the end is arbitrary. My conjecture is that the determinant of $C$ is always odd. I do wish some one would like to help me.  Thank you.  
 A: This is only an answer in the case where there are no zeros at the end (i.e., one zero in each row and column). Perhaps you can extend the method.
In this case, let $C=(c_{i,j})$ be the matrix in question. Of course $C$ is $n\times n$ where $n=2k+2$. We want to show that $\det(C)\equiv 1 \pmod 2$.
By definition, $$\det(C) \equiv \sum_{\sigma\in S_n} c_{1,\sigma(1)}c_{2,\sigma(2)}\cdots c_{n,\sigma(n)} \pmod 2.$$
Note each summand is $0$ or $1$ of course, and the summand corresponding to permutation $\sigma$ is $0$ if and only if at least one of the $c_{i,\sigma(i)}=0$. Let $\Omega$ be the set of such $\sigma$. We'll use inclusion-exclusion to count $|\Omega|$. 
Let $D$ be the coordinates of the entries of $C$ which are zero (so $|D|=n$). Of course, for any $k$ coordinates $(i_1,j_1),\ldots,(i_k,j_k)\in D$, there are $(n-k)!$ permutations $\sigma\in S_n$ with $\sigma(i_\ell)=j_\ell$ for all $\ell=1,\ldots,k$. It follows that \begin{eqnarray*} |\Omega| &=& n(n-1)! - {n\choose 2}(n-2)!+{n\choose 3}(n-3)!-\cdots + (-1)^n {n \choose n}(n-n)!\\&=&n!-\frac{n!}{2!}+\frac{n!}{3!}-\cdots + (-1)^n.\end{eqnarray*}
But $n=2k+2$ is even, so for $1\leq p <n$, $\frac{n!}{p!}$ is even, and thus $|\Omega|$ is odd. Therefore, $n!-|\Omega|$ is odd and we see that in $\det(C)$ there are an odd number of $1$'s, and so $\det(C)\equiv 1 \pmod 2$ as desired.
