(Easier) way to determine determinant of a almost-circulant matrix Consider the matrix:
$$\begin{pmatrix} \color{red}{-x+y} &-1&0&-1 \\ -1&\color{blue}{x+y}&-1&0 
\\ 0&-1&\color{red}{-x+y}&-1 \\ -1&0&-1&\color{blue}{x+y} \end{pmatrix}$$
Is there any easy way to calculate the determinant of this matrix?
I have read up on circulant matrices and there is some nice theory on them, however this is not quite circulant, the only difference is that the main diagonal contains 2 different values. 
So I was wondering if there is a 'short cut' to find the determinant of these kind of almost-circulant matrices.
Thanks.

Substract line two from line four and develop by first column (factoring out $\;x+y\;$ from last row):
$$\begin{vmatrix}-x+y&-1&0&-1\\-1&x+y&-1&0\\0&-1&-x+y&-1\\0&-x-y&0&x+y\end{vmatrix}=(-x+y)(x+y)\begin{vmatrix}x+y&-1&0\\-1&-x+y&-1\\-1&0&1\end{vmatrix}+$$
$$+(x+y)\begin{vmatrix}-1&0&-1\\-1&-x+y&-1\\-1&0&1\end{vmatrix}=$$$${}$$
$$=(y^2-x^2)\left(y^2-x^2-2\right)+(x+y)\left(2x-2y\right)=(y^2-x^2)(y^2-x^2-2)-2(y^2-x^2)=$$
$$=(y^2-x^2)(y^2-x^2-4)$$
