(Easier) way to determine determinant of a almost-circulant matrix

Question

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.

Answer 1

There are (at least) a few options:

1. For this particular matrix, row-reduction (to produce an upper-triangular matrix) is efficient.
2. This is a rank-$2$ update of a circulant matrix, so one can apply the Matrix Determinant Lemma.
3. Because of the special form of the matrix, one can use the special block matrix identity $$\det \pmatrix{A&B\\B&A} = \det(A + B) \det(A - B) .$$

Answer 2

Substract line two from line four and develop by first column (factoring out $\;x+y\;$ from last row):

$$\begin{vmatrix}-x+y&-1&0&-1\\-1&x+y&-1&0\\0&-1&-x+y&-1\\0&-x-y&0&x+y\end{vmatrix}=(-x+y)(x+y)\begin{vmatrix}x+y&-1&0\\-1&-x+y&-1\\-1&0&1\end{vmatrix}+$$

$$+(x+y)\begin{vmatrix}-1&0&-1\\-1&-x+y&-1\\-1&0&1\end{vmatrix}=$$$${}$$

$$=(y^2-x^2)\left(y^2-x^2-2\right)+(x+y)\left(2x-2y\right)=(y^2-x^2)(y^2-x^2-2)-2(y^2-x^2)=$$

$$=(y^2-x^2)(y^2-x^2-4)$$