# How to compute determinant (or eigenvalues) of this matrix?

Let us have the $n \times n$ circulant matrix given by

$$C(c_0,c_1,\cdots, c_{n-1}) =\begin{bmatrix} c_0 & c_1 & c_2 &\cdots & c_{n-1}\\ c_{n-1} & c_0 & c_1 &\cdots & c_{n-2}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ c_1 & c_2 & c_3 & \cdots & c_0 \end{bmatrix}.$$ Also we have a $n \times n$ matrix

$$\Delta = \begin{bmatrix} x & a & a & a& \cdots &a\\ b & x & a & a & \cdots & a\\ b & b & x &a & \cdots & a\\ \vdots &\vdots &\vdots &\vdots &\ddots &\vdots \\ b & b& b& b & \cdots & x \end{bmatrix};$$ i.e $\Delta$ is a matrix, where diagonal entries are $x$, each entry of the upper triangular matrix is $a$ and each element of lower triangular matrix is $b$.

Now I want to compute the determinant of the matrix

$$S= \begin{bmatrix} C(c_0,c_1,\cdots, c_{n-1}) & \Delta \\ \Delta^T & C(c_0',c_1',\cdots, c_{n-1}') \end{bmatrix},$$ where $\Delta^T$ denotes the transpose.

If $\Delta$ were also a circulant matrix, then the problem is easy. Circulant matrices of same orders commute. hence we can use the formula $\det\begin{bmatrix} A & B \\ C& D\end{bmatrix} = \det(AD -BC)$. However, I am looking for the case when these are non commutative.

Is there a method to exploit the symmetry and determine the determinant? Also, can we compute the eigenvalues of this matrix $S$? Please help. By the way, none of the parameters is zero.