This is a variant of another MO question. Consider the matrix $$M_n:=\begin{bmatrix}c_1& a & b&a& \ddots & a \\ b & c_2 & a& b&\ddots & b\\ a & b & c_3&a & \ddots & a \\b& a & b & c_4 & \ddots & \ddots\\ \ddots & \ddots & \ddots & \ddots & \ddots &a \\ b & a & b & \ddots &b& c_n \end{bmatrix}_{n\times n}$$ (This is for even $n$; for odd $n$, the NE corner will be $b$ and the SW $a$.)

I couldn't resist to choose the title as is, even though of course the stripes don't run the American way (because we want matrices of full rank...) but rather along the diagonals, consisting alternatingly of $a$'s or of $b$'s.

Question 1.Is there still a nice closed formula for the determinant $\det(M_n)$?

Question 2.Is there still a nice expression for the inverse $M_n^{-1}$?

Things do not fall as nicely into place as in the original question. If $v $ denotes the all-1-vector, the matrices $M_n-av^Tv$ and $M_n-bv^Tv$ still have half of their off-diagonal entries vanishing; but their inverses don't. Also for $n\ge4$, the determinant is no more symmetric in all the $c_i$, but at least in $c_i\leftrightarrow c_{n+1-i}$. Even in the case of all $c_i$ equal, i.e. a special kind of Toeplitz matrix, this seems tricky for general $n$.