Inverse of this block matrix

Question

I have three square real matrices $A, B$ and $C$ of the same order, say $n$. I know that $A+B$ and $C$ are invertible. Then I built a new $nN \times nN$ big block matrix as follows:
$$M = \begin{pmatrix}
A+B & C & C &\cdots & C \\
C & A+B & C & \cdots & C \\
C & C & A+B & \cdots & C \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
C & C & C & \cdots & A+B
\end{pmatrix}
$$
Is there an expression for the inverse $M^{-1}$? I tried toeplitz inverse and also block inverse but both attempts did not work.

Answer 1

Let $X = A + B - C$. Suppose we know a priori that $X$ is invertible. The matrix can be expressed in the form $$ M = I_N \otimes X + (ee^T) \otimes C $$ where $\otimes$ denotes a Kronecker product and $e \in \Bbb R^N$ is the vector $e = (1,\dots,1)^T$. With the Woodbury matrix identity, we can express $M^{-1}$ in the form $$ M^{-1} = I_N \otimes X^{-1} - (e \otimes X^{-1}C)(I_n + N \cdot X^{-1}C)^{-1}(e^T \otimes X^{-1}). $$ This requires only the computation of an $n\times n$ inverse