Is there a name for this particular type of matrix? Consider the following matrix structure:
$$
M = \begin{pmatrix}
a & d & c & c \\
d & a & b & b \\
c & b & a & e \\
c & b & e & a
\end{pmatrix}
$$
It is a real, symmetric matrix.
I'm looking into physical properties (spin correlations) of a model that are described by a matrix of this structure.
Qn: Does anyone recognize this matrix as any special type of matrix?

It has the following (weird) property:
Consider its inverse, $M^{-1}$. Its elements are extremely complicated, for example the (3,3) element is:
$$
(M^{-1})_{3,3} = \frac{a \left(a^{2} - d^{2}\right) \left(\left(a^{2} - c^{2}\right) \left(a^{2} - d^{2}\right) - \left(a b - c d\right)^{2}\right)}{\left(\left(a^{2} - c^{2}\right) \left(a^{2} - d^{2}\right) - \left(a b - c d\right)^{2}\right)^{2} - \left(\left(a^{2} - d^{2}\right) \left(a e - c^{2}\right) - \left(a b - c d\right)^{2}\right)^{2}}
$$
and the (3,4) element is:
$$
(M^{-1})_{3,4} = - \frac{a \left(a^{2} - d^{2}\right) \left(\left(a^{2} - d^{2}\right) \left(a e - c^{2}\right) - \left(a b - c d\right)^{2}\right)}{\left(\left(a^{2} - c^{2}\right) \left(a^{2} - d^{2}\right) - \left(a b - c d\right)^{2}\right)^{2} - \left(\left(a^{2} - d^{2}\right) \left(a e - c^{2}\right) - \left(a b - c d\right)^{2}\right)^{2}}
$$
but their difference is extremely simple:
$$
(M^{-1})_{3,3} - (M^{-1})_{3,4} = \boxed{\frac{1}{a-e}}
$$
You can use Mathematica etc. to verify.
But I realized, this is just the same as considering the (3,4) subblock:
$$
M' = \begin{pmatrix} a & e\\e & a \end{pmatrix}
$$
and its inverse is trivial:
$$
(M')^{-1} = \frac{1}{(a+e)(a-e)} \begin{pmatrix} a & -e\\-e & a \end{pmatrix}
$$
And the difference between the diagonal and off-diagonal elements:
$$
(M'^{-1})_{1,1} - (M'^{-1})_{1,2} = \boxed{\frac{1}{a-e}}
$$
is exactly the same as before!
Is this just a coincidence or is there some justification to just considering a subblock $M'$ instead of the full $M$ when taking the inverse?
 A: I don't know the physics, but the phenomenon you described can be explained mathematically as follows. Suppose $M$ is an invertible matrix of the form $\pmatrix{X&pv^T\\ uq^T&Z}$, where $X,Z$ are invertible matrices of the same sizes and $p,q,u,v$ are some vectors. Then $M^{-1}$ is in the form of $\pmatrix{\ast&\ast\\ \ast&S^{-1}}$, where
$$
S=Z-(uq^T)X^{-1}(pv^T)=Z-\beta uv^T
$$
is the Schur complement of $Z$ in $X$ and $\beta\ (=q^TX^{-1}p)$ is some scalar. Since $S$ is a rank-$1$ modifiction of $Z$, by Sherman-Morrison formula, $S^{-1}$ is also a rank-$1$ modification of $Z^{-1}$, with
$$
S^{-1}
=Z^{-1}+\frac{\beta Z^{-1}uv^TZ^{-1}}{1+\beta v^TZ^{-1}u}
=Z^{-1}+\gamma Z^{-1}uv^TZ^{-1}\tag{1}
$$
for some scalar $\gamma\ (=\beta/(1+\beta v^TZ^{-1}u))$.
In your case, $Z=\pmatrix{a&e\\ e&a}$ and $u=v=\pmatrix{1\\ 1}$. Therefore all elements of $Z^{-1}uv^TZ^{-1}$ are identical to each other. It follows from $(1)$ that $S^{-1}$ is obtained by shifting the entries of $Z^{-1}$ by the same constant. Therefore $
(M^{-1})_{33}-(M^{-1})_{34}=(S^{-1})_{11}-(S^{-1})_{12}=(Z^{-1})_{11}-(Z^{-1})_{12}$.
