Consider the symmetric block matrix

\[M = \begin{pmatrix} A & C \\ C^T & B \\ \end{pmatrix}\]

with $A = A^T$ and $B = B^T$ and $A, B$ are invertible.

We want to calculate the inverse:

\[M^{-1} = \begin{pmatrix} E & G \\ G^T & F \\ \end{pmatrix}\]

Note that the inverse of an invertible symmetric matrix is symmetric. This is used in the general form for $M^{-1}$ above. Quick proof: $A = A^T$. So (multiplying on the left and right by $A^{-1}$), $A^{-1}A = A^{-1}A^T = I$ and $AA^{-1} = A^T A^{-1} = I$. By uniquess of inverses, $(A^T)^{-1} = A^{-1}$. Since $(B^T)^{-1} = (B^{-1})^T$ for any general invertible $B$, $(A^{-1})^T = A^{-1}$ i.e. $A^{-1}$ is symmetric.

\[\begin{pmatrix} A & C \\ C^T & B \\ \end{pmatrix} \begin{pmatrix} E & G \\ G^T & F \\ \end{pmatrix} = \begin{pmatrix} I_1 & 0 \\ 0 & I_2 \\ \end{pmatrix}\]

where the two identity matrices will be of different sizes generally to be consistent with the block structure. This results in 4 identities for the three unknowns: $E, F, G$.

\[\begin{equation} AE + CG^T = I_1 \\ C^T G + BF = I_2 \\ AG + CF = 0 \\ C^T E + B G^T = 0 \\ \end{equation}\]

\(AG + CF = 0\): Solve for $G$ to get $G = -A^{-1}CF$.

\(C^T G + BF = I_2\):

Substituting for $G$, we get

\[-C^TA^{-1}CF + BF = I_2\] \[\implies (B-C^TA^{-1}C)F = I_2\]

to get

\[\boxed{F = (B-C^TA^{-1}C)^{-1}}\]

Using $G = -A^{-1}CF$, we get

\[\boxed{G = -A^{-1}C(B-C^TA^{-1}C)^{-1}}\]

\(AE + CG^T = I_1\):

Solving for $E$,

\[E = A^{-1}(I_1 - CG^T) = A^{-1}(I_1 - C[-A^{-1}C(B-C^TA^{-1}C)^{-1}]^T)\]

Simplifying, $\boxed{E = A^{-1}(I_1 + C[B-C^TA^{-1}C]^{-1}C^TA^{-1})}$

where we have freely used the symmetry of $A, B$ and the identity $(A^T)^{-1} = (A^{-1})^T$.

So,

\[\boxed{M^{-1} = \begin{pmatrix} A^{-1}(I_1 + C[B-C^TA^{-1}C]^{-1}C^TA^{-1}) & -A^{-1}C(B-C^TA^{-1}C)^{-1} \\ -(B-C^TA^{-1}C)^{-1}C^TA^{-1} & (B-C^TA^{-1}C)^{-1} \\ \end{pmatrix}}\]

Sanity Check 1:

If $M$ is block diagonal ($C = 0$), we get,

\[M^{-1} = \begin{pmatrix} A^{-1}(I_1 + C[B-C^TA^{-1}C]^{-1}C^TA^{-1}) & -A^{-1}C(B-C^TA^{-1}C)^{-1} \\ -(B-C^TA^{-1}C)^{-1}C^TA^{-1} & (B-C^TA^{-1}C)^{-1} \\ \end{pmatrix} = \begin{pmatrix} A^{-1} & 0 \\ 0 & B^{-1} \\ \end{pmatrix}\]

as expected.

Sanity Check 2:

We used three equations to derive $M^{-1}$. Check to see our solution is consistent with the fourth equation:

\[C^T E + B G^T = 0\] \[\begin{equation} \begin{split} LHS &= C^T E + B G^T \\ &= C^T A^{-1}(I_1 + C[B-C^TA^{-1}C]^{-1}C^TA^{-1}) + B [-A^{-1}C(B-C^TA^{-1}C)^{-1}]^T \\ &= C^T A^{-1} + C^T A^{-1}C[B-C^TA^{-1}C]^{-1}C^TA^{-1} - B(B-C^TA^{-1}C)^{-1}]C^TA^{-1}\\ &= C^T A^{-1} + [C^T A^{-1} C - B][B-C^TA^{-1}C]^{-1}C^TA^{-1}\\ &= C^T A^{-1} - C^T A^{-1} \\ &= 0 \\ &= RHS \end{split} \end{equation}\]