Inverting a $3\times 3$ block matrix Suppose that $a$ and $b$ below are scalars, $F$ a square matrix, $v$ a column vector. I'm trying to invert the matrix $M$ of the form
$$
M=\begin{pmatrix}
a & v' & 0\\
v & F  & 0\\
0 & 0  & b
\end{pmatrix}
$$
When $v$ and $F$ are actually scalars, we can use $M^{-1}=\frac{1}{\det M}\text{adj}(M)$ but how does one handle the general case above please?
 A: Let us define the matrix
$$
X = \begin{pmatrix} c & w^T & 0\\ w & G & 0 \\ 0 & 0 &b^{-1}\end{pmatrix},
$$
that will be our candidate for $M^{-1}$.
Then,
$$
M X = 
\begin{pmatrix}
ac + v^Tw & aw^T + v^TG & 0\\
cv + Fw & vw^T + FG & 0 \\
0 & 0 & 1.
\end{pmatrix}
$$
In order to have $MX = I$, we solve the linear system
$$
\tag{1}
\begin{cases}
ac + v^Tw = 1,\\
aw^T + v^TG = 0,\\
cv +  Fw = 0,\\
vw^T + FG = I.
\end{cases}
$$
This can be solved by substitution, provided $F$ invertible and $a-v^TF^{-1}v \neq 0$. Indeed we get
$$
\begin{cases}
c = (a-v^TF^{-1}v)^{-1},\\
w = -c F^{-1}v,\\
G = F^{-1}(I-vw^T),
\end{cases}
$$
using only the $1^{st}$, $3^{rd}$ and $4^{th}$ equations in $(1)$. If you plug the obtained expressions of $b,w$ and $G$ in the $2^{nd}$ equation, you get an additional hypothesis for the invertibility of $M$.
A: By a block inversion formula, we can write
$$
M^{-1}=\begin{pmatrix}N & 0 \\ 0 & b^{-1}
\end{pmatrix}
\quad\text{with}\quad N=\begin{pmatrix}a & v' \\ v & F\end{pmatrix}^{-1}.
$$
Using the block inversion formula once more, we have
$$
N=\begin{pmatrix}
(a-v'F^{-1}v)^{-1}              & -v'(aF-vv')^{-1}\\
-F^{-1}v(a-v'F^{-1}v)^{-1}      & a(aF-vv')^{-1}
\end{pmatrix}\cdot
$$
In sum,
$$
M^{-1}=\begin{pmatrix}
(a-v'F^{-1}v)^{-1}              & -v'(aF-vv')^{-1} & 0\\
-F^{-1}v(a-v'F^{-1}v)^{-1}      & a(aF-vv')^{-1}   & 0\\
0                               & 0                & b^{-1}
\end{pmatrix}\cdot
$$
One can verify that
$$
MM^{-1}=\begin{pmatrix}1&0&0\\0&I&0\\0&0&1\end{pmatrix}\cdot
$$
And because $M$ and $M^{-1}$ as defined above are square matrices, we infer that $MM^{-1}=I$ implies $M^{-1}M=I$. The verification that $M^{-1}$ is symmetric can be done along the line shown here.
