Inverse of a matrix! Let $I_n$ be the $n$ by $n$ identity matrix and $b$ and $c$ be two vectors in ${\mathbb R}^n$ such that $b^Tc\ne 0$.
Then one can easily see that the $n+1$ by $n+1$ matrix $X$ defined as following
$$
X =
\begin{bmatrix}
I_n&b\\c^T& 0
\end{bmatrix}
$$
has an inverse matrix $X^{-1}$. Is the any close and simple formula for $X^{-1}$?
 A: Following the link in the comment, we have
$$
X^{-1} = 
\frac{1}{c^Tb}\pmatrix{
(c^Tb)I - bc^T & b\\
c^T & -1}
$$
A: Wiki does not show how to get the solution. Here I give a simple way to prove it using the idea of row operations to get the inverse
$$
(X,I_{n+1})\longrightarrow(I_{n+1}, X^{-1})
$$
First multiply a left matrix whose role is similar to elementary row operation of adding the 1st row multiplied by $-c^T$ to the 2nd row. So we have
$$
\pmatrix{I_n&0\\-c^T&1}\pmatrix{I_n&b&I_n&0\\c^T&0&0&1}=\pmatrix{I_n&b&I_n&0\\0&-c^Tb&-c^T&1}
$$
Then multiply a left matrix whose role is similar to elementary row operation of adding the 2nd row multiplied by $b/(c^Tb)$ to 2nd row. So we get
$$
\pmatrix{I_n&b/(c^Tb)\\0&1}\pmatrix{I_n&b&I_n&0\\0&-c^Tb&-c^T&1}=\pmatrix{I_n&0&I_n-bc^T/(c^Tb)&b/(c^Tb)\\0&-c^Tb&-c^T&1}
$$
Finally multiply a left matrix whose role is similar to elementary row operation of multiplying $-1/(c^Tb)$ to the 2nd row. So we have
$$
\pmatrix{I_n&0\\0&-1/(c^Tb)}\pmatrix{I_n&0&I_n-bc^T/(c^Tb)&b/(c^Tb)\\0&-c^Tb&-c^T&1}=\pmatrix{I_n&0&I_n-bc^T/(c^Tb)&b/(c^Tb)\\0&1&c^T/(c^Tb)&-1/(c^Tb)}
$$
So we have
$$
X^{-1}=\frac1{(c^Tb)}\pmatrix{(c^Tb)I_n-bc^T&b\\c^T&-1}
$$
A: Let's look for an inverse of the form
$$
\begin{bmatrix}
A & x \\
y^T & z
\end{bmatrix}
$$
and do block multiplication:
$$
\begin{bmatrix}
I_n & b \\
c^T & 0
\end{bmatrix}
\begin{bmatrix}
A & x \\
y^T & z
\end{bmatrix}
=
\begin{bmatrix}
I_nA+by^T & I_nx+bz \\
c^TA & c^Tx
\end{bmatrix}
$$
Then we get
$$
\begin{cases}
A+by^T = I_n\\[3px]
x+bz = 0 \\[3px]
c^TA = 0 \\[3px]
c^Tx = 1
\end{cases}
$$
The first gives $A=I_n-by^T$, so $c^T-c^Tby^T=0$ and therefore $y^T=d^{-1}c^T$ (where $d=c^Tb=b^Tc\ne0$, by hypothesis).
This also determines $A=I-b(d^{-1})c^T=I-d^{-1}bc^T$.
Now multiply the second equation by $c^T$, so $c^Tx+c^Tbz=0$, or $1+dz=0$, so $z=-d^{-1}$ and $x=-d^{-1}b$. Thus
$$
X^{-1}=
\begin{bmatrix}
I-d^{-1}bc^T & -d^{-1}b \\
d^{-1}c^T & -d^{-1}
\end{bmatrix}
$$
Note also that the condition $c^Tb\ne0$ is necessary for invertibility, because
$$
\begin{bmatrix}
I_n & b \\
c^T & 0
\end{bmatrix}
\begin{bmatrix}
b\\
-1
\end{bmatrix}
=
\begin{bmatrix}
0\\
c^Tb
\end{bmatrix}
$$
