<s>[INST] Find the inverse of $A$ Let
$$
A=
\begin{bmatrix}
n & n_1 & n_2 & \cdots & n_s & 0 \\
n_1 & n_1 & 0 & \cdots & 0 & 1  \\
n_2 & 0 & n_2 & \cdots & 0 & 1 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
n_s & 0 & 0 & \cdots & n_s & 1 \\
0 & 1 & 1 & \cdots & 1 & 0
\end{bmatrix},
$$
where $n=\sum_{i=1}^sn_i$.
My questions: What is $A^{-1}$? Does it have an elegant expression?
My attempts: I tried some small matrices and found that $A^{-1}$ has the following form:
$$
A^{-1}=\begin{bmatrix}
\cdots & \cdots & \cdots & \cdots & \cdots & -\frac{1}{s} \\
\cdots & \cdots & \cdots & \cdots & \cdots & \frac{1}{s}  \\
\cdots & \cdots & \cdots & \cdots & \cdots & \frac{1}{s} \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
\cdots & \cdots & \cdots & \cdots & \cdots & \frac{1}{s} \\
-\frac{1}{s} & \frac{1}{s} & \frac{1}{s} & \cdots & \frac{1}{s} & 0
\end{bmatrix}
$$
While I faild to find patterns in those "dots" parts.
Thanks a lot.
 [/INST] Partial Answer: Here's a strategy you might find helpful.
Let $M$ denote the submatrix of $A$ given by
$$
M = \pmatrix{n & n_1 & n_2 & \cdots & n_s  \\
n_1 & n_1 & 0 & \cdots & 0   \\
n_2 & 0 & n_2 & \cdots & 0  \\
\vdots & \vdots & \vdots & \ddots & \vdots  \\
n_s & 0 & 0 & \cdots & n_s}.
$$
we see that $M$ can be expressed in the form
$$
M = \sum_{i=1}^s (e_1 + e_{1+i})(e_1 + e_{1+i})^T,
$$
where $e_1,e_2,\dots,e_{s+1}$ denotes the canonical basis of $\Bbb R^n$. This gives us the decomposition $M = BDB^T$, where
$$
B = \pmatrix{e_1 + e_2 & e_1 + e_3 & \cdots & e_1 + e_{s+1}} = 
\pmatrix{1&1&\cdots&1\\1\\&1\\ && \ddots \\&&&1},
\quad D = \pmatrix{n_1 \\ & \ddots \\ && n_s}.
$$
So, the matrix $A$ can be written in the form
$$
A = \pmatrix{BDB^T & x\\ x^T & 0},
$$
where $x = (0,1,1,\dots,1)^T$.
Let $C$ denote the matrix
$$
C = \pmatrix{1&-1&\cdots & -1\\
&1\\
&&\ddots\\
&&&1}.
$$
We note that
$$
CB = \pmatrix{0\\&I_{s}}.
$$
With that in mind, we find that
$$
\tilde C A \tilde C ^T = \overbrace{\pmatrix{C&0\\0 & 1}}^{\tilde C} \pmatrix{BDB^T & x\\ x^T & 0} \pmatrix{C&0\\0 & 1}^T = 
\pmatrix{D & Cx\\ (Cx)^T & 0}.
$$
It would suffice to find the inverse of this "nicer" matrix $\tilde C A \tilde C ^T$, then compute
$$
A^{-1} = \tilde C^T[\tilde C A \tilde C ^T]^{-1}\tilde C.
$$
</s>