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.
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 $$ A^{-1} = \tilde C^T[\tilde C A \tilde C ^T]^{-1}\tilde C. $$