Let $m,n\in\mathbb{Z^+}$. I want to calculate inverse of the matrix $ A_{(m+1)\times (m+1)}$. $ \begin{equation} A=\begin{bmatrix} 1& n&n^2&\cdot&n^m\\ 1& n+1&(n+1)^2&\cdot &(n+1)^m\\ \vdots &\vdots &\vdots &\vdots\\ 1&n+m&(n+m)^2&\cdot&(n+m)^m\\ \end{bmatrix} \end{equation} $
I worked on the problem and found that:
If $m=1$, then we have $$ \begin{equation} A=\begin{bmatrix} 1& n\\ 1& n+1\\ \end{bmatrix}, A^{-1}=\begin{bmatrix} n+1& -n\\ -1& 1\\ \end{bmatrix} \end{equation} $$
If $m=2$, then
$ A^{-1}=(\frac{1}{2!}) \begin{equation} \begin{bmatrix} (n+1)(n+2)& -n(n+2)&-n(n+1)\\ -(2n+3)&2(2n+2)&-(2n+1)\\ 1&-2&1 \end{bmatrix} \end{equation} $
for $m=3$, we set $ A^{-1}=\frac{1}{3!} \begin{equation} \begin{bmatrix} (n+1)(n+2)(n+3)& -n(n+2)(n+3)&-n(n+1)(n+3)&-n(n+1)(n+2)\\ 3(n+\frac{6}{3})^2-1&3(n+\frac{5}{3})^2-\frac{7}{3}&3(n+\frac{4}{3})^2-\frac{7}{3}&3(n+\frac{3}{3})^2-1\\ 3(n+\frac{6}{3})&-3(n+\frac{5}{3})&3(n+\frac{4}{3})&-3(n+\frac{3}{3})\\ -1&3&-3&1\\ \end{bmatrix} \end{equation} $
I guess explicate form for the first and last rows of the matrix $A^{-1}$are as follows\ $ a_{1j}=\frac{1}{2!}\times(-1)^{j-1}\times\prod_{k=0,k\neq i}^m (n+k)\times\dbinom {m}{j-1};1 \leq j \leq m+1 $ and
$ a_{(m+1)j}=\frac{1}{m!}\times(-1)^{j-1}\times\dbinom {m}{j-1}; 1\leq j \leq m+1 $
Now, can some body give an explicit formula for all $a_{ij}?$ Thanks in advance.
