Inverse of a general term matrix with term $\frac{(i-1-n)^{j-1}}{(j-1)!}$ I need to compute the determinant for the square matrix A, defined as follows:

*

*its dimension $N$ is odd : $N=2.n+1$

*its coefficients are

$$\forall (i,j) \in [ 1..N], a_{i,j}=\frac{(i-1-n)^{j-1}}{(j-1)!}$$
For $n=2$, we have:
$$A=\begin{bmatrix}1&-2&2&-4/3&2/3\\1&-1&1/2&-1/6&1/24\\1&0&0&0&0\\1&1&1/2&1/6&1/24\\1&2&2&4/3&2/3\end{bmatrix}$$
I conjecture that $det(A)=1$, but I cannot prove it.
Can you help me find a general expression of the inverse of A ?
I have tried to develop along the middle row, but it does not seem simpler after that ...
Thank yo very much
 A: The matrix you defined here is a scaled version of a Vandermonde matrix. Recall that a Vandermonde matrix $V$ is defined by $V_{i, j} = x_i^{j - 1}$ and has determinant
$$
\det(V) = \prod_{1 \leq i < j \leq n} (x_j - x_i)\ .
$$
Now your matrix is essentially a Vandermonde matrix with $x_i = i - 1 - n$, but the $j$-th column is scaled by the factor $1/(j-1)!$. In the determinant, we can collect all such factors into $1/\prod_{j=1}^N (j-1)!$. Thus in order to prove your hypothesis that $\det{A} = 1$, you need only show that the corresponding Vandermonde matrix has determinant $\prod_{j=1}^N (j-1)!$.
This can be computed from the determinant formula above by plugging in $x_i = i - 1 - n$,
\begin{align}
& \quad \prod_{1\leq p < q \leq N} [(q - 1 - n) - (p - 1 - n)]\\
& = \prod_{p = 1}^{N-1}\prod_{q=p+1}^{N} (q - p) \\
& = \prod_{p=1}^{N-1}\prod_{r = 1}^{N - p} r \\
& = \prod_{p=1}^{N-1} (N-p)!
\end{align}
After reindexing the last product from $N-p\mapsto j - 1$, and noting that the $j=1$ factor in the product is just $1$, we can conclude that this determinant is indeed equal to $\prod_{j=1}^N(j-1)!$.
As for the inverse of the matrix, you can also compute it by first finding the inverse of the Vandermonde matrix, then scaling the rows of the matrix correspondingly. More formally, we can write your matrix $A$ in terms of the corresponding Vandermonde matrix $V$ as $V \operatorname{diag}(1, 1, \dots, 1/(j-1)!, \dots, 1/(N-1)!)$, where the diagonal matrix is used to scale the $j$-th column of $V$ by $1/(j-1)!$. Now if we know the inverse of $V$, then the inverse of $A$ should be $\operatorname{diag}(1, 1, \dots, (j-1)!, \dots, (N-1)!) V^{-1}$. Finally, the inverse formula of a Vandermonde matrix can be found here,
$$
V^{-1}_{i,j} = \begin{cases}
(-1)^{j-1}\left(\frac{\displaystyle\sum_{\substack{1\leq m_1 < \dots < m_{n-1} \leq n\\m_1, \dots,m_{n-j}\neq i}} x_{m_1},\dots x_{m_{n-j}}}{\displaystyle x_i \prod_{\substack{1\leq m\leq n \\ m\neq i}}(x_m - x_i)}\right) & 1 \leq j < n\ , \\
\frac{\displaystyle 1}{\displaystyle x_i \prod_{\substack{1\leq m\leq n \\ m\neq i}}(x_i - x_m)} & j = n\ .
\end{cases}
$$
This formula can be adapted to your specific case, but the final result probably still won't look very pretty.
