Let be $A=(a_i,_j)\in M_{n+1}(\mathbb{R})$ defined for all $(i,j)\in [\![ 1,n+1 ]\!]^2$, by $a_{i,j}={\dbinom{j-1}{i-1}}$. Let's show this is invertible and determine its inverted matrix.
To my mind, one had to start like this:
A is invertible if its determinant is different from 0.
I tried to implement Sarrus'rule, trying to create a general formula:: $$\sum\limits_{i=j=1}^{n}\prod_{i=j=1}^n \dbinom{j-1}{i-1}- \sum\limits_{i=j=1}^{n}\prod_{i-j=0}^n \dbinom{j-1}{i-1}\neq0 $$
but this one is definitely flase...
An hint was to use the basis $(1,X,...,X^n)$ and $\phi$
\begin{cases} \phi:&\mathbb R_n[X]\rightarrow R_n[X]\\ &P\rightarrow P(X+1) \end{cases} The idea is to prove that $\phi$ has $A$ as matrice in the canonical basis $(1,X,...,X^n)$ and then invert $\phi$.
I know that $\sum\limits_{k=0}^{n}\dbinom{n}{k}X^k=(X+1)^k$, but that's all... I was said that $\phi (X^j)$ should be equal to $\sum\limits_{k=0}^{n}\dbinom{n}{k}X^k$, but I don't understand why...
Can you help me? I'm really fond of simple examples that goes with demonstrations to understand the whole. :)