TITLE: Is there a shorter way to prove this?
QUESTION [11 upvotes]: Let $n$ a natural number and $A=(a_{ij})$, where $a_{ij}=\left(\begin{array}{c}i+j \\i\end{array} \right)$, for $0≤i,j<n$. Prove that A has an inverse matrix and that all the entries of $A^{−1}$ are integers.
I tried to prove this like this: (but I'm not sure if it's correct or formal enough)


Also if you could give me a hint to prove the second aasertion it would be great. 
Thank you.

REPLY [4 votes]: Let $B=\left[b_{ij}=\binom{i}{j}\right]$ and $C=BB^t$. We have $c_{ij}=\sum_{k=0}^i\binom{i}{k}\binom{j}{k}=\binom{i+j}{i}=a_{ij}$, thus $C=A$. So it suffices to show that the inverse of $B$, has all integer entries. Because $B$ is lower-triangular $B^{-1}$ is also lower-triangular and can be found column by column. The first column of $B^{-1}$ is a vector of alternating  $+1$ and $-1$s. I think it should be easy from this point to show the entries in the next columns are also integers one by one. To be precise you need to inspect the linear equations corresponding to the entry of $B^{-1}$ being inspected.