Is there a shorter way to prove this?

Solution 1:

By the rules of Pascal's Triangle, $$ \binom{i+1+j}{j}-\binom{i+j}{j}=\binom{i+j}{j-1} $$ which means that row $i+1$ minus row $i$ gives row $i+1$ shifted to the right. $$ a_{i+1,j}-a_{i,j}=a_{i+1,j-1} $$ This can be done to shift row $n$ to the right, then shift row $n-1$, up to row $2$. We can repeat the process to shift rows $n$ through $3$ to the right. We can continue this until we have all the $1$s on the diagonal and $0$s in the lower left triangle.

Subtracting one row from another does not change the determinant, so the original determinant was $1$.

Using Cramer's Rule, we get that the inverse is an integer matrix.

Subtract row $3$ from row $4$: $$\small \begin{bmatrix} \binom{0}{0}&\binom{1}{1}&\binom{2}{2}&\binom{3}{3}\\ \binom{1}{0}&\binom{2}{1}&\binom{3}{2}&\binom{4}{3}\\ \binom{2}{0}&\binom{3}{1}&\binom{4}{2}&\binom{5}{3}\\ \binom{3}{0}&\binom{4}{1}&\binom{5}{2}&\binom{6}{3}\\ \end{bmatrix} \rightarrow \begin{bmatrix} \binom{0}{0}&\binom{1}{1}&\binom{2}{2}&\binom{3}{3}\\ \binom{1}{0}&\binom{2}{1}&\binom{3}{2}&\binom{4}{3}\\ \binom{2}{0}&\binom{3}{1}&\binom{4}{2}&\binom{5}{3}\\ 0&\binom{3}{0}&\binom{4}{1}&\binom{5}{2}\\ \end{bmatrix} $$ Subtract row $2$ from row $3$: $$\small \begin{bmatrix} \binom{0}{0}&\binom{1}{1}&\binom{2}{2}&\binom{3}{3}\\ \binom{1}{0}&\binom{2}{1}&\binom{3}{2}&\binom{4}{3}\\ \binom{2}{0}&\binom{3}{1}&\binom{4}{2}&\binom{5}{3}\\ 0&\binom{3}{0}&\binom{4}{1}&\binom{5}{2}\\ \end{bmatrix} \rightarrow \begin{bmatrix} \binom{0}{0}&\binom{1}{1}&\binom{2}{2}&\binom{3}{3}\\ \binom{1}{0}&\binom{2}{1}&\binom{3}{2}&\binom{4}{3}\\ 0&\binom{2}{0}&\binom{3}{1}&\binom{4}{2}\\ 0&\binom{3}{0}&\binom{4}{1}&\binom{5}{2}\\ \end{bmatrix} $$ Subtract row $1$ from row $2$: $$\small \begin{bmatrix} \binom{0}{0}&\binom{1}{1}&\binom{2}{2}&\binom{3}{3}\\ \binom{1}{0}&\binom{2}{1}&\binom{3}{2}&\binom{4}{3}\\ 0&\binom{2}{0}&\binom{3}{1}&\binom{4}{2}\\ 0&\binom{3}{0}&\binom{4}{1}&\binom{5}{2}\\ \end{bmatrix} \rightarrow \begin{bmatrix} \binom{0}{0}&\binom{1}{1}&\binom{2}{2}&\binom{3}{3}\\ 0&\binom{1}{0}&\binom{2}{1}&\binom{3}{2}\\ 0&\binom{2}{0}&\binom{3}{1}&\binom{4}{2}\\ 0&\binom{3}{0}&\binom{4}{1}&\binom{5}{2}\\ \end{bmatrix} $$ Subtract row $3$ from row $4$: $$\small \begin{bmatrix} \binom{0}{0}&\binom{1}{1}&\binom{2}{2}&\binom{3}{3}\\ 0&\binom{1}{0}&\binom{2}{1}&\binom{3}{2}\\ 0&\binom{2}{0}&\binom{3}{1}&\binom{4}{2}\\ 0&\binom{3}{0}&\binom{4}{1}&\binom{5}{2}\\ \end{bmatrix} \rightarrow \begin{bmatrix} \binom{0}{0}&\binom{1}{1}&\binom{2}{2}&\binom{3}{3}\\ 0&\binom{1}{0}&\binom{2}{1}&\binom{3}{2}\\ 0&\binom{2}{0}&\binom{3}{1}&\binom{4}{2}\\ 0&0&\binom{3}{0}&\binom{4}{1}\\ \end{bmatrix} $$ Subtract row $2$ from row $3$: $$\small \begin{bmatrix} \binom{0}{0}&\binom{1}{1}&\binom{2}{2}&\binom{3}{3}\\ 0&\binom{1}{0}&\binom{2}{1}&\binom{3}{2}\\ 0&\binom{2}{0}&\binom{3}{1}&\binom{4}{2}\\ 0&0&\binom{3}{0}&\binom{4}{1}\\ \end{bmatrix} \rightarrow \begin{bmatrix} \binom{0}{0}&\binom{1}{1}&\binom{2}{2}&\binom{3}{3}\\ 0&\binom{1}{0}&\binom{2}{1}&\binom{3}{2}\\ 0&0&\binom{2}{0}&\binom{3}{1}\\ 0&0&\binom{3}{0}&\binom{4}{1}\\ \end{bmatrix} $$ Subtract row $3$ from row $4$: $$\small \begin{bmatrix} \binom{0}{0}&\binom{1}{1}&\binom{2}{2}&\binom{3}{3}\\ 0&\binom{1}{0}&\binom{2}{1}&\binom{3}{2}\\ 0&0&\binom{2}{0}&\binom{3}{1}\\ 0&0&\binom{3}{0}&\binom{4}{1}\\ \end{bmatrix} \rightarrow \begin{bmatrix} \binom{0}{0}&\binom{1}{1}&\binom{2}{2}&\binom{3}{3}\\ 0&\binom{1}{0}&\binom{2}{1}&\binom{3}{2}\\ 0&0&\binom{2}{0}&\binom{3}{1}\\ 0&0&0&\binom{3}{0}\\ \end{bmatrix} $$ The determinant is $\binom{0}{0}\binom{1}{0}\binom{2}{0}\binom{3}{0}=1$.

Solution 2:

There are (at least) two approaches to showing that the inverse has integer entries. The first is Cramer's rule, which gives the entries of the inverse of $A$ as quotients of the form "determinants of certain submatrices of $A$ divided by the determinant of $A$". Since the entries of your $A$ are integers, and since your row-reduction shows that the determinant of $A$ is $1$, it follows that these quotients are integers.

Alternatively, do you know the algorithm for inverting a square matrix by (1) writing an identity matrix of the same size next to it (so that the combination is an $n\times 2n$ matrix), (2) applying elementary row operations to convert the original matrix into the identity matrix, and (3) reading off the inverse in the region where you originally put the identity matrix? If so, then you've done most of the work for showing that the inverse of your matrix has integer entries. You've row-reduced your matrix to upper triangular form with $1$'s (not just non-zero entries, but $1$'s) on the diagonal, without ever having to do any dividing. You could complete the row-reduction to the identity without any division. The whole reduction process, if applied to an identity matrix, would never produce anything but integers.

Solution 3:

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.