Evaluate a determinant of square matrix of complex binomial values

This can be proved a little more simply than is done in the linked paper. If $A(n,x)$ is the $n \times n$ matrix with components $A(n,x)_{i,j} \; = \; \binom{j+x}{i-1} \hspace{2cm} 1 \le i,j \le n$ Then it is clear that $A(n,x)_{1,j} = 1$ for all $1 \le j \le n$ . If we apply the following elementary column operations to $A(n,x)$ in turn:

• Subtract column $n-1$ from column $n$ ,
• Subtract column $n-2$ from column $n-1$ ,
• Subtract column $n-3$ from column $n-2$ ,
• ....
• Subtract column $1$ from column $2$

then $A(n,x)$ is transformed into a matrix $B(n,x)$ of the form $B(n,x) \; = \; \left(\begin{array}{cccccc} 1 & 0 & 0 & \cdots & 0 & 0 \\ A(n,x)_{2,1} & & & & & \\ A(n,x)_{3,1} & & & & & \\ \vdots & & & C(n,x) & & \\ A(n,x)_{n-1,1} & & & & \\ A(n,x)_{n,1} & & & & \end{array}\right)$ so that $d(x,n) \; = \; \mathrm{det}\,A(n,x) \; = \; \mathrm{det}\,C(n,x)$ , where $C(n,x)$ is an $(n-1) \times (n-1)$ matrix with coefficients $C(n,x)_{i,j} \; = \; A(n,x)_{i+1,j+1} - A(n,x)_{i+1,j} \; = \; \binom{j+x+1}{i} - \binom{j+x}{i} \; = \; \binom{j+x}{i-1} \hspace{2cm} 1 \le i,j \le n-1$ so that $C(n,x) \; = \; A(n-1,x)$ . Hence we deduce that $\mathrm{det}\,A(n,x) = \mathrm{det}\,A(n-1,x)$ for any integer $n \ge 2$ , which implies that $\mathrm{det}\,A(n,x) = \mathrm{det}\,A(1,x) = 1$ for all $n \in \mathbb{N}$ .

The actual matrix of the question asks for one where either the rows and columns are both even permutations of the rows and columns of $A(n,x)$ , or else they are both odd permutations. In either case the determinant of the matrix does not change under these permutations.

Thus we have shown that $d(x,n) = \boxed{1}$ for all $x \in \mathbb{C}$ and all $n\in \mathbb{N}$ .

The complex value $x$ has no effect on the result. $d(0,n)\ \{\forall n|n\in \mathbb{N}_+\}=1$ .

Please, see this document .