\[ \det A=\left| \begin{matrix} 1& 1& \cdots& 1\ 1& C_{2}^{1}& \cdots& C_{n}^{1}\ 1& C_{3}^{2}& \cdots& C_{n+1}^{2}\ \vdots& \vdots& & \vdots\ 1& C_{n}^{n-1}& \cdots& C_{2n-2}^{n-1}\\end{matrix} \right| \]
\(solution:\)
AS根据
\[
\left( \begin{array}{c}
n\ m\\end{array} \right) =\left( \begin{array}{c}
n-1\ m-1\\end{array} \right) +\left( \begin{array}{c}
n-1\ m\\end{array} \right),
\]
第\(n\)行减去第\(n-1\)行，第\(n-1\)行减去第\(n-2\)行，以此类推，可完成对\(\det A\)的依次降阶，一直做下取，可得\(\det A\)=1.
原文：https://www.cnblogs.com/lagrange/p/8854413.html