一道行列式计算
2018.04.10
\[\det A=\left| \begin{matrix}
1& 1& \cdots& 1\\
1& C_{2}^{1}& \cdots& C_{n}^{1}\\
1& C_{3}^{2}& \cdots& C_{n+1}^{2}\\
\vdots& \vdots& & \vdots\\
1& C_{n}^{n-1}& \cdots& C_{2n-2}^{n-1}\\
\end{matrix} \right|
\]
\(solution:\)
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.
