高代绿皮第四版课后习题复习题一T19

本文主要是介绍高代绿皮第四版课后习题复习题一T19，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

原题

计算下列n阶行列式的值

$|A|=\left| \begin{matrix} 1 & {{x}_{1}} & \cdots & x_{1}^{i-1} & x_{1}^{i+1} & \cdots & x_{1}^{n} \\ 1 & {{x}_{2}} & \cdots & x_{2}^{i-1} & x_{2}^{i+1} & \cdots & x_{2}^{n} \\ \vdots & \vdots & {} & \vdots & \vdots & {} & \vdots \\ 1 & {{x}_{n}} & \cdots & x_{n}^{i-1} & x_{n}^{i+1} & \cdots & x_{n}^{n} \\ \end{matrix} \right|\,\,(1\leqslant i\leqslant n-1)$

解析

思路:

注意到 $|A|$ 与标准的Vandermonde行列式缺少了 i 次幂的列,故可构造n+1阶Vandermonde行列式

即

$|B|=\left| \begin{matrix} 1 & {{x}_{1}} & \cdots & x_{1}^{i-1} &x_{1}^{i}& x_{1}^{i+1} & \cdots & x_{1}^{n} \\ 1 & {{x}_{2}} & \cdots & x_{2}^{i-1} &x_{2}^{i}& x_{2}^{i+1} & \cdots & x_{2}^{n} \\ \vdots & \vdots & {} & \vdots &\vdots& \vdots & {} & \vdots \\ 1 & {{x}_{n}} & \cdots & x_{n}^{i-1} &x_{n}^{i}& x_{n}^{i+1} & \cdots & x_{n}^{n} \\1&y&\cdots&y^{i-1}&y^{i}&y^{i+1}&\cdots&y^{n} \end{matrix} \right|$

将 $|B|$ 按第n+1行展开得

$|B|=1\cdot A_{n+1,1}+y\cdot A_{n+1,2}+\cdots+y^{i}\cdot A_{n+1,i+1}+\cdots+y^{n}\cdot A_{n+1,n+1}$

由于

$A_{n+1,i+1}=(-1)^{n+1+i+1}|A|=(-1)^{n+i}|A|$

则只需求出 $y^{i}$ 的系数,由于 $|B|$ 为Vandermonde行列式,根据公式求得

$|B|=(y-x_{1})(y-x_{2})\cdots(y-x_{n})\prod\limits_{1\leqslant i\leqslant j\leqslant n}{(x_{j}-x_{i})}$

根据多项式知识可得 $y^{i}$ 的系数为

$\sum\limits_{1\leqslant k_{1}<k_{2}<\cdots<k_{n-i}\leqslant n}{(-1)^{n-i}x _{k_{1}}x_{k_{2}}\cdots x_{k_{n-i}}}\prod\limits_{1\leqslant i\leqslant j\leqslant n}{(x_{j}-x_{i})}$

于是

$(-1)^{n+i}\,|A|=\sum\limits_{1\leqslant k_{1}<k_{2}<\cdots<k_{n-i}\leqslant n}{(-1)^{n-i}x _{k_{1}}x_{k_{2}}\cdots x_{k_{n-i}}}\prod\limits_{1\leqslant i\leqslant j\leqslant n}{(x_{j}-x_{i})}$

故求得

$|A|=\sum\limits_{1\leqslant k_{1}<k_{2}<\cdots<k_{n-i}\leqslant n}{x _{k_{1}}x_{k_{2}}\cdots x_{k_{n-i}}}\prod\limits_{1\leqslant i\leqslant j\leqslant n}{(x_{j}-x_{i})}$