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习题一27:
设$f_i(x)$是数域$K$上的$i$次多项式,其首项系数为$a_i(i=0,1,2,\cdots,n-1)$。又设$b_1,b_2,\cdots,b_n$是$K$内一组数。试计算下列$n$阶行列式:
$$\left| {\begin{array}{*{20}{c}}
{f_0(b_1)}&{f_0(b_2)}&{\cdots}&{f_0(b_n)}\\
{f_1(b_1)}&{f_1(b_2)}&{\cdots}&{f_1(b_n)}\\
{\vdots}&{\vdots}&{}&{\vdots}\\
{f_{n-1}(b_1)}&{f_{n-1}(b_2)}&{\cdots}&{f_{n-1}(b_n)} \end{array}}
\right|$$
解:
记
$$D_n=\left| {\begin{array}{*{20}{c}}
{f_0(b_1)}&{f_0(b_2)}&{\cdots}&{f_0(b_n)}\\
{f_1(b_1)}&{f_1(b_2)}&{\cdots}&{f_1(b_n)}\\
{\vdots}&{\vdots}&{}&{\vdots}\\
{f_{n-1}(b_1)}&{f_{n-1}(b_2)}&{\cdots}&{f_{n-1}(b_n)} \end{array}}
\right|$$
构造多项式
$$F(x)=\left| {\begin{array}{*{20}{c}}
{f_0(b_1)}&{f_0(b_2)}&{\cdots}&{f_0(x)}\\
{f_1(b_1)}&{f_1(b_2)}&{\cdots}&{f_1(x)}\\
{\vdots}&{\vdots}&{}&{\vdots}\\
{f_{n-1}(b_1)}&{f_{n-1}(b_2)}&{\cdots}&{f_{n-1}(x)} \end{array}}
\right|$$
易知$F(x)$为数域$K$上的$n-1$次多项式,且
$$F(b_1)=F(b_2)=\cdots=F(b_{n-1})=0$$
令
$$F(x)=A_{n-1}\prod\limits_{k=1}^{n-1}(x-b_k)$$
其中$A_{n-1}$为$F(x)$的首项系数
按最后一列展开后可知
$$A_{n-1}=D_{n-1}a_{n-1}$$
所以
$$\begin{eqnarray*}
D_n&=&F(b_n)=D_{n-1}a_{n-1}\prod\limits_{k=1}^{n-1}(b_n-b_k)\\
&=&\prod\limits_{i=0}^{n-1}a_i\prod\limits_{1 \le k<l \le n}(b_l-b_k)
\end{eqnarray*}$$ |
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