蓝以中上册 行列式 210页 习题三7 解答

castelu

习题三7：
　　计算下面$n$阶行列式：
$$\left| {\begin{array}{*{20}{c}}
{a_1-b_1}&{a_1-b_2}&{\cdots}&{a_1-b_n}\\
{a_2-b_1}&{a_2-b_2}&{\cdots}&{a_2-b_n}\\
{\vdots}&{\vdots}&{}&{\vdots}\\
{a_n-b_1}&{a_n-b_2}&{\cdots}&{a_n-b_n}
\end{array}} \right|$$

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解：
　　记
$$D_n=\left| {\begin{array}{*{20}{c}}
{a_1-b_1}&{a_1-b_2}&{\cdots}&{a_1-b_n}\\
{a_2-b_1}&{a_2-b_2}&{\cdots}&{a_2-b_n}\\
{\vdots}&{\vdots}&{}&{\vdots}\\
{a_n-b_1}&{a_n-b_2}&{\cdots}&{a_n-b_n}
\end{array}} \right|$$
　　根据矩阵乘积公式
$$D_n=\left| \left( {\begin{array}{*{20}{c}}
{a_1}&{1}\\
{a_2}&{1}\\
{\vdots}&{\vdots}\\
{a_n}&{1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{1}&{1}&{\cdots}&{1}\\
{-b_1}&{-b_2}&{\cdots}&{-b_n}
\end{array}} \right) \right|$$
　　由$Binet-Cauchy$公式可知
$$D_n=\sum\limits_{1 \le i<j \le n}\left| {\begin{array}{*{20}{c}}
{a_i}&1\\
{a_j}&1
\end{array}} \right|\left| {\begin{array}{*{20}{c}}
1&1\\
{-b_i}&{-b_j}
\end{array}} \right|=\sum\limits_{1 \le i<j \le n}(a_i-a_j)(b_i-b_j)$$
　　所以
$$D_n=\left\{ \begin{array}{l}
a_1-b_1, n=1\\
(a_1-a_2)(b_1-b_2), n=2\\
0, n \ge 3
\end{array} \right.$$