12 级高代 I 期中 03

依赖于

• 例 1.21

被以下题目直接调用

• 无

12 级高代 I 期中 03

设 $n$ 阶行列式

$$D_k=\left|\begin{array}{cccccccc}1& 1& \dots & 1& 0& 1& \dots & 1\\ x_1& x_2& \dots & x_{k-1}& 1& x_{k+1}& \dots & x_n\\ x_1^2& x_2^2& \dots & x_{k-1}^2& 2x_k& x_{k+1}^2& \dots & x_n^2\\ x_1^3& x_2^3& \dots & x_{k-1}^3& 3x_k^2& x_{k+1}^3& \dots & x_n^3\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ x_1^{n-1}& x_2^{n-1}& \dots & x_{k-1}^{n-1}& (n-1)x_k^{n-2}& x_{k+1}^{n-1}& \dots & x_n^{n-1}\end{array}\right|,1\le k\le n.$$

试求 ${\sum }_{k=1}^n{D}_k$ 的值.

解答

记 $D={\prod }_{1\le i<j\le n}(x_j-x_i)$ 为 Vandermonde 行列式的值, 则由行列式的求导公式 (例 1.21) 并经简单计算可得 $D_k=\frac{\partial D}{\partial x_k}=D{\sum }_{i\ne k}\frac{1}{x_k-x_i}$, 于是

$$\sum _{k=1}^n{D}_k=D\sum _{k=1}^n\sum _{i\ne k}\frac{1}{x_k-x_i}=D\sum _{1\le i<k\le n}\left(\frac{1}{x_k-x_i}+\frac{1}{x_i-x_k}\right)=0.$$