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$$D_n=\left|\begin{array}{cccc}1 & 1 & \cdots & 1 \\ a_{1} & a_{2} & \cdots & a_{n} \\ \ldots & \ldots& \ldots &\ldots \\ a_{1}^{n-2} & a_{2}^{n-2} & \cdots & a_{n}^{n-2} \\ a_{1}^{n} & a_{2}^{n} & \cdots & a_{n}^{n}\end{array}\right|$$解法一. 利用线性方程组的解的方法计算。$D_n$与范德蒙行列式$$H_n=\left|\begin{array}{cccc}1 & 1 & \cdots & 1 \\ a_{1} & a_{2} & \cdots & a_{n} \\ \cdots& \cdots & \cdots & \cdots \\ a_{1}^{n-1} & a_{2}^{n-1} & \cdots & a_{n}^{n-1}\end{array}\right|=\prod_{1\le i<j\le n}(a_i-a_j)$$的比值可以看作线性方程组$$\cases{x_{1}+a_{1} x_{2}+a_{1}^{2} x_{3}+\cdots+a_{1}^{n-1} x_{n}=a_{1}^{n}\\\cdots\cdots\cdots\cdots\\x_{1}+a_{n} x_{2}+a_{n}^{2} x_{3}+\cdots+a_{n}^{n-1} x_{n}=a_{n}^{n}}\tag1$$的解$x_n$. 如解出$x_n$乘以$H_n$，即得$D_n$。
但(1)又可看作$n$次方程$$t^{n}-x_{n} t^{n-1}-x_{n-1} t^{n-2} \cdots-x_{2} t-x_{1}=0\tag2$$($t$是未知数,$x_{1}, x_{2}, \cdots ,x_{n}$看作系数)有$n$个根$a_1,\dots,a_n$.用根与系数的关系即得$x_{n}=a_{1}+a_{2}+\cdots+a_{n}$.故$D_{n}=x_n H_{n}=\left(a_{1}+a_{2} \cdots+a_{n}\right) \prod_{1\leqslant i\lt j\leqslant n}\left(a_i-a_j\right)$.
解法二. 升阶法. 考虑$n+1$阶范德蒙行列式,$$g(x)=\left|\begin{array}{c}1 & 1 & \cdots & 1&x \\ a_{1} & a_{2} & \cdots & a_{n} &x^2\\ \cdots& \cdots & \cdots & \cdots & \cdots \\ a_{1}^{n-2} & a_{2}^{n-2} & \cdots & a_{n}^{n-2}&x^{n-2} \\ a_{1}^{n-1} & a_{2}^{n-1} & \cdots & a_{n}^{n-1}&x^{n-1} \\ a_{1}^{n} & a_{2}^{n} & \cdots & a_{n}^{n}&x^{n}\end{array}\right|=\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right) \prod_{1 \leqslant i<j\leqslant n}\left(a_{i}-a_{j}\right)$$从上式左端看，多项式$g(x)$的$x^{n-1}$的系数为$(-1)^{2n+1}D_n=-D_n$。但从上式右端看$x^{n-1}$的系数为$-(a_1+a_2+\cdots+a_n)\prod_{1\leqslant i<j \leqslant n}\left(a_i-a_j\right)$.二者应相等. 故$D_n=\left(a_{1}+a_{2} \cdots+a_{n}\right) \prod_{1\leqslant i\lt j\leqslant n}\left(a_i-a_j\right)$.
类似于此题, 如$\left|\begin{array}{cccc}1 & 1 & \cdots & 1 \\ a_{1}^2 & a_{2} ^2& \cdots & a_{n}^2 \\ a_{1}^3 & a_{2}^3 & \cdots & a_{n}^3 \\ \ldots & \ldots& \ldots &\ldots \\ a_{1}^{n} & a_{2}^{n} & \cdots & a_{n}^{n}\end{array}\right|$学生便会计算了.
高等代数教学点滴体会_周立仁.pdf
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