计算行列式通常用降阶法, 但有时候也可反其道而行之. 升阶法常常用于一些 “缺少” 某行 (列) 的行列式, 加上适当的行列后反而可以简化问题. 下面是 3 个典型的例子.

例 1.34

计算下列行列式:

∣ A ∣ = 1 + x_{1} 1 + x_{2} ⋮ 1 + x_{n} 1 + x_{1}^{2} 1 + x_{2}^{2} ⋮ 1 + x_{n}^{2} \dots \dots \dots 1 + x_{1}^{n} 1 + x_{2}^{n} ⋮ 1 + x_{n}^{n}

解

将行列式升阶为

∣ A ∣ = 111 ⋮ 1 0 1 + x_{1} 1 + x_{2} ⋮ 1 + x_{n} 0 1 + x_{1}^{2} 1 + x_{2}^{2} ⋮ 1 + x_{n}^{2} \dots \dots \dots \dots 0 1 + x_{1}^{n} 1 + x_{2}^{n} ⋮ 1 + x_{n}^{n} = 111 ⋮ 1 - 1 x_{1} x_{2} ⋮ x_{n} - 1 x_{1}^{2} x_{2}^{2} ⋮ x_{n}^{2} \dots \dots \dots \dots - 1 x_{1}^{n} x_{2}^{n} ⋮ x_{n}^{n} .

将第一行拆开, 得

∣ A ∣ = 211 ⋮ 1 0 x_{1} x_{2} ⋮ x_{n} 0 x_{1}^{2} x_{2}^{2} ⋮ x_{n}^{2} \dots \dots \dots \dots 0 x_{1}^{n} x_{2}^{n} ⋮ x_{n}^{n} + - 1 11 ⋮ 1 - 1 x_{1} x_{2} ⋮ x_{n} - 1 x_{1}^{2} x_{2}^{2} ⋮ x_{n}^{2} \dots \dots \dots \dots - 1 x_{1}^{n} x_{2}^{n} ⋮ x_{n}^{n} .

后面一个行列式的第一行提出公因子 -1 后是一个关于 $1, x_{1}, x_{2}, \dots, x_{n}$ 的 Vander Monde 行列式, 从而可得

∣ A ∣ = (2 x_{1} x_{2} \dots x_{n} - (x_{1} - 1) (x_{2} - 1) \dots (x_{n} - 1)) 1 \leq i < j \leq n \prod (x_{j} - x_{i}) .

计算下列行列式:

∣ A ∣ = 11 ⋮ 1 x_{1} x_{2} ⋮ x_{n} x_{1}^{2} x_{2}^{2} ⋮ x_{n}^{2} \dots \dots \dots x_{1}^{n - 2} x_{2}^{n - 2} ⋮ x_{n}^{n - 2} x_{1}^{n} x_{2}^{n} ⋮ x_{n}^{n} .

注意这个行列式和 Vander Monde 行列式的区别在于它们的最后一列. 现添上一行一列使之成为 Vander Monde 行列式,再求出 $y^{n - 1}$ 的系数即可解出答案:

∣ B ∣ = 11 ⋮ 11 x_{1} x_{2} ⋮ x_{n} y x_{1}^{2} x_{2}^{2} x_{n}^{2} y^{2} \dots \dots ⋮ \dots \dots x_{1}^{n - 1} x_{2}^{n - 1} ⋮ x_{n}^{n - 1} y^{n - 1} x_{1}^{n} x_{2}^{n} ⋮ x_{n}^{n} y^{n}

则

∣ B ∣ = (y - x_{1}) (y - x_{2}) \dots (y - x_{n}) 1 \leq i < j \leq n \prod (x_{j} - x_{i}) .

因此 $y^{n - 1}$ 的系数是

- (x_{1} + x_{2} + \dots + x_{n}) 1 \leq i < j \leq n \prod (x_{j} - x_{i}) .

而 $∣ B ∣$ 中元素 $y^{n - 1}$ 的代数余子式为 $(- 1)^{n + 1 + n} ∣ A ∣ = - ∣ A ∣$ ,因此

∣ A ∣ = (x_{1} + x_{2} + \dots + x_{n}) 1 \leq i < j \leq n \prod (x_{j} - x_{i}) .

计算下列行列式:

∣ A ∣ = 11 ⋮ 1 x_{1}^{2} x_{2}^{2} ⋮ x_{n}^{2} \dots \dots \dots x_{1}^{n} x_{2}^{n} ⋮ x_{n}^{n}

将这个行列式添上一行一列使之成为 Vander Monde 行列式,再求出 $y$ 的系数即可解出答案:

∣ B ∣ = 11 ⋮ 11 x_{1} x_{2} ⋮ x_{n} y x_{1}^{2} x_{2}^{2} ⋮ x_{n}^{2} y^{2} \dots \dots \dots \dots x_{1}^{n} x_{2}^{n} ⋮ x_{n}^{n} y^{n}

则

∣ B ∣ = (y - x_{1}) (y - x_{2}) \dots (y - x_{n}) 1 \leq i < j \leq n \prod (x_{j} - x_{i}) .

因此 $y$ 的系数是

(- 1)^{n - 1} (i = 1 \sum n x_{1} \dots x_{i - 1} x_{i + 1} \dots x_{n}) 1 \leq i < j \leq n \prod (x_{j} - x_{i}) .

而 $∣ B ∣$ 中元素 $y$ 的代数余子式为 $(- 1)^{n + 1 + 2} ∣ A ∣ = (- 1)^{n - 1} ∣ A ∣$ ,因此

∣ A ∣ = (i = 1 \sum n x_{1} \dots x_{i - 1} x_{i + 1} \dots x_{n}) 1 \leq i < j \leq n \prod (x_{j} - x_{i}) .