例 1.9

设 $f_{ij} (t)$ 是可微函数,

F (t) = f_{11} (t) f_{21} (t) ⋮ f_{n 1} (t) f_{12} (t) f_{22} (t) ⋮ f_{n 2} (t) \dots \dots \dots f_{1 n} (t) f_{2 n} (t) ⋮ f_{nn} (t)

求证: $\frac{d}{d t} F (t) = j = 1 \sum n F_{j} (t)$ ,其中

F_{j} (t) = f_{11} (t) f_{21} (t) ⋮ f_{n 1} (t) f_{12} (t) f_{22} (t) ⋮ f_{n 2} (t) \dots \dots \dots \frac{d}{d t} f_{1 j} (t) \frac{d}{d t} f_{2 j} (t) ⋮ \frac{d}{d t} f_{nj} (t) \dots \dots \dots f_{1 n} (t) f_{2 n} (t) ⋮ f_{nn} (t) .

证明

对行列式的阶 $n$ 用数学归纳法. $n = 1$ 时显然成立. 设结论对 $n - 1$ 正确,现证明对 $n$ 结论也对. 将 $F (t)$ 按第一列展开:

F (t) = f_{11} (t) A_{11} (t) + f_{21} (t) A_{21} (t) + \dots + f_{n 1} (t) A_{n 1} (t),

其中 $A_{i 1} (t)$ 是元素 $f_{i 1} (t)$ 的代数余子式. 对上式两边求导数并记 $A_{ij}^{k} (t)$ 为对 $A_{ij} (t)$ 的第 $k$ 列元素求导数后得到的行列式,则

\frac{d}{d t} F (t) = \frac{d}{d t} (i = 1 \sum n f_{i 1} (t) A_{i 1} (t)) = i = 1 \sum n f_{i 1}^{'} (t) A_{i 1} (t) + i = 1 \sum n f_{i 1} (t) A_{i 1}^{'} (t)

= F_{1} (t) + i = 1 \sum n f_{i 1} (t) (k = 1 \sum n - 1 A_{i 1}^{k} (t)) = F_{1} (t) + k = 1 \sum n - 1 i = 1 \sum n f_{i 1} (t) A_{i 1}^{k} (t) .

而

F_{2} (t) = f_{11} (t) A_{11}^{1} (t) + f_{21} (t) A_{21}^{1} (t) + \dots + f_{n 1} (t) A_{n 1}^{1} (t),

…

F_{n} (t) = f_{11} (t) A_{11}^{n - 1} (t) + f_{21} (t) A_{21}^{n - 1} (t) + \dots + f_{n 1} (t) A_{n 1}^{n - 1} (t),

因此

\frac{d}{d t} F (t) = F_{1} (t) + F_{2} (t) + \dots + F_{n} (t)