例 1.21

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• 12 级高代 I 期中 03
• 问题 2018A01

例 1.21

设 $f_{ij}(t)$ 是可微函数，

$$F(t)=\left|\begin{array}{cccc}{f}_{11}(t)& f_{12}(t)& \cdots & f_{1n}(t)\\ f_{21}(t)& f_{22}(t)& \cdots & f_{2n}(t)\\ ⋮& ⋮& & ⋮\\ f_{n1}(t)& f_{n2}(t)& \cdots & f_{nn}(t)\end{array}\right|,$$

求证：

$$\frac{d}{dt}F(t)=\sum _{j=1}^n{F}_j(t),$$

其中

$$F_j(t)=\left|\begin{array}{cccccc}{f}_{11}(t)& f_{12}(t)& \cdots & \frac{d}{dt}{f}_{1j}(t)& \cdots & f_{1n}(t)\\ f_{21}(t)& f_{22}(t)& \cdots & \frac{d}{dt}{f}_{2j}(t)& \cdots & f_{2n}(t)\\ ⋮& ⋮& & ⋮& & ⋮\\ f_{n1}(t)& f_{n2}(t)& \cdots & \frac{d}{dt}{f}_{nj}(t)& \cdots & f_{nn}(t)\end{array}\right|.$$

解答

证明 对阶数 $n$ 进行归纳，$n=1$ 时显然成立。设结论对 $n-1$ 阶行列式成立，现证 $n$ 阶行列式的情形。将 $F(t)$ 按第一列展开：

$$F(t)=f_{11}(t)A_{11}(t)+f_{21}(t)A_{21}(t)+\cdots +f_{n1}(t)A_{n1}(t),$$

其中 $A_{i1}(t)$ 是元素 $f_{i1}(t)$ 的代数余子式。对上式两边求导并记 $A_{ij}^k(t)$ 为对 $A_{ij}(t)$ 的第 $k$ 列元素求导后得到的行列式，则

$$\begin{array}{rl}\frac{d}{dt}F(t)& =\frac{d}{dt}\left(\sum _{i=1}^n{f}_{i1}(t)A_{i1}(t)\right)=\sum _{i=1}^n{f}_{i1}^{\prime }(t)A_{i1}(t)+\sum _{i=1}^n{f}_{i1}(t)A_{i1}^{\prime }(t)\\ & =F_1(t)+\sum _{i=1}^n{f}_{i1}(t)\left(\sum _{k=1}^{n-1}{A}_{i1}^k(t)\right)=F_1(t)+\sum _{k=1}^{n-1}\sum _{i=1}^n{f}_{i1}(t)A_{i1}^k(t).\end{array}$$

由行列式的定义可知

$$\sum _{i=1}^n{f}_{i1}(t)A_{i1}^k(t)=F_{k+1}(t)(1\le k\le n-1),$$

故

$$\frac{d}{dt}F(t)=F_1(t)+F_2(t)+\cdots +F_n(t).\square$$