行列式求导公式与一类行列式的计算

• 行列式求导公式与一类行列式的计算

(1) 证明：

$\dfrac{d}{dt}\left| \begin{matrix} a_{11}\left( t \right)& a_{12}\left( t \right)& \cdots& a_{1n}\left( t \right)\\ a_{21}\left( t \right)& a_{22}\left( t \right)& \cdots& a_{2n}\left( t \right)\\ \vdots& \vdots& & \vdots\\ a_{n1}\left( t \right)& a_{n2}\left( t \right)& \cdots& a_{nn}\left( t \right)\\ \end{matrix} \right|=\sum_{j=1}^n{\left| \begin{matrix} a_{11}\left( t \right)& \cdots& \dfrac{d}{dt}a_{1j}\left( t \right)& \cdots& a_{1n}\left( t \right)\\ a_{21}\left( t \right)& \cdots& \dfrac{d}{dt}a_{2j}\left( t \right)& \cdots& a_{2n}\left( t \right)\\ \vdots& & \vdots& & \vdots\\ a_{n1}\left( t \right)& \cdots& \dfrac{d}{dt}a_{nj}\left( t \right)& \cdots& a_{nn}\left( t \right)\\ \end{matrix} \right|}$

(2) 计算如下行列式：

$\left| \begin{matrix} \lambda& a& a& a& \cdots& a\\ b& \alpha& \beta& \beta& \cdots& \beta\\ b& \beta& \alpha& \beta& \cdots& \beta\\ b& \beta& \beta& \alpha& \cdots& \beta\\ \vdots& \vdots& \vdots& \vdots& & \vdots\\ b& \beta& \beta& \beta& \cdots& \alpha\\ \end{matrix} \right|$

(1) 证明： 按照行列式展开定义证明。

$\text{左边}=\dfrac{d}{dt}\sum\limits_{i_1i_2\cdots i_n}a_{i_11}(t)a_{i_22}(t)\cdots a_{i_nn}(t)\\ =\sum\limits_{i_1i_2\cdots i_n}\dfrac{d}{dt}(a_{i_11}(t)a_{i_22}(t)\cdots a_{i_nn}(t))\\ =\sum\limits_{i_1i_2\cdots i_n}\sum \limits_{j=1}^{n}a_{i_11}(t)\cdots\dfrac{d}{dt}a_{i_jj}(t)\cdots a_{i_nn}(t)\\ =\sum\limits_{j=1}^{n}\sum\limits_{i_1i_2\cdots i_n}a_{i_11}(t)\cdots\dfrac{d}{dt}a_{i_jj}(t)\cdots a_{i_nn}(t) =\text{右边}.$

(2) 思路: 第二行起到第 $n-1$ 行都减去第 $n$ 行,然后将第二列及以后的各列都加到最后一列,再按第一列来展开行列式.
解 先设 $n \geqslant 2$. 这时将行列式按提示的步骤去做, 逐步得到

$\text { 原行列式 }=\left|\begin{array}{cccccc} \lambda & a & a & \cdots & a & a \\ 0 & \alpha-\beta & 0 & \cdots & 0 & \beta-\alpha \\ 0 & 0 & \alpha-\beta & \cdots & 0 & \beta-\alpha \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & \alpha-\beta & \beta-\alpha \\ b & \beta & \beta & \cdots & \beta & \alpha \end{array}\right|$

$\begin{aligned} &=\left|\begin{array}{cccccc} \lambda & a & a & \cdots & a & (n-1) a \\ 0 & \alpha-\beta & 0 & \cdots & 0 & 0 \\ 0 & 0 & \alpha-\beta & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & \alpha-\beta & 0 \\ b & \beta & \beta & \cdots & \beta & \alpha+(n-2) \beta \end{array}\right|\\ &=\lambda\left|\begin{array}{ccccc} \alpha-\beta & 0 & \cdots & 0 & 0 \\ 0 & \alpha-\beta & \cdots & 0 & 0 \\ \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & \cdots & \alpha-\beta & 0 \\ \beta & \beta & \cdots & \beta & \alpha+(n-2) \beta \end{array}\right|+\\ &(-1)^{n+1} b\left|\begin{array}{ccccc} a & a & \cdots & a & (n-1) a \\ \alpha-\beta & 0 & \cdots & 0 & 0 \\ 0 & \alpha-\beta & \cdots & 0 & 0 \\ \vdots & \vdots & & \vdots & \vdots \\ 0 & 0 & \cdots & \alpha-\beta & 0 \end{array}\right|\\ &=\lambda(\alpha-\beta)^{n-2}[(n-2) \beta+\alpha]+(-1)^{n+1} b(-1)^{n}(n-1) a(\alpha-\beta)^{n-2}\\ &=[\lambda \alpha+(n-2) \lambda \beta-(n-1) a b](\alpha-\beta)^{n-2} \text {. }\\ \end{aligned}$

$\text { 又 } n=1 \text { 时, 原行列式 }=\lambda \text {. }$