zhangzujin 发表于 2023-1-28 16:27:40

张祖锦常用结论01行列式的迅速降阶法

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# 行列式的迅速降阶法
设 $\displaystyle A$ 是 $\displaystyle m\times n$ 矩阵, $\displaystyle B$ 是 $\displaystyle n\times m$ 矩阵, $\displaystyle E\_k$ 表示 $\displaystyle k$ 阶单位矩阵, 则
\begin\{aligned\} \lambda^n\cdot |\lambda E\_m-AB|=\lambda^m\cdot |\lambda E\_n-BA|,\left(\mbox\{$\lambda$ 是复数\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}

取 $\displaystyle \lambda=1$, 得到如下计算行列式的常用形式:
\begin\{aligned\} |E\_m+AB|=|E\_n+BA|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}

而如果 $\displaystyle m=1$, 则立马得到 $\displaystyle n$ 阶行列式的值!



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(1)、 当 $\displaystyle \lambda=0$ 时, 结论自明.



(2)、 当 $\displaystyle \lambda\neq 0$ 时, 由
\begin\{aligned\} \left(\begin\{array\}\{cccccccccc\} E\_m&-A\\\\ 0&E\_n\end\{array\}\right)\left(\begin\{array\}\{cccccccccc\} \lambda E\_m&A\\\\ B&E\_n\end\{array\}\right)\left(\begin\{array\}\{cccccccccc\} E\_m&0\\\\ -B&E\_n\end\{array\}\right)&=\left(\begin\{array\}\{cccccccccc\} \lambda E\_m-AB&0\\\\ 0&E\_n\end\{array\}\right),\\\\ \left(\begin\{array\}\{cccccccccc\} E\_m&0\\\\ -\frac\{1\}\{\lambda\}B&E\_n\end\{array\}\right)\left(\begin\{array\}\{cccccccccc\} \lambda E\_m&A\\\\ B&E\_n\end\{array\}\right)\left(\begin\{array\}\{cccccccccc\} E\_m&-\frac\{1\}\{\lambda\}A\\\\ 0&E\_n\end\{array\}\right)&=\left(\begin\{array\}\{cccccccccc\} \lambda E\_m&0\\\\ 0&E\_n-\frac\{1\}\{\lambda \}BA\end\{array\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}


\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\} \lambda E\_m-AB&0\\\\ 0&E\_n\end\{array\}\right|=\left|\begin\{array\}\{cccccccccc\} \lambda E\_m&A\\\\ B&E\_n\end\{array\}\right|=\left|\begin\{array\}\{cccccccccc\} \lambda E\_m&0\\\\0&E\_n-\frac\{1\}\{\lambda \}BA\end\{array\}\right|, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}


\begin\{aligned\} |\lambda E\_m-AB| =\lambda ^m \left|E\_n-\frac\{1\}\{\lambda \}BA\right|=\lambda ^\{m-n\}|\lambda E\_n-BA|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}



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