行列式展开及应用
| 序号 | 题型 | 秒杀步骤 | 出题概率 | |-------|--------------------------------------|----------|----------| | 考点1 | 1.行列式展开<br>2.余子式、代数余子式 | 3 | 100% | | 考点2 | 3.克拉默法则 | 2步 | 70% |
行列式转置
$ D= \left| \begin{array}{cccc} a_{11} & a_{12} & ... & a_{1n} \\ a_{21} & a_{22} & ... & a_{2n} \\ ...&...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{array} \right| , D^T= \left| \begin{array}{cccc} a_{11} & a_{21} & ... & a_{n1} \\ a_{12} & a_{22} & ... & a_{n2} \\ ...&...&...&...\\ a_{1n} & a_{2n} & ... & a_{nn}\\ \end{array} \right| =D $
行列式的值：=det(a_{ij})$
例
$ D= \left| \begin{array}{cc} 5&1\\ 8&2\\ \end{array} \right| =2 \Longleftrightarrow D^T= \left| \begin{array}{cc} 5&8\\ 1&2\\ \end{array} \right| =2 $
行列式的展开
性质3
> [!summary] 性质3 > 如果行列式的某行(列）的各元素是两个元素之和，那么这个行列式等于两个行列式的和。 > $ > D= > \left| > \begin{array}{cccc} > a_{11} & \textcolor{lightgreen}{a_{12}}\textcolor{magenta}{+}\textcolor{pink}{b_{12}} & a_{13} \\ > a_{21} & \textcolor{lightgreen}{a_{22}}\textcolor{magenta}{+}\textcolor{pink}{b_{22}} & a_{23} \\ > a_{31} & \textcolor{lightgreen}{a_{32}}\textcolor{magenta}{+}\textcolor{pink}{b_{32}} & a_{33}\\ > \end{array} > \right| > = > \left| > \begin{array}{cccc} > a_{11} & \textcolor{lightgreen}{a_{12}} & a_{13} \\ > a_{21} & \textcolor{lightgreen}{a_{22}} & a_{23} \\ > a_{31} & \textcolor{lightgreen}{a_{32}} & a_{33}\\ > \end{array} > \right| > \textcolor{red}{+} > \left| > \begin{array}{cccc} > a_{11} & \textcolor{pink}{b_{12}} & a_{13} \\ > a_{21} & \textcolor{pink}{b_{22}} & a_{23} \\ > a_{31} & \textcolor{pink}{b_{32}} & a_{33}\\ > \end{array} > \right| > $
例题
> [!question] 题干 > 已知 > > $ > \left| > \begin{array}{ccc} > \textcolor{pink}{a_1} & \textcolor{lightgreen}{b_1} & \textcolor{cyan}{c_1}\\ > \textcolor{pink}{a_2} & \textcolor{lightgreen}{b_2} & \textcolor{cyan}{c_2}\\ > \textcolor{pink}{a_3} & \textcolor{lightgreen}{b_3} & \textcolor{cyan}{c_3}\\ > \end{array} > \right| > =1 > $ > 求行列式 > $ > \left| > \begin{array}{ccc} > \textcolor{pink}{a_1} + \textcolor{cyan}{c_1} & \textcolor{lightgreen}{b_1} & \textcolor{pink}{a_1} + \textcolor{lightgreen}{b_1}\\ > \textcolor{pink}{a_2} + \textcolor{cyan}{c_2} & \textcolor{lightgreen}{b_2} & \textcolor{pink}{a_2} + \textcolor{lightgreen}{b_2}\\ > \textcolor{pink}{a_3} + \textcolor{cyan}{c_3} & \textcolor{lightgreen}{b_3} & \textcolor{pink}{a_3} + \textcolor{lightgreen}{b_3}\\ > \end{array} > \right| > $ > 的值
> [!done]- 解 > 第一步：把行列式的第一组加号拆解为两个行列式 > > $ > \left| > \begin{array}{ccc} > \textcolor{pink}{a_1} \textcolor{magenta}{+} \textcolor{cyan}{c_1} & \textcolor{lightgreen}{b_1} & \textcolor{pink}{a_1} + \textcolor{lightgreen}{b_1}\\ > \textcolor{pink}{a_2} \textcolor{magenta}{+} \textcolor{cyan}{c_2} & \textcolor{lightgreen}{b_2} & \textcolor{pink}{a_2} + \textcolor{lightgreen}{b_2}\\ > \textcolor{pink}{a_3} \textcolor{magenta}{+} \textcolor{cyan}{c_3} & \textcolor{lightgreen}{b_3} & \textcolor{pink}{a_3} + \textcolor{lightgreen}{b_3}\\ > \end{array} > \right| > \xlongequal[]{性质3} > \left| > \begin{array}{ccc} > \textcolor{pink}{a_1} & \textcolor{lightgreen}{b_1} & \textcolor{pink}{a_1} + \textcolor{lightgreen}{b_1}\\ > \textcolor{pink}{a_2} & \textcolor{lightgreen}{b_2} & \textcolor{pink}{a_2} + \textcolor{lightgreen}{b_2}\\ > \textcolor{pink}{a_3} & \textcolor{lightgreen}{b_3} & \textcolor{pink}{a_3} + \textcolor{lightgreen}{b_3}\\ > \end{array} > \right| > \textcolor{red}{+} > \left| > \begin{array}{ccc} > \textcolor{cyan}{c_1} & \textcolor{lightgreen}{b_1} & \textcolor{pink}{a_1} + \textcolor{lightgreen}{b_1}\\ > \textcolor{cyan}{c_2} & \textcolor{lightgreen}{b_2} & \textcolor{pink}{a_2} + \textcolor{lightgreen}{b_2}\\ > \textcolor{cyan}{c_3} & \textcolor{lightgreen}{b_3} & \textcolor{pink}{a_3} + \textcolor{lightgreen}{b_3}\\ > \end{array} > \right| > $ > > > 第二步：把两个行列式的加号拆解成2个行列式，变成四个。 > > $ > \left| > \begin{array}{ccc} > \textcolor{pink}{a_1} & \textcolor{lightgreen}{b_1} & \textcolor{pink}{a_1} \textcolor{magenta}{+} \textcolor{lightgreen}{b_1}\\ > \textcolor{pink}{a_2} & \textcolor{lightgreen}{b_2} & \textcolor{pink}{a_2} \textcolor{magenta}{+} \textcolor{lightgreen}{b_2}\\ > \textcolor{pink}{a_3} & \textcolor{lightgreen}{b_3} & \textcolor{pink}{a_3} \textcolor{magenta}{+} \textcolor{lightgreen}{b_3}\\ > \end{array} > \right| > \textcolor{black}{+} > \left| > \begin{array}{ccc} > \textcolor{cyan}{c_1} & \textcolor{lightgreen}{b_1} & \textcolor{pink}{a_1} \textcolor{magenta}{+} \textcolor{lightgreen}{b_1}\\ > \textcolor{cyan}{c_2} & \textcolor{lightgreen}{b_2} & \textcolor{pink}{a_2} \textcolor{magenta}{+} \textcolor{lightgreen}{b_2}\\ > \textcolor{cyan}{c_3} & \textcolor{lightgreen}{b_3} & \textcolor{pink}{a_3} \textcolor{magenta}{+} \textcolor{lightgreen}{b_3}\\ > \end{array} > \right| > > \\\xlongequal[]{性质3} > > \left| > \begin{array}{ccc} > \textcolor{pink}{a_1} & \textcolor{lightgreen}{b_1} & \textcolor{pink}{a_1}\\ > \textcolor{pink}{a_2} & \textcolor{lightgreen}{b_2} & \textcolor{pink}{a_2}\\ > \textcolor{pink}{a_3} & \textcolor{lightgreen}{b_3} & \textcolor{pink}{a_3}\\ > \end{array} > \right| > \textcolor{red}{+} > \left| > \begin{array}{ccc} > \textcolor{pink}{a_1} & \textcolor{lightgreen}{b_1} & \textcolor{lightgreen}{b_1}\\ > \textcolor{pink}{a_2} & \textcolor{lightgreen}{b_2} & \textcolor{lightgreen}{b_2}\\ > \textcolor{pink}{a_3} & \textcolor{lightgreen}{b_3} & \textcolor{lightgreen}{b_3}\\ > \end{array} > \right| > \textcolor{black}{+} > \left| > \begin{array}{ccc} > \textcolor{cyan}{c_1} & \textcolor{lightgreen}{b_1} & \textcolor{pink}{a_1}\\ > \textcolor{cyan}{c_2} & \textcolor{lightgreen}{b_2} & \textcolor{pink}{a_2}\\ > \textcolor{cyan}{c_3} & \textcolor{lightgreen}{b_3} & \textcolor{pink}{a_3}\\ > \end{array} > \right| > \textcolor{red}{+} > \left| > \begin{array}{ccc} > \textcolor{cyan}{c_1} & \textcolor{lightgreen}{b_1} & \textcolor{lightgreen}{b_1}\\ > \textcolor{cyan}{c_2} & \textcolor{lightgreen}{b_2} & \textcolor{lightgreen}{b_2}\\ > \textcolor{cyan}{c_3} & \textcolor{lightgreen}{b_3} & \textcolor{lightgreen}{b_3}\\ > \end{array} > \right| > \\\xlongequal[]{性质5} > 0+0+ > \left| > \begin{array}{ccc} > \textcolor{cyan}{c_1} & \textcolor{lightgreen}{b_1} & \textcolor{pink}{a_1}\\ > \textcolor{cyan}{c_2} & \textcolor{lightgreen}{b_2} & \textcolor{pink}{a_2}\\ > \textcolor{cyan}{c_3} & \textcolor{lightgreen}{b_3} & \textcolor{pink}{a_3}\\ > \end{array} > \right| > +0 > \\= > \left| > \begin{array}{ccc} > \textcolor{cyan}{c_1} & \textcolor{lightgreen}{b_1} & \textcolor{pink}{a_1}\\ > \textcolor{cyan}{c_2} & \textcolor{lightgreen}{b_2} & \textcolor{pink}{a_2}\\ > \textcolor{cyan}{c_3} & \textcolor{lightgreen}{b_3} & \textcolor{pink}{a_3}\\ > \end{array} > \right| > \xlongequal[性质7]{\textcolor{magenta}{r_1}\leftrightarrow \textcolor{magenta}{r_3}} > \textcolor{red}{-} > \left| > \begin{array}{ccc} > \textcolor{pink}{a_1} & \textcolor{lightgreen}{b_1} & \textcolor{cyan}{c_1} \\ > \textcolor{pink}{a_2} & \textcolor{lightgreen}{b_2} & \textcolor{cyan}{c_2} \\ > \textcolor{pink}{a_3} & \textcolor{lightgreen}{b_3} & \textcolor{cyan}{c_3} \\ > \end{array} > \right| > \xlongequal[]{代入题干给的值}-1 > $ > > 答：目标行列式的值为-1
余子式和代数余子式
$ 元素\textcolor{magenta}{a_{ij}}的 \textcolor{green}{余子式}为 \textcolor{red}{M_{ij}}，
\textcolor{green}{代数余子式}为 \textcolor{red}{A_{ij}=(-1)^{i+j}M_{ij}} $
> [!note] 例1 > $ > D= > \left| > \begin{array}{cccc} > \textcolor{pink}{a_{11}} & \textcolor{pink}{a_{12}} & \textcolor{gray}{a_{13}} & \textcolor{pink}{a_{14}} \\ > \textcolor{gray}{a_{21}} & \textcolor{gray}{a_{22}} & \textcolor{cyan}{a_{23}} & \textcolor{gray}{a_{24}} \\ > \textcolor{pink}{a_{31}} & \textcolor{pink}{a_{32}} & \textcolor{gray}{a_{33}} & \textcolor{pink}{a_{34}} \\ > \textcolor{pink}{a_{41}} & \textcolor{pink}{a_{42}} & \textcolor{gray}{a_{43}} & \textcolor{pink}{a_{44}} \\ > > \end{array} > \right| > \\ > \Longrightarrow > \begin{cases} > \textcolor{red}{M_{23}}= > \left| > \begin{array}{cccc} > \textcolor{pink}{a_{11}} & \textcolor{pink}{a_{12}} & \textcolor{pink}{a_{14}} \\ > \textcolor{pink}{a_{31}} & \textcolor{pink}{a_{32}} & \textcolor{pink}{a_{34}} \\ > \textcolor{pink}{a_{41}} & \textcolor{pink}{a_{42}} & \textcolor{pink}{a_{44}} \\ > \end{array} > \right|\\ > \textcolor{red}{A_{23}}=(-1)^{2+3}M_{23}=-M_{23} > \end{cases} > $
> [!note] 例2 > > $ > D= > \left| > \begin{array}{ccc} > \textcolor{pink}{1}&\textcolor{pink}{2}&\textcolor{gray}{3}\\ > \textcolor{gray}{1}&\textcolor{gray}{3}&\textcolor{cyan}{1}\\ > \textcolor{pink}{1}&\textcolor{pink}{3}&\textcolor{gray}{2}\\ > \end{array} > \right| > > \Longrightarrow \\ > \begin{cases} > a_{23}的余子式M_{23}= > \left| > \begin{array}{ccc} > \textcolor{pink}{1}&\textcolor{pink}{2}\\ > \textcolor{pink}{1}&\textcolor{pink}{3}\\ > \end{array} > \right| > =1\times3-1\times2=1\\ > a_{23}的代数余子式A_{23}=(-1)^{2+3}M_{23}=-1 > \end{cases} > $
利用代数余子式求行列式的值，代数余子式公式
> [!hint] 代数余子式公式 > $ > \left| > \begin{array}{cccc} > \textcolor{yellow}{a_{11}} & \textcolor{yellow}{a_{12}} & \textcolor{yellow}{a_{13}} & \textcolor{yellow}{a_{14}} \\ > a_{21} & a_{22} & a_{23} & a_{24} \\ > a_{31} & a_{32} & a_{33} & a_{34} \\ > a_{41} & a_{42} & a_{43} & a_{44} \\ > \end{array} > \right| > = > \textcolor{yellow}{a_{11}}A_{11}+\textcolor{yellow}{a_{12}}A_{12}+\textcolor{yellow}{a_{13}}A_{13}+\textcolor{yellow}{a_{14}}A_{14} > $
降阶
> [!example] 例1:推导出推论1 > > $ > D= > \left| > \begin{array}{cccc} > \textcolor{pink}{a_{11}} & \textcolor{pink}{a_{12}} & \textcolor{gray}{a_{13}} & \textcolor{pink}{a_{24}} \\ > \textcolor{pink}{a_{21}} & \textcolor{pink}{a_{22}} & \textcolor{gray}{a_{23}} & \textcolor{pink}{a_{24}} \\ > \textcolor{yellow}{0} & \textcolor{yellow}{0} & \textcolor{cyan}{a_{33}} & \textcolor{yellow}{0} \\ > \textcolor{pink}{a_{41}} & \textcolor{pink}{a_{42}} & \textcolor{gray}{a_{43}} & \textcolor{pink}{a_{44}} \\ > \end{array} > \right|\\ > =\textcolor{yellow}{0}\times A_{31}+\textcolor{yellow}{0}\times A_{32}+\textcolor{cyan}{a_{33}}A_{31}+\textcolor{yellow}{0}\times A_{33}\\ > =\textcolor{cyan}{a_{33}}A_{31} > =(-1)^{\textcolor{cyan}{3}+\textcolor{cyan}{3}}\textcolor{cyan}{a_{33}}M_{31}\\ > =(-1)^{\textcolor{cyan}{3}+\textcolor{cyan}{3}}\textcolor{cyan}{a_{33}} > \left| > \begin{array}{cccc} > \textcolor{pink}{a_{11}} & \textcolor{pink}{a_{12}} & \textcolor{pink}{a_{24}} \\ > \textcolor{pink}{a_{21}} & \textcolor{pink}{a_{22}} & \textcolor{pink}{a_{24}} \\ > \textcolor{pink}{a_{41}} & \textcolor{pink}{a_{42}} & \textcolor{pink}{a_{44}} \\ > \end{array} > \right| > $
整理后得到推论1：
> [!important] 推论1 > $ > D= > \left| > \begin{array}{cccc} > \textcolor{pink}{a_{11}} & \textcolor{pink}{a_{12}} & \textcolor{gray}{a_{13}} & \textcolor{pink}{a_{24}} \\ > \textcolor{pink}{a_{21}} & \textcolor{pink}{a_{22}} & \textcolor{gray}{a_{23}} & \textcolor{pink}{a_{24}} \\ > \textcolor{yellow}{0} & \textcolor{yellow}{0} & \textcolor{cyan}{a_{33}} & \textcolor{yellow}{0} \\ > \textcolor{pink}{a_{41}} & \textcolor{pink}{a_{42}} & \textcolor{gray}{a_{43}} & \textcolor{pink}{a_{44}} \\ > \end{array} > \right| > > =(-1)^{\textcolor{cyan}{3}+\textcolor{cyan}{3}}\textcolor{cyan}{a_{33}} > > \left| > \begin{array}{cccc} > \textcolor{pink}{a_{11}} & \textcolor{pink}{a_{12}} & \textcolor{pink}{a_{24}} \\ > \textcolor{pink}{a_{21}} & \textcolor{pink}{a_{22}} & \textcolor{pink}{a_{24}} \\ > \textcolor{pink}{a_{41}} & \textcolor{pink}{a_{42}} & \textcolor{pink}{a_{44}} \\ > \end{array} > \right| > $
----
> [!example] 例2:直接套用例1的推论1 > > $ > D= > \left| > \begin{array}{ccccc} > \textcolor{pink}{5} & \textcolor{pink}{3} & \textcolor{pink}{-1} & \textcolor{pink}{2} & \textcolor{gray}{0} \\ > \textcolor{gray}{1} & \textcolor{gray}{7} & \textcolor{gray}{2} & \textcolor{gray}{5} & \textcolor{cyan}{2} \\ > \textcolor{pink}{0} & \textcolor{pink}{-2} & \textcolor{pink}{3} & \textcolor{pink}{1} & \textcolor{gray}{0} \\ > \textcolor{pink}{0} & \textcolor{pink}{-4} & \textcolor{pink}{-1} & \textcolor{pink}{4} & \textcolor{gray}{0} \\ > \textcolor{pink}{0} & \textcolor{pink}{2} & \textcolor{pink}{3} & \textcolor{pink}{5} & \textcolor{gray}{0} \\ > \end{array} > \right| > \xlongequal[]{推论1} > (-1)^{\textcolor{cyan}{2}+\textcolor{cyan}{5}}\times \textcolor{cyan}{2}\times > \left| > \begin{array}{ccccc} > \textcolor{pink}{5} & \textcolor{pink}{3} & \textcolor{pink}{-1} & \textcolor{pink}{2} \\ > \textcolor{pink}{0} & \textcolor{pink}{-2} & \textcolor{pink}{3} & \textcolor{pink}{1} \\ > \textcolor{pink}{0} & \textcolor{pink}{-4} & \textcolor{pink}{-1} & \textcolor{pink}{4} \\ > \textcolor{pink}{0} & \textcolor{pink}{2} & \textcolor{pink}{3} & \textcolor{pink}{5} \\ > \end{array} > \right| > = > -2\times > \left| > \begin{array}{ccccc} > \textcolor{black}{5} & \textcolor{black}{3} & \textcolor{black}{-1} & \textcolor{black}{2} \\ > \textcolor{black}{0} & \textcolor{black}{-2} & \textcolor{black}{3} & \textcolor{black}{1} \\ > \textcolor{black}{0} & \textcolor{black}{-4} & \textcolor{black}{-1} & \textcolor{black}{4} \\ > \textcolor{black}{0} & \textcolor{black}{2} & \textcolor{black}{3} & \textcolor{black}{5} \\ > \end{array} > \right| > $ > > $ > D= > -2\times > \left| > \begin{array}{ccccc} > \textcolor{cyan}{5} & \textcolor{gray}{3} & \textcolor{gray}{-1} & \textcolor{gray}{2} \\ > \textcolor{gray}{0} & \textcolor{pink}{-2} & \textcolor{pink}{3} & \textcolor{pink}{1} \\ > \textcolor{gray}{0} & \textcolor{pink}{-4} & \textcolor{pink}{-1} & \textcolor{pink}{4} \\ > \textcolor{gray}{0} & \textcolor{pink}{2} & \textcolor{pink}{3} & \textcolor{pink}{5} \\ > \end{array} > \right| > \xlongequal[]{推论1} > -2\times \textcolor{cyan}{5} > \left| > \begin{array}{ccccc} > \textcolor{pink}{-2} & \textcolor{pink}{3} & \textcolor{pink}{1} \\ > \textcolor{pink}{-4} & \textcolor{pink}{-1} & \textcolor{pink}{4} \\ > \textcolor{pink}{2} & \textcolor{pink}{3} & \textcolor{pink}{5} \\ > \end{array} > \right| > $ > > $ > D= > -10\times > \left| > \begin{array}{ccccc} > \textcolor{green}{-2} & \textcolor{yellow}{3} & \textcolor{yellow}{1} \\ > \textcolor{magenta}{-4} & \textcolor{magenta}{-1} & \textcolor{magenta}{4} \\ > \textcolor{magenta}{2} & \textcolor{magenta}{3} & \textcolor{magenta}{5} \\ > \end{array} > \right| > \xlongequal[\textcolor{magenta}{r_3}+\textcolor{yellow}{r_1}\rightarrow \textcolor{red}{r_3}]{\textcolor{magenta}{r_2}-2\textcolor{yellow}{r_1}\rightarrow \textcolor{red}{r_2}} > -10\times > \left| > \begin{array}{ccccc} > \textcolor{green}{-2} & \textcolor{yellow}{3} & \textcolor{yellow}{1} \\ > \textcolor{red}{0} & \textcolor{red}{-7} & \textcolor{red}{2} \\ > \textcolor{red}{0} & \textcolor{red}{6} & \textcolor{red}{6} \\ > \end{array} > \right| > $ > > $ > D= > -10\times > \left| > \begin{array}{ccccc} > \textcolor{cyan}{-2} & \textcolor{gray}{3} & \textcolor{gray}{1} \\ > \textcolor{gray}{0} & \textcolor{pink}{-7} & \textcolor{pink}{2} \\ > \textcolor{gray}{0} & \textcolor{pink}{6} & \textcolor{pink}{6} \\ > \end{array} > \right| > \xlongequal[]{推论1} > -10\times(\textcolor{cyan}{-2}) > \left| > \begin{array}{ccccc} > \textcolor{pink}{-7} & \textcolor{pink}{2} \\ > \textcolor{pink}{6} & \textcolor{pink}{6} \\ > \end{array} > \right| > =20\times (-42-12)=-1080 > $ -----
题型
> [!question] 题干 > 已知 > > $ > D= > \left| > \begin{array}{cccc} > \textcolor{black}{3} & \textcolor{black}{-5} & \textcolor{black}{2} & \textcolor{black}{1}\\ > \textcolor{black}{1} & \textcolor{black}{1} & \textcolor{black}{0} & \textcolor{black}{-5}\\ > \textcolor{black}{-1} & \textcolor{black}{3} & \textcolor{black}{1} & \textcolor{black}{3}\\ > \textcolor{black}{2} & \textcolor{black}{-4} & \textcolor{black}{-1} & \textcolor{black}{-3}\\ > \end{array} > \right| > > $ > 求{11}+A_{12}+A_{13}+A_{14}$和{11}+M_{12}+M_{13}+M_{14}$
> [!hint]- 思路 > 把M改造成A，再改变行列式的系数
> [!done]- 解 > > 由代数余子式定义得到： > > $ > D=a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}+a_{14}A_{14}= > \left| > \begin{array}{cccc} > a_{11}&a_{12}&a_{13}&a_{14}\\ > a_{21}&a_{22}&a_{23}&a_{24}\\ > a_{31}&a_{32}&a_{33}&a_{34}\\ > a_{41}&a_{42}&a_{43}&a_{44}\\ > \end{array} > \right| > $ > > 由于代数余子式{ij}$与{ij}$的值无关，因此替换{11}、a_{12}、a_{13}、a_{14}$为1，A不变。 > > 因此， > > $ > A_{11}+A_{12}+A_{13}+A_{14}\xlongequal[]{A_{ij}与a_{ij}的值无关} > \left| > \begin{array}{cccc} > \textcolor{red}{1} & \textcolor{red}{1} & \textcolor{red}{1} & \textcolor{red}{1}\\ > \textcolor{black}{1} & \textcolor{black}{1} & \textcolor{black}{0} & \textcolor{black}{-5}\\ > \textcolor{black}{-1} & \textcolor{black}{3} & \textcolor{black}{1} & \textcolor{black}{3}\\ > \textcolor{black}{2} & \textcolor{black}{-4} & \textcolor{black}{-1} & \textcolor{black}{-3}\\ > \end{array} > \right| > $ > > > [!hint] 思路 > > 把行列式构造成可以利用代数余子式求值的形式 > > $ > A_{11}+A_{12}+A_{13}+A_{14}= > \left| > \begin{array}{cccc} > \textcolor{green}{1} & \textcolor{yellow}{1} & \textcolor{yellow}{1} & \textcolor{yellow}{1}\\ > \textcolor{black}{1} & \textcolor{black}{1} & \textcolor{black}{0} & \textcolor{black}{-5}\\ > \textcolor{green}{-1} & \textcolor{magenta}{3} & \textcolor{magenta}{1} & \textcolor{magenta}{3}\\ > \textcolor{magenta}{2} & \textcolor{magenta}{-4} & \textcolor{magenta}{-1} & \textcolor{magenta}{-3}\\ > \end{array} > \right| > \xlongequal[\textcolor{magenta}{r_3}-\textcolor{yellow}{r_1}\rightarrow \textcolor{red}{r_3}]{\textcolor{magenta}{r_4}+\textcolor{yellow}{r3} \rightarrow \textcolor{red}{r_4}} > \left| > \begin{array}{cccc} > \textcolor{black}{1} & \textcolor{black}{1} & \textcolor{black}{1} & \textcolor{black}{1}\\ > \textcolor{black}{1} & \textcolor{black}{1} & \textcolor{black}{0} & \textcolor{black}{-5}\\ > \textcolor{red}{-2} & \textcolor{red}{2} & \textcolor{red}{0} & \textcolor{red}{2}\\ > \textcolor{red}{1} & \textcolor{red}{-1} & \textcolor{red}{0} & \textcolor{red}{0}\\ > \end{array} > \right| > $ > > $ > A_{11}+A_{12}+A_{13}+A_{14}= > \left| > \begin{array}{cccc} > \textcolor{gray}{1} & \textcolor{gray}{1} & \textcolor{cyan}{1} & \textcolor{gray}{1}\\ > \textcolor{pink}{1} & \textcolor{pink}{1} & \textcolor{gray}{0} & \textcolor{pink}{-5}\\ > \textcolor{pink}{-2} & \textcolor{pink}{2} & \textcolor{gray}{0} & \textcolor{pink}{2}\\ > \textcolor{pink}{1} & \textcolor{pink}{-1} & \textcolor{gray}{0} & \textcolor{pink}{0}\\ > \end{array} > \right| > \xlongequal[]{推论1} > (-1)^{\textcolor{cyan}{1}+\textcolor{cyan}{3}}\times\textcolor{cyan}{1}\times > \left| > \begin{array}{cccc} > \textcolor{pink}{1} & \textcolor{pink}{1} & \textcolor{pink}{-5}\\ > \textcolor{pink}{-2} & \textcolor{pink}{2} & \textcolor{pink}{2}\\ > \textcolor{pink}{1} & \textcolor{pink}{-1} & \textcolor{pink}{0}\\ > \end{array} > \right| > = > \left| > \begin{array}{cccc} > \textcolor{black}{1} & \textcolor{black}{1} & \textcolor{black}{-5}\\ > \textcolor{black}{-2} & \textcolor{black}{2} & \textcolor{black}{2}\\ > \textcolor{black}{1} & \textcolor{black}{-1} & \textcolor{black}{0}\\ > \end{array} > \right| > $ > > $ > A_{11}+A_{12}+A_{13}+A_{14}= > \left| > \begin{array}{cccc} > \textcolor{yellow}{1} & \textcolor{magenta}{1} & \textcolor{black}{-5}\\ > \textcolor{yellow}{-2} & \textcolor{magenta}{2} & \textcolor{black}{2}\\ > \textcolor{yellow}{1} & \textcolor{magenta}{-1} & \textcolor{black}{0}\\ > \end{array} > \right| > \xlongequal[]{\textcolor{magenta}{c_2}+\textcolor{yellow}{c_1}\rightarrow \textcolor{red}{c_2}} > \left| > \begin{array}{cccc} > \textcolor{black}{1} & \textcolor{red}{2} & \textcolor{black}{-5}\\ > \textcolor{black}{-2} & \textcolor{red}{0} & \textcolor{black}{2}\\ > \textcolor{black}{1} & \textcolor{red}{0} & \textcolor{black}{0}\\ > \end{array} > \right| > $ > > $ > A_{11}+A_{12}+A_{13}+A_{14}= > \left| > \begin{array}{cccc} > \textcolor{gray}{1} & \textcolor{cyan}{2} & \textcolor{gray}{-5}\\ > \textcolor{pink}{-2} & \textcolor{gray}{0} & \textcolor{pink}{2}\\ > \textcolor{pink}{1} & \textcolor{gray}{0} & \textcolor{pink}{0}\\ > \end{array} > \right| > \xlongequal[]{推论1} > (-1)^{\textcolor{cyan}{1}+\textcolor{cyan}{2}}\times > \textcolor{cyan}{2} > \times > \left| > \begin{array}{ccc} > \textcolor{pink}{-2} & \textcolor{pink}{2}\\ > \textcolor{pink}{1} & \textcolor{pink}{0}\\ > \end{array} > \right| > =-2\times(0-2)=4 > $ > > 下一步求解{11}+M_{12}+M_{13}+M_{14}$ > > > [!hint] 思路 > > 把M构造成A加减的形式 > $ > M_{11}+M_{12}+M_{13}+M_{14} > \xlongequal[]{根据A的定义构造} > (-1)^{1+1}M_{11}- > (-1)^{2+1}M_{21}+ > (-1)^{3+1}M_{31}- > (-1)^{4+1}M_{41}\\ > =A_{11}-A_{21}+A_{31}-A_{41} > =\textcolor{yellow}{1}\times A_{11}+\textcolor{yellow}{(-1)}\times A_{21}+\textcolor{yellow}{1}\times A_{31}+\textcolor{yellow}{(-1)}\times A_{41} > $ > 由于代数余子式{ij}$与{ij}$的值无关，因此替换{11}、a_{21}、a_{31}、a_{41}$为1，A不变。 > $ > M_{11}+M_{12}+M_{13}+M_{14}= > =\textcolor{yellow}{1}\times A_{11}+\textcolor{yellow}{(-1)}\times A_{21}+\textcolor{yellow}{1}\times A_{31}+\textcolor{yellow}{(-1)}\times A_{41}\\ > \xlongequal[]{A_{ij}与a_{ij}的值无关} > \left| > \begin{array}{cccc} > \textcolor{yellow}{1} & \textcolor{black}{-5} & \textcolor{black}{2} & \textcolor{black}{1}\\ > \textcolor{yellow}{-1} & \textcolor{black}{1} & \textcolor{black}{0} & \textcolor{black}{-5}\\ > \textcolor{yellow}{1} & \textcolor{black}{3} & \textcolor{black}{1} & \textcolor{black}{3}\\ > \textcolor{yellow}{-1} & \textcolor{black}{-4} & \textcolor{black}{-1} & \textcolor{black}{-3}\\ > \end{array} > \right| > $ > > $ > M_{11}+M_{12}+M_{13}+M_{14}= > \left| > \begin{array}{cccc} > \textcolor{black}{1} & \textcolor{black}{-5} & \textcolor{black}{2} & \textcolor{black}{1}\\ > \textcolor{black}{-1} & \textcolor{black}{1} & \textcolor{black}{0} & \textcolor{black}{-5}\\ > \textcolor{yellow}{1} & \textcolor{yellow}{3} & \textcolor{yellow}{1} & \textcolor{yellow}{3}\\ > \textcolor{magenta}{-1} & \textcolor{magenta}{-4} & \textcolor{magenta}{-1} & \textcolor{magenta}{-3}\\ > \end{array} > \right| > \xlongequal[]{\textcolor{magenta}{r_4}+\textcolor{yellow}{r_3}\rightarrow \textcolor{red}{r_4}} > \left| > \begin{array}{cccc} > \textcolor{black}{1} & \textcolor{black}{-5} & \textcolor{black}{2} & \textcolor{black}{1}\\ > \textcolor{black}{-1} & \textcolor{black}{1} & \textcolor{black}{0} & \textcolor{black}{-5}\\ > \textcolor{black}{1} & \textcolor{black}{3} & \textcolor{black}{1} & \textcolor{black}{3}\\ > \textcolor{red}{0} & \textcolor{red}{-1} & \textcolor{red}{0} & \textcolor{red}{0}\\ > \end{array} > \right| > $ > > $ > M_{11}+M_{12}+M_{13}+M_{14}= > \left| > \begin{array}{cccc} > \textcolor{pink}{1} & \textcolor{gray}{-5} & \textcolor{pink}{2} & \textcolor{pink}{1}\\ > \textcolor{pink}{-1} & \textcolor{gray}{1} & \textcolor{pink}{0} & \textcolor{pink}{-5}\\ > \textcolor{pink}{1} & \textcolor{gray}{3} & \textcolor{pink}{1} & \textcolor{pink}{3}\\ > \textcolor{gray}{0} & \textcolor{cyan}{-1} & \textcolor{gray}{0} & \textcolor{gray}{0}\\ > \end{array} > \right| > \xlongequal[]{推论1} > (-1)^{\textcolor{cyan}{4}+\textcolor{cyan}{2}}\times\textcolor{cyan}{-1}\times > \left| > \begin{array}{cccc} > \textcolor{pink}{1} & \textcolor{pink}{2} & \textcolor{pink}{1}\\ > \textcolor{pink}{-1} & \textcolor{pink}{0} & \textcolor{pink}{-5}\\ > \textcolor{pink}{1} & \textcolor{pink}{1} & \textcolor{pink}{3}\\ > \end{array} > \right| > = > - > \left| > \begin{array}{cccc} > \textcolor{black}{1} & \textcolor{black}{2} & \textcolor{black}{1}\\ > \textcolor{black}{-1} & \textcolor{black}{0} & \textcolor{black}{-5}\\ > \textcolor{black}{1} & \textcolor{black}{1} & \textcolor{black}{3}\\ > \end{array} > \right| > $ > > $ > M_{11}+M_{12}+M_{13}+M_{14}= > - > \left| > \begin{array}{cccc} > \textcolor{magenta}{1} & \textcolor{magenta}{2} & \textcolor{magenta}{1}\\ > \textcolor{black}{-1} & \textcolor{black}{0} & \textcolor{black}{-5}\\ > \textcolor{yellow}{1} & \textcolor{yellow}{1} & \textcolor{yellow}{3}\\ > \end{array} > \right| > \xlongequal[]{\textcolor{magenta}{r_1}-2\textcolor{yellow}{r_3}\rightarrow \textcolor{red}{r_3}} > - > \left| > \begin{array}{cccc} > \textcolor{red}{-1} & \textcolor{red}{0} & \textcolor{red}{-5}\\ > \textcolor{black}{-1} & \textcolor{black}{0} & \textcolor{black}{-5}\\ > \textcolor{black}{1} & \textcolor{black}{1} & \textcolor{black}{3}\\ > \end{array} > \right| > $ > > $ > M_{11}+M_{12}+M_{13}+M_{14}= > - > \left| > \begin{array}{cccc} > \textcolor{pink}{-1} & \textcolor{gray}{0} & \textcolor{pink}{-5}\\ > \textcolor{pink}{-1} & \textcolor{gray}{0} & \textcolor{pink}{-5}\\ > \textcolor{gray}{1} & \textcolor{cyan}{1} & \textcolor{gray}{3}\\ > \end{array} > \right| > \xlongequal[]{推论1} > -1\times\textcolor{cyan}{1}\times(-1)^{\textcolor{cyan}{3}+\textcolor{cyan}{2}}\times > \left| > \begin{array}{cccc} > \textcolor{pink}{-1} & \textcolor{pink}{-5}\\ > \textcolor{pink}{-1} & \textcolor{pink}{-5}\\ > \end{array} > \right| > =0 > $ > 答： > $ > A_{11}+A_{12}+A_{13}+A_{14}=4\\ > M_{11}+M_{12}+M_{13}+M_{14}=0 > $
克拉默法则
> [!question] 例：求解二元线性方程组 > > $ > \begin{cases} > \textcolor{pink}{3}x_1-\textcolor{lightgreen}{2}x_2=\textcolor{cyan}{12}\\ > \textcolor{pink}{2}x_1+x_2=\textcolor{cyan}{1} > \end{cases} > $
> [!done]- 解 > $ > D= > \left| > \begin{array}{cc} > \textcolor{pink}{3}&\textcolor{lightgreen}{-2}\\ > \textcolor{pink}{2}&\textcolor{lightgreen}{1} > \end{array} > \right| > =3-(-4)=7\ne0 > $ > > $ > D_1= > \left| > \begin{array}{cc} > \textcolor{cyan}{12}&\textcolor{lightgreen}{-2}\\ > \textcolor{cyan}{1}&\textcolor{lightgreen}{1} > \end{array} > \right| > =12-(-2)=14 > $ > > $ > D_2= > \left| > \begin{array}{cc} > \textcolor{pink}{3}&\textcolor{cyan}{12}\\ > \textcolor{pink}{2}&\textcolor{cyan}{1} > \end{array} > \right| > =3-24=-21 > $ > > $ > x_1=\frac{D_1}{D}=\frac{14}{7}=2\\ > x_2=\frac{D_2}{D}=\frac{-21}{7}=-3\\ > $
> [!hint]- 推广 > =\frac{D_i}{D}$
题型
• 【概念】齐次方程：常数项都为0的方程
> [!question] 题干 > 问$\lambda$取何值时，齐次方程组$ > \begin{cases} > (1-\lambda)x_1-2x_2+4x_3=0\\ > 2x_1+(3-\lambda)x_2+x_3=0\\ > x_1+x_2+(1-\lambda)x_3=0 > \end{cases} > $有非零解
> [!done]- 解（通过该方法“反证”推论，只是帮助记忆，没有数学道理） > > $ > D= > \left| > \begin{array}{ccc} > 1-\lambda&-2&4\\ > 2&3-\lambda&1\\ > 1&1&1-\lambda > \end{array} > \right| > $ > > $ > D_1= > \left| > \begin{array}{ccc} > 0&-2&4\\ > 0&3-\lambda&1\\ > 0&1&1-\lambda > \end{array} > \right| > \xlongequal[]{代数余子式公式}0 > $ > 同理$和$都为0 > > 则根据克拉默法则，推出=x_2=x_3=0$ > > 影出推论：如果齐次方程组有非0解，则必有D=0 （这个应该是克拉默法则的一个规则） > $ > D= > \left| > \begin{array}{ccc} > 1-\lambda&-2&4\\ > 2&3-\lambda&1\\ > 1&1&1-\lambda > \end{array} > \right| > = > \left| > \begin{array}{ccc} > 1-\lambda&-2&4\\ > 0&1-\lambda&2\lambda-1\\ > 1&1&1-\lambda > \end{array} > \right| > =-\lambda(\lambda-2)(\lambda-3) > $ > > $ > \Longrightarrow > 当\lambda为0、2、3时，齐次方程组有非零解x_i=\frac{D_i}{D} > $
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[[线性代数：课时1]]