1. 矩阵乘法

1.1 常规方法

$$\underset{A_{m\ast n}}{\underbrace{\left[\begin{array}{cccc}...& ...& ...& ...\\ a_{31}& a_{32}& a_{33}& a_{34}\\ ...& ...& ...& ...\end{array}\right]}}\underset{B_{n\ast p}}{\underbrace{\left[\begin{array}{cccc}...& ...& ...& b_{14}\\ ...& ...& ...& b_{24}\\ ...& ...& ...& b_{34}\\ ...& ...& ...& b_{44}\end{array}\right]}}=\underset{C_{m\ast p}}{\underbrace{\left[\begin{array}{cccc}...& ...& ...& ...\\ ...& ...& ...& C_{34}\\ ...& ...& ...& ...\end{array}\right]}}$$
$$C_{34}=A_{ro{w}_3}\ast B_{co{l}_4}=\sum _{i=1}^n{a}_{3i}\ast b_{i4}$$

1.2 列向量组合

已知
$$\begin{array}{rl}\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]\left[\begin{array}{c}{B}_{11}\\ B_{21}\\ B_{31}\end{array}\right]& =B_{11}\ast A_{col1}+B_{21}\ast A_{col2}+B_{31}\ast A_{col3}\\ & =\left[\begin{array}{c}{B}_{11}\ast A_{11}+B_{21}\ast A_{12}+B_{31}\ast A_{13}\\ B_{11}\ast A_{21}+B_{21}\ast A_{22}+B_{31}\ast A_{23}\\ B_{11}\ast A_{31}+B_{21}\ast A_{32}+B_{31}\ast A_{33}\end{array}\right]\end{array}$$
那么
$$\begin{array}{rl}\underset{A}{\underbrace{\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]}}\underset{B}{\underbrace{\left[\begin{array}{cc}{B}_{11}& B_{12}\\ B_{21}& B_{22}\\ B_{31}& B_{32}\end{array}\right]}}& =\underset{C}{\underbrace{\left[\begin{array}{cc}{B}_{11}\ast A_{col1}+B_{21}\ast A_{col2}+B_{31}\ast A_{col3}& B_{12}\ast A_{col1}+B_{22}\ast A_{col2}+B_{32}\ast A_{col3}\end{array}\right]}}\\ & =\left[\begin{array}{cc}{B}_{11}\ast A_{11}+B_{21}\ast A_{12}+B_{31}\ast A_{13}& B_{12}\ast A_{11}+B_{22}\ast A_{12}+B_{32}\ast A_{13}\\ B_{11}\ast A_{21}+B_{21}\ast A_{22}+B_{31}\ast A_{23}& B_{12}\ast A_{21}+B_{22}\ast A_{22}+B_{32}\ast A_{23}\\ B_{11}\ast A_{31}+B_{21}\ast A_{32}+B_{31}\ast A_{33}& B_{12}\ast A_{31}+B_{22}\ast A_{32}+B_{32}\ast A_{33}\end{array}\right]\end{array}$$
C矩阵是A矩阵的列向量组合

1.3 行向量组合

已知
$$\begin{array}{rl}\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\end{array}\right]\left[\begin{array}{cc}{B}_{11}& B_{12}\\ B_{21}& B_{22}\\ B_{31}& B_{32}\end{array}\right]& =A_{11}\ast B_{row1}+A_{12}\ast B_{row2}+A_{13}\ast B_{row3}\\ & =\left[\begin{array}{cc}{A}_{11}\ast B_{11}& A_{11}\ast B_{12}\\ +& +\\ A_{12}\ast B_{21}& A_{12}\ast B_{22}\\ +& +\\ A_{13}\ast B_{31}& A_{13}\ast B_{32}\end{array}\right]\end{array}$$
那么
$$\begin{array}{rl}\underset{A}{\underbrace{\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]}}\underset{B}{\underbrace{\left[\begin{array}{cc}{B}_{11}& B_{12}\\ B_{21}& B_{22}\\ B_{31}& B_{32}\end{array}\right]}}& =\underset{C}{\underbrace{\left[\begin{array}{c}{A}_{11}\ast B_{row1}+A_{12}\ast B_{row2}+A_{13}\ast B_{row3}\\ A_{21}\ast B_{row1}+A_{22}\ast B_{row2}+A_{23}\ast B_{row3}\\ A_{31}\ast B_{row1}+A_{32}\ast B_{row2}+A_{33}\ast B_{row3}\end{array}\right]}}\end{array}$$
C矩阵是B矩阵的行向量组合

1.4 单行和单列的乘积和

$$\begin{array}{rl}\left[\begin{array}{cc}2& 7\\ 3& 8\\ 4& 9\end{array}\right]\left[\begin{array}{cc}1& 6\\ 1& 1\end{array}\right]& =\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]\left[\begin{array}{cc}1& 6\end{array}\right]+\left[\begin{array}{c}7\\ 8\\ 9\end{array}\right]\left[\begin{array}{cc}1& 1\end{array}\right]\\ & =\left[\begin{array}{cc}9& 19\\ 11& 26\\ 13& 33\end{array}\right]\end{array}$$

1.5 块乘法

$$\left[\begin{array}{ccc}{A}_1& |& A_2\\ ——& ——& ——\\ A_3& |& A_4\end{array}\right]\left[\begin{array}{ccc}{B}_1& |& B_2\\ ——& ——& ——\\ B_3& |& B_4\end{array}\right]=\left[\begin{array}{ccc}{A}_1\ast B_1+A_2\ast B_3& |& A_1\ast B_2+A_2\ast B_4\\ ————————& ——& ————————\\ A_3\ast B_1+A_4\ast B_3& |& A_3\ast B_2+A_4\ast B_4\end{array}\right]$$

2. 逆矩阵

2.1 逆矩阵的定义

存在
$$A^{-1}A=I$$
那么，称$A^{-1}$为A的逆矩阵，A是可逆的，记为非奇异矩阵

当A为方阵（行数=列数）时，左逆矩阵=右逆矩阵
$$A^{-1}A=I=A{A}^{-1}$$

2.2 奇异矩阵

存在$Ax=0(x非零向量)\Rightarrow A不可逆$
证明如下
$$\begin{array}{rl}& Ax=0\\ & \stackrel{A^{-1}A=I}{\Rightarrow }{A}^{-1}Ax=0\\ & \Rightarrow x=0（与x为非零向量冲突）\end{array}$$

延伸（学习了后面的列向量等）：

• $Ax$是A的列向量的线性组合，$Ax=0有解$说明，存在A的列向量的组合为0，A不是满秩矩阵。
• 那么奇异矩阵不是满秩矩阵
• 那能不能说明由此推导出满秩矩阵可逆？
• 好像不是很充分，除非能推导出$Ax=0(x非零向量)无解\Rightarrow A可逆$

2.3 Gauss-Jordan 求逆矩阵

2.3.1 求逆矩阵$⟺$解方程组

$$\underset{A}{\underbrace{\left[\begin{array}{cc}1& 3\\ 2& 7\end{array}\right]}}\underset{A^{-1}}{\underbrace{\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]}}=\underset{I}{\underbrace{\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]}}⟺\{\begin{array}{l}a+3b=1\\ 2c+7d=1\end{array}$$

2.3.2 Gauss-Jordan求逆矩阵

$A{A}^{-1}=I$ 可写为：
$$\{\begin{array}{l}\left[\begin{array}{cc}1& 3\\ 2& 7\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \\ \left[\begin{array}{cc}1& 3\\ 2& 7\end{array}\right]\left[\begin{array}{c}c\\ d\end{array}\right]=\left[\begin{array}{c}0\\ 1\end{array}\right]\end{array}$$

$$\begin{array}{rl}\underset{\text{增广矩阵[A|I]}}{\underbrace{\left[\begin{array}{cccc}1& 3& 1& 0\\ 2& 7& 0& 1\end{array}\right]}}& \stackrel{ro{w}_2-2ro{w}_1}{\Rightarrow }\left[\begin{array}{cccc}1& 3& 1& 0\\ 0& 1& -2& 1\end{array}\right]\\ & \stackrel{ro{w}_1-3ro{w}_2}{\Rightarrow }\underset{[I|E]}{\underbrace{\left[\begin{array}{cccc}1& 0& 7& -3\\ 0& 1& -2& 1\end{array}\right]}}\end{array}$$
第一种，老师上课讲的，公式推导
$$E\left[\begin{array}{cc}A& I\end{array}\right]=\left[\begin{array}{cc}I& E\end{array}\right]\Rightarrow EA=I\Rightarrow E=A^{-1}$$

第二种，回代到方程组中，也能求出解
$$\{\begin{array}{l}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=\left[\begin{array}{c}7\\ -2\end{array}\right]\\ \\ \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}c\\ d\end{array}\right]=\left[\begin{array}{c}-3\\ 1\end{array}\right]\end{array}\Rightarrow \{\begin{array}{l}a=7\\ b=-2\\ c=-3\\ d=1\end{array}$$