1. 矩陣乘法

1.1 常規方法

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

1.2 列向量組合

已知
$$\begin{array}{rl}\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]\left[\begin{array}{c}{B}_{11}\\ B_{21}\\ B_{31}\end{array}\right]& =B_{11}?A_{col1}+B_{21}?A_{col2}+B_{31}?A_{col3}\\ & =\left[\begin{array}{c}{B}_{11}?A_{11}+B_{21}?A_{12}+B_{31}?A_{13}\\ B_{11}?A_{21}+B_{21}?A_{22}+B_{31}?A_{23}\\ B_{11}?A_{31}+B_{21}?A_{32}+B_{31}?A_{33}\end{array}\right]\end{array}$$
那么
$$\begin{array}{rl}\underset{A}{\underbrace{\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]}}\underset{B}{\underbrace{\left[\begin{array}{cc}{B}_{11}& B_{12}\\ B_{21}& B_{22}\\ B_{31}& B_{32}\end{array}\right]}}& =\underset{C}{\underbrace{\left[\begin{array}{cc}{B}_{11}?A_{col1}+B_{21}?A_{col2}+B_{31}?A_{col3}& B_{12}?A_{col1}+B_{22}?A_{col2}+B_{32}?A_{col3}\end{array}\right]}}\\ & =\left[\begin{array}{cc}{B}_{11}?A_{11}+B_{21}?A_{12}+B_{31}?A_{13}& B_{12}?A_{11}+B_{22}?A_{12}+B_{32}?A_{13}\\ B_{11}?A_{21}+B_{21}?A_{22}+B_{31}?A_{23}& B_{12}?A_{21}+B_{22}?A_{22}+B_{32}?A_{23}\\ B_{11}?A_{31}+B_{21}?A_{32}+B_{31}?A_{33}& B_{12}?A_{31}+B_{22}?A_{32}+B_{32}?A_{33}\end{array}\right]\end{array}$$
C矩陣是A矩陣的列向量組合

1.3 行向量組合

已知
$$\begin{array}{rl}\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\end{array}\right]\left[\begin{array}{cc}{B}_{11}& B_{12}\\ B_{21}& B_{22}\\ B_{31}& B_{32}\end{array}\right]& =A_{11}?B_{row1}+A_{12}?B_{row2}+A_{13}?B_{row3}\\ & =\left[\begin{array}{cc}{A}_{11}?B_{11}& A_{11}?B_{12}\\ +& +\\ A_{12}?B_{21}& A_{12}?B_{22}\\ +& +\\ A_{13}?B_{31}& A_{13}?B_{32}\end{array}\right]\end{array}$$
那么
$$\begin{array}{rl}\underset{A}{\underbrace{\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]}}\underset{B}{\underbrace{\left[\begin{array}{cc}{B}_{11}& B_{12}\\ B_{21}& B_{22}\\ B_{31}& B_{32}\end{array}\right]}}& =\underset{C}{\underbrace{\left[\begin{array}{c}{A}_{11}?B_{row1}+A_{12}?B_{row2}+A_{13}?B_{row3}\\ A_{21}?B_{row1}+A_{22}?B_{row2}+A_{23}?B_{row3}\\ A_{31}?B_{row1}+A_{32}?B_{row2}+A_{33}?B_{row3}\end{array}\right]}}\end{array}$$
C矩陣是B矩陣的行向量組合

1.4 單行和單列的乘積和

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

1.5 塊乘法

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

2. 逆矩陣

2.1 逆矩陣的定義

存在
$$A^{?1}A=I$$
那么，稱$A^{?1}$為A的逆矩陣，A是可逆的，記為非奇異矩陣

當A為方陣（行數=列數）時，左逆矩陣=右逆矩陣
$$A^{?1}A=I=A{A}^{?1}$$

2.2 奇異矩陣

存在$Ax=0(x非零向量)?A不可逆$
證明如下
$$\begin{array}{rl}& Ax=0\\ & \stackrel{A^{?1}A=I}{\Rightarrow }{A}^{?1}Ax=0\\ & \Rightarrow x=0（與x為非零向量沖突）\end{array}$$

延伸（學習了后面的列向量等）：

• $Ax$是A的列向量的線性組合，$Ax=0有解$說明，存在A的列向量的組合為0，A不是滿秩矩陣。
• 那么奇異矩陣不是滿秩矩陣
  那能不能說明由此推導出滿秩矩陣可逆？
  好像不是很充分，除非能推導出$Ax=0(x非零向量)無解?A可逆$

2.3 Gauss-Jordan 求逆矩陣

2.3.1 求逆矩陣$?$解方程組

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

2.3.2 Gauss-Jordan求逆矩陣

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

$$\begin{array}{rl}\underset{\text{增廣矩陣[A|I]}}{\underbrace{\left[\begin{array}{cccc}1& 3& 1& 0\\ 2& 7& 0& 1\end{array}\right]}}& \stackrel{ro{w}_2?2ro{w}_1}{\Rightarrow }\left[\begin{array}{cccc}1& 3& 1& 0\\ 0& 1& ?2& 1\end{array}\right]\\ & \stackrel{ro{w}_1?3ro{w}_2}{\Rightarrow }\underset{[I|E]}{\underbrace{\left[\begin{array}{cccc}1& 0& 7& ?3\\ 0& 1& ?2& 1\end{array}\right]}}\end{array}$$
第一種，老師上課講的，公式推導
$$E\left[\begin{array}{cc}A& I\end{array}\right]=\left[\begin{array}{cc}I& E\end{array}\right]?EA=I?E=A^{?1}$$
ps:

• 從矩陣A經過消元變成了單位矩陣, 那么A滿秩，不然變不成單位矩陣。
• 所以說，如果A可逆，那么A一定是滿秩矩陣。
• 如果A滿秩，那么A一定可逆。

第二種，回代到方程組中，也能求出解
$$?\begin{array}{l}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=\left[\begin{array}{c}7\\ ?2\end{array}\right]\\ \\ \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}c\\ d\end{array}\right]=\left[\begin{array}{c}?3\\ 1\end{array}\right]\end{array}??\begin{array}{l}a=7\\ b=?2\\ c=?3\\ d=1\end{array}$$