矩阵的乘法和逆

2024-04-19 19:20| 来源: 网络整理| 查看: 265

一、矩阵乘法一）矩阵乘法的四种方法

设A= \begin{gathered} \begin{equation} \left[ \begin{array}{cccc} 1&2\cr2& 3\cr3&4 \end{array} \right ] _{3\times2} B= \left[ \begin{array}{cccc}2&1\cr 3 &2 \end{array}\right ] _{2\times2} \quad\quad row_n(A)代表A的第n行,col_n(B)代表B的第n列 \end{equation} \end{gathered}

1、行向量乘以列向量

通过A行向量和B的列向量的点乘计算

AB \begin{array}{c} = \end{array} \left[ \begin{array}{ccccc} R_{1}(A) \times C_{1}(B) & R_{1}(A) \times C_{2}(B) \cr R_{2}(A)\times C_{1}(B) & R_{2}(A)\times C_{2}(B) \cr R_{3}(A)\times C_{1}(B) & R_{3}(A)\times C_{2}(B) \cr \end{array} \right ] \begin{array}{c} \end{array} \left[ \begin{array}{ccccc} 1\times2+2\times3 & 1\times1+2\times2 \cr 2\times2+3\times3 & 2\times1+3\times2 \cr 3\times2+4\times3 & 3\times1+4\times2 \cr \end{array} \right ] \begin{array}{c} = \end{array} \left[ \begin{array}{cccc} 8 & 5 \cr13 & 8 \cr18 & 11 \end{array}\right ]\tag{式1}

2、AB的每一列是A的每一列的线性组合

AB \begin{array}{c} = \end{array} \left[ \left[ \begin{array}{cccc} 1 \cr 2 \cr 3 \cr \end{array} \right ] 2+ \left[ \begin{array}{cccc} 2 \cr 3 \cr 4 \cr \end{array} \right ] 3 \quad\quad \left[ \begin{array}{cccc} 1 \cr 2 \cr 3 \cr \end{array} \right ] 1+ \left[ \begin{array}{cccc} 2 \cr 3 \cr 4\end{array} \right ] 2 \right ] \begin{array}{c} = \end{array} \left[ \begin{array}{cccc} 8 & 5 \cr 13 & 8 \cr 18 & 11\end{array} \right ] \tag{式2}

3、AB的每一行是B的每一行的线性组合

AB \begin{array}{c} = \end{array} \left[ \begin{array}{cccc} 1\times \left[ \begin{array}{cccc} 2 &1 \end{array} \right ]+ 2\times \left[ \begin{array}{cccc} 3 &2 \end{array} \right ] \cr 2\times \left[ \begin{array}{cccc} 2 &1 \end{array} \right ]+ 3\times \left[ \begin{array}{cccc} 3 &2 \end{array} \right ] \cr 3\times \left[ \begin{array}{cccc} 2 & 1 \end{array} \right ]+ 4\times \left[ \begin{array}{cccc} 3 & 2 \end{array} \right ] \cr \end{array} \right ] \begin{array}{c} \end{array} \left[ \begin{array}{cccc} 8 & 5 \cr 13 & 8 \cr 18 & 11 \cr \end{array} \right ] \tag{式3}

4、A的第n列与B的第n行的点乘，再求和

AB \begin{array}{c} = \end{array} \left[ \begin{array}{cccccc} 1 \cr 2 \cr 3 \cr \end{array} \right ] \left[ \begin{array}{ccccc} 2 &1 \cr \end{array} \right ] + \left[ \begin{array}{ccccc} 2 \cr 3 \cr 4 \cr \end{array} \right ] \left[ \begin{array}{cccc} 3 &2 \cr \end{array} \right ]= \begin{array}{c} \end{array} \left[ \begin{array}{ccccc} 2 &1 \cr 4&2 \cr 6&3 \cr \end{array} \right ]+ \left[ \begin{array}{cccc} 6 &4 \cr 9&6 \cr 12&8 \cr \end{array} \right ] \begin{array}{c} \end{array}= \left[ \begin{array}{ccccc} 8 & 5 \cr 13 & 8 \cr 18 & 11 \cr \end{array} \right ]\tag{式4}

二）矩阵运算的规则

设A，B，C均为矩阵，d为常数

1、加法性质

两个矩阵相加，A和B的行数和列数必须相等

\begin{array}{} 1、A+B=B+A \cr 2、d(A+B) = dA+dB \cr 3、A+B+C = A+(B+C) \end{array}

2、乘法性质

两个矩阵相乘，A的列数必须等于B的行数

\begin{array}{} 1、AB\ne BA \cr 2、A(B+C) = AB+AC \cr 3、(A+B)C = AC+AB \cr 4、A(BC) = (AB)C \end{array}

3、单位阵

把对角线上全为1，其余全为0的方阵叫做单位阵（Identity Matrix），记做$I$

I_{3\times3} = \left[ \begin{array}{cccc} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \end{array} \right ] _{3\times3} 若有矩阵A，则 AI=A

三）分块矩阵（Block Matrix）

设有矩阵A和B，把A和B切块如下，如所示：

A= \left[ \begin{array}{cc|cc} 1 & 0& 1 &0\cr 0& 1& 0 &1\cr \hline 1& 0 & 1 &0\cr 0& 1 & 0 &1\cr \end{array} \right], B= \left[ \begin{array}{cc|cc} 1 & 2& 5 & 3\cr 2& 1& 1 & 3\cr \hline2& 1 & 1 & 3\cr 1& 6 & 3 & 1\end{array} \right]

设 A_{11},A_{12},A_{21},A_{22} 分别为矩阵A的四块， B_{11},B_{12},B_{21},B_{22} 分别为矩阵B的四块： A_{11}=\left[ \begin{array}{cccc} 1 &0 \cr 0&1 \end{array} \right ], A_{12}=\left[ \begin{array}{cccc} 1 &0 \cr 0&1 \end{array} \right ], A_{21}=\left[ \begin{array}{cccc} 1 &0 \cr 0&1 \end{array} \right ], A_{22}=\left[ \begin{array}{cccc} 1 & 0 \cr 0&1 \end{array} \right ]

B_{11}=\left[ \begin{array}{cccc} 1 & 2 \cr 2& 1 \end{array} \right ], B_{12}=\left[ \begin{array}{cccc} 5 &3 \cr 1& 3 \end{array} \right ], B_{21}=\left[\begin{array}{cccc} 2 & 1 \cr 1 & 6 \end{array} \right ],B_{22}=\left[ \begin{array}{cccc} 1 & 3 \cr 3 & 1 \end{array}\right ]

AB= \left[ \begin{array}{cccc} A_{11} & A_{12} \cr A_{21} & A_{22} \end{array} \right ] \left[ \begin{array}{cccc} B_{11} & B_{12} \cr B_{21} & B_{22} \end{array} \right ] = \left[ \begin{array}{cccc} A_{11}B_{11}+A_{12}B_{21} & A_{11}B_{12}+A_{12}B_{22}\cr A_{21}B_{11}+A_{22}B_{21} & A_{21}B_{12}+A_{22}B_{22}\cr \end{array} \right ] = \left[ \begin{array}{cccc} 6 & 5 & 3 & 4 \cr 3 & 4 & 4 & 7 \cr 6 & 5 & 3 & 4 \cr 3 & 4 & 4 & 7 \cr \end{array} \right ]

在矩阵乘法的式(4)中，就是按分块矩阵的思想来处理的。

二、矩阵的逆一）定义

当A为方阵， AA^{-1}=A^{-1}A=I ，则 A^{-1} 和A互为逆矩阵

二）关于逆矩阵的问题不是所有的矩阵都有逆矩阵。若 A_{n \times n} 的矩阵有逆矩阵，则通过消元，必定有n个主元。矩阵A若有逆矩阵，则逆矩阵是唯一的。若有一个非0向量，使得 Ax=0 ，那么A必定是不可逆矩阵。若有一个对角矩阵（diagonal matrix），对角线为 d_{1},d_{2},d_{3},...d_{n} ，则其逆矩阵为对角线为 1/d_{1},1/d_{2},1/d_{3},...1/d_{n} 若A和B都是可逆的，则AB也是可逆的，其逆矩阵： (AB)^{-1}=B^{-1}A^{-1}

若A,B,C都是可逆的，则ABC也是可逆的，其逆矩阵： (ABC)^{-1}=C^{-1}B^{-1}A^{-1}

三）初等行变换和逆矩阵

将运用到矩阵的LU分解中

1、行交换（Row Exchange）

若A = \left [ \begin{array}{cccc} 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0\end{array} \right ] , 则A^{-1} = \left [ \begin{array}{cccc} 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0\end{array} \right ]

A的第二行和第三行交换

2、数乘运算

若A = \left [ \begin{array}{cccc} 5 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{array} \right ] , 则A^{-1} = \left [ \begin{array}{cccc} 1/5 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr \end{array} \right ]

A的第一行多乘了5，则在 A^{-1} 中的第一行除以5

3、倍加运算

若A = \left [ \begin{array}{cccc} 1 & 0 & 0 \cr 5 & 1 & 0 \cr 0 & 0 & 1 \cr \end{array} \right ] , 则A^{-1} = \left [ \begin{array}{cccc} 1 & 0 & 0 \cr -5 & 1 & 0 \cr 0 & 0 & 1 \cr \end{array} \right ]

A表示把A的第一行乘以5加到第二行，其逆操作 A^{-1} 就是把第一行乘以-5加到第二行

四）求解逆矩阵1、求解证明

一个分块矩阵 [A \quad I] ，且 A有逆矩阵=A^{-1} ，当A^{-1}[A \quad I]=[AA^{-1}\quad A^{-1}]=[I \quad A^{-1}]

2、求解举例

有可逆矩阵 A=\left [ \begin{array}{cccc} 1&2\cr2&3\end{array} \right ] ，求 A^{-1}

[A \quad I] = \left [ \begin{array}{cccc} 1 & 2 & 1 & 0 \cr 2 & 3 & 0 & 1\cr \end{array} \right ] \begin{array}{cccc} E_{21} \cr -2r_1+r_2 \cr ====> \end{array} \left [ \begin{array}{cccc} 1 & 2 & 1 & 0 \cr 0 & -1 & -2 & 1\cr \end{array} \right ] \begin{array}{cccc} E_{22} \cr -r_2 \cr ====> \end{array} \left [ \begin{array}{cccc} 1 & 2 & 1 & 0 \cr 0 & 1 & 2 & -1 \end{array} \right ] \begin{array}{cccc} E_{12} \cr -2r_2+r_1 \cr ====> \end{array} \left [ \begin{array}{cccc} 1 & 0&-3 & 2 \cr 0 & 1 & 2 & -1\end{array} \right ]

得：A^{-1} = \left [ \begin{array}{cc}-3&2\cr2& -1\end{array} \right ]

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矩阵的乘法和逆

矩阵的乘法和逆

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