名词解析
以下说法等价:非奇异矩阵⇔满秩矩阵⇔可逆矩阵⇔标准型为E⇔可表示为有限个初等方阵的乘积
\begin{aligned}
A=&
\begin{pmatrix}
\begin{array}{ccc|c}
\textcolor{red}{a_{11}} & \textcolor{red}{a_{12}} & \textcolor{red}{a_{13}} & \textcolor{green}{a_{14}} \\
\textcolor{red}{a_{21}} & \textcolor{red}{a_{22}} & \textcolor{red}{a_{23}} & \textcolor{green}{a_{24}} \\
\hline
\textcolor{yellow}{a_{31}} & \textcolor{yellow}{a_{32}} & \textcolor{yellow}{a_{33}} & a_{34} \\
\end{array}
\end{pmatrix}
\\
=&
\begin{pmatrix}
\textcolor{red}{A_{11}} & \textcolor{green}{A_{12}} \\
\textcolor{yellow}{A_{21}} & A_{22}
\end{pmatrix}
\end{aligned}
A=
\begin{pmatrix}
A_1 & 0 & \cdots & 0 \\
0 & A_2 & \cdots & 0 \\
\vdots & \vdots & & \vdots \\
0 & 0 & \cdots & A_n \\
\end{pmatrix}
例如选取7阶行列式的1、4、7行和列
\\
\begin{pmatrix}
\begin{array}{c|cc|c|cc|c}
\textcolor{red}{a_{11}} & \textcolor{green}{a_{12}} & \textcolor{green}{a_{13}} & \textcolor{red}{a_{14}} & \textcolor{green}{a_{15}} & \textcolor{green}{a_{16}} & \textcolor{red}{a_{17}} \\
\hline
\textcolor{green}{a_{21}} & a_{22} & a_{23} & \textcolor{green}{a_{24}} & a_{25} & a_{26} & \textcolor{green}{a_{27}} \\
\textcolor{green}{a_{31}} & a_{32} & a_{33} & \textcolor{green}{a_{34}} & a_{35} & a_{36} & \textcolor{green}{a_{37}} \\
\hline
\textcolor{red}{a_{41}} & \textcolor{green}{a_{42}} & \textcolor{green}{a_{43}} & \textcolor{red}{a_{44}} & \textcolor{green}{a_{45}} & \textcolor{green}{a_{46}} & \textcolor{red}{a_{47}} \\
\hline
\textcolor{green}{a_{51}} & a_{52} & a_{53} & \textcolor{green}{a_{54}} & a_{55} & a_{56} & \textcolor{green}{a_{57}} \\
\textcolor{green}{a_{61}} & a_{62} & a_{63} & \textcolor{green}{a_{64}} & a_{65} & a_{66} & \textcolor{green}{a_{67}} \\
\hline
\textcolor{red}{a_{71}} & \textcolor{green}{a_{72}} & \textcolor{green}{a_{73}} & \textcolor{red}{a_{74}} & \textcolor{green}{a_{75}} & \textcolor{green}{a_{76}} & \textcolor{red}{a_{77}}
\end{array}
\end{pmatrix}
\\组成的行列式:\\
\begin{vmatrix}
a_{11} & a_{14} & a_{17} \\
a_{41} & a_{44} & a_{47} \\
a_{71} & a_{74} & a_{77}
\end{vmatrix}
A = \begin{pmatrix}
1 & 2 & 4 & 0 & 3 \\
\color{blue}0 & \color{red}-3 & 2 & 4 & 0 \\
\color{blue}0 & \color{blue}0 & 0 & 0 & 5 \\
\color{blue}0 & \color{blue}0 & \color{green}0 & \color{green}0 & \color{green}0
\end{pmatrix}
A = \begin{pmatrix}
1 & \color{blue}0 & \frac{16}{3} & \frac{8}{3} & 0 \\
\color{blue}0 & \color{red}1 & -\frac{2}{3} & -\frac{4}{3} & 0 \\
\color{blue}0 & \color{blue}0 & 0 & 0 & 1 \\
\color{blue}0 & \color{blue}0 & 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
E_{r×r} & 0 \\
0 & 0
\end{pmatrix}=
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
A=(a_{ij})_{m×n} \\
A_{ij} = (-1)^{i+j}M_{ij} \\
A^* = \begin{pmatrix}
A_{11} & A_{21} & \cdots & A_{n1} \\
A_{12} & A_{22} & \cdots & A_{n2} \\
\vdots & \vdots & \ddots & \vdots \\
A_{1n} & A_{2n} & \cdots & A_{nn}
\end{pmatrix}
分块矩阵
性质
分块矩阵满足普通矩阵运算:
\begin{aligned}
A=&
\begin{pmatrix}
A_{11} & \cdots & A_{1r} \\
\vdots & & \vdots \\
A_{s1} & \cdots & A_{sr}
\end{pmatrix}
\\
B=&
\begin{pmatrix}
B_{11} & \cdots & B_{1r} \\
\vdots & & \vdots \\
B_{s1} & \cdots & B_{sr}
\end{pmatrix}
\\
A±B=&
\begin{pmatrix}
A_{11}±B_{11} & \cdots & A{1r}±B_{1r} \\
\vdots & & \vdots \\
A_{s1}±B_{s1} & \cdots & A_{sr}±B_{sr}
\end{pmatrix}
\end{aligned}
λA=
\begin{pmatrix}
λA_{11} & \cdots & λA_{1r} \\
\vdots & & \vdots \\
λA_{s1} & \cdots & λA_{sr}
\end{pmatrix}
\\
\begin{aligned}
C_{ij}&=\sum\limits_{k=1}^s A_{ik}B_{kj}
\\
&=A_{i1}B_{k1}+A_{i2}B_{2j}+\cdots+A_{is}B_{sj}
\\例&如:\\
AB&=
\begin{pmatrix}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
\end{pmatrix}
\begin{pmatrix}
B_{11} & B_{12} \\
B_{21} & B_{22} \\
B_{31} & B_{32}
\end{pmatrix}
=
\end{aligned}
\\
\begin{pmatrix}
A_{11}B_{11}+A_{12}B_{21}+A_{13}B_{31} &
A_{11}B_{12}+A_{12}B_{22}+A_{13}B_{32}\\
A_{21}B_{11}+A_{22}B_{21}+A_{23}B_{31} &
A_{21}B_{12}+A_{22}B_{22}+A_{23}B_{32} \\
\end{pmatrix}
分块对角矩阵的方阵有:\begin{vmatrix}
A
\end{vmatrix}
=
\begin{vmatrix}
A_{1}
\end{vmatrix}
\begin{vmatrix}
A_{2}
\end{vmatrix}
\cdots
\begin{vmatrix}
A_{n}
\end{vmatrix}
分块反对角矩阵有:\begin{vmatrix}
A
\end{vmatrix}
=±
\begin{vmatrix}
A_{1}
\end{vmatrix}
\begin{vmatrix}
A_{2}
\end{vmatrix}
\cdots
\begin{vmatrix}
A_{n}
\end{vmatrix}
,正负号计算转换到分块对角矩阵的方阵的步骤数。
矩阵的秩
矩阵等价
性质
可与平行线类比
反身性:矩阵A与它本身等价
对称性:A与B等价,则B与A等价(互相等价)
传递性:A与B等价,B与C等价,则A与C等价
相互等价的A、B有R(A)=R(B)
初等方阵
类型
E(i,j) = \begin{pmatrix}
1 & & & & & & \\
& \ddots & & & & & \\
& & 1 & & & & \\
& & & 0 & \cdots & 1 & \\
& & & \vdots & \ddots & \vdots & \\
& & & 1 & \cdots & 0 & \\
& & & & & & \ddots \\
& & & & & & & 1
\end{pmatrix}
\begin{matrix}
\\
\\
\longleftarrow \text{第}i\text{行} \\
\\
\longleftarrow \text{第}j\text{行} \\
\\
\end{matrix}
E(i(k)) = \begin{pmatrix}
1 \\
& \ddots \\
& & 1 \\
& & & k & & \\
& & & & 1 & \\
& & & & & \ddots \\
& & & & & & 1
\end{pmatrix}
\begin{matrix}
\\
\\
\longleftarrow \text{第}i\text{行} \\
\\
\\
\\
\end{matrix}
E(j(k),i) = \begin{pmatrix}
1 & & & & & & \\
& \ddots & & & & & \\
& & 1 & & & & \\
& & & 1 & \cdots & k & \\
& & & & \ddots & \vdots & \\
& & & & & 1 & \\
& & & & & & \ddots \\
& & & & & & & 1
\end{pmatrix}
\begin{matrix}
\\
\\
\longleftarrow \text{第}i\text{行} \\
\\
\longleftarrow \text{第}j\text{行} \\
\\
\end{matrix}
由于E的对称性,行变换和列变换得到的结果是一样的,所以不需要区分行/列
性质
用初等方阵与矩阵A相乘相当于对A进行初等变换(E左行右列):
行变换
E_xA形式
E(i,j)A为对调第i、j行
E(i(k))A为第i行乘一个系数
E(j(k),i)A为第j行乘系数k加到i行
列变换
AE_x形式
AE(i,j)为对调第i、j列
AE(i(k))为第i列乘一个系数
AE(j(k),i)为第j列乘系数k加到i列
对A=(a_{ij})_{m×n}做初等变换,行变换为左乘一个m阶初等方阵,列变换为右乘一个n阶初等方阵
逆矩阵
性质
判断可逆的充要条件为\begin{vmatrix}
A
\end{vmatrix}
≠0
若A可逆,则A的逆矩阵唯一
若A可逆,则λA可逆且(λA)^{-1}=\frac 1 λ A^{-1}
若A、B为同阶可逆方阵,则AB可逆且(AA)^{-1}=B^{-1}A^{-1}
若A可逆,则A^T可逆,且(A^T)^{-1}=(A^{-1})^T
初等方阵可逆,且逆矩阵为同类矩阵(可看作逆矩阵为反向初等变换的初等矩阵)
伴随矩阵重要公式:
AA^*=A^*=
\begin{vmatrix}
A
\end{vmatrix}
E
若A可逆,则上条可推出
\begin{aligned}
A^{-1}=&\frac 1
{\begin{vmatrix}
A
\end{vmatrix}}
A^*
\\
{\begin{vmatrix}
A^{-1}
\end{vmatrix}}
=&
{\begin{vmatrix}
A
\end{vmatrix}}^{-1}
\end{aligned}
求解逆矩阵
对可逆且满秩矩阵A求逆矩阵A^{-1},先生成一个新矩阵\begin{pmatrix}
A & E
\end{pmatrix},之后进行初等行变换把左侧变成单位矩阵,得到\begin{pmatrix}
E & A^{-1}
\end{pmatrix}
对其他矩阵,只能通过求伴随矩阵,再通过逆矩阵的关系求解
正交矩阵
正交矩阵A有\begin{vmatrix}
A
\end{vmatrix}
=±1
克拉默法则解
解线性方程组:
\begin{cases}
a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = b_1, \\
a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = b_2, \\
\qquad \dots \dots \dots \dots \\
a_{n1}x_1 + a_{n2}x_2 + \dots + a_{nn}x_n = b_n.
\end{cases}
转换为矩阵
\begin{pmatrix}
a_{11} & a_{12} & \dots & a_{1n} \\
a_{21} & a_{22} & \dots & a_{2n} \\
\vdots & \vdots & & \vdots \\
a_{n1} & a_{n2} & \dots & a_{nn}
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2 \\
\vdots \\
x_n
\end{pmatrix}= \begin{pmatrix}
b_1 \\
b_2 \\
\vdots \\
b_n
\end{pmatrix}
分别赋值AX=B:
\begin{aligned}
A &= (a_{ij})_{n \times n}
\\
&= \begin{pmatrix}
a_{11} & a_{12} & \dots & a_{1n} \\
a_{21} & a_{22} & \dots & a_{2n} \\
\vdots & \vdots & & \vdots \\
a_{n1} & a_{n2} & \dots & a_{nn}
\end{pmatrix}
\\
\quad X &= \begin{pmatrix}
x_1 \\
x_2 \\
\vdots \\
x_n
\end{pmatrix}, \quad B = \begin{pmatrix}
b_1 \\
b_2 \\
\vdots \\
b_n
\end{pmatrix}
\end{aligned}
克拉默法则:如果\begin{vmatrix}
A
\end{vmatrix}
≠0方程组为唯一解,且解为
X = \begin{pmatrix}
x_1 \\
x_2 \\
\vdots \\
x_n
\end{pmatrix} = A^{-1}B
等价于行列式中的形式(D=|A|):
x_1 = \frac{D_1}{|A|}, \quad x_2 = \frac{D_2}{|A|}, \quad \dots, \quad x_n = \frac{D_n}{|A|}
特殊情况:如B=0且\begin{vmatrix}
A
\end{vmatrix}
≠0,则AX=0只有0解,即X=\mathbf{0}
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