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\begin_layout Section
矩阵的初等变换
\end_layout
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矩阵的初等变换
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
矩阵的初等变换
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
在计算行列式时, 利用行列式的性质可以将给定的行列时化为上 (下) 三角形行列式, 从而简化行列式的计算, 把行列式的某些性质引用到矩阵上, 会给我们研究矩阵带
来很大的方便, 这些性质反映到矩阵上就是矩阵的初等变换.
\end_layout
\begin_layout Definition
矩阵的下列三种变换称为
\series bold
矩阵的初等行变换
\series default
:
\end_layout
\begin_layout Definition
(1) 交换矩阵的两行 (交换
\begin_inset Formula $i,j$
\end_inset
两行, 记作
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\end_inset
);
\end_layout
\begin_layout Definition
(2) 以一个非零的数
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\end_inset
乘矩阵的某一行 (第
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\end_inset
行乘数
\begin_inset Formula $k$
\end_inset
, 记作
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\end_inset
);
\end_layout
\begin_layout Definition
(3) 把矩阵的某一行的
\begin_inset Formula $k$
\end_inset
倍加到另一行 (第
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\end_inset
行乘
\begin_inset Formula $k$
\end_inset
加到
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\end_inset
行, 记为
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\end_inset
).
\end_layout
\begin_layout Definition
把定义中的 “行” 换成 “列”, 即得
\series bold
矩阵的初等列变换
\series default
的定义 (相应记号中把
\begin_inset Formula $r$
\end_inset
换成
\begin_inset Formula $c$
\end_inset
).
\end_layout
\begin_layout Definition
初等行变换与初等列变换统称为
\series bold
初等变换
\series default
.
\end_layout
\begin_layout Remark*
初等变换的逆变换仍是初等变换, 且变换类型相同.
\end_layout
\begin_layout Standard
例如, 变换
\begin_inset Formula $r_{i}\leftrightarrow r_{j}$
\end_inset
的逆变换即为其本身; 变换
\begin_inset Formula $r_{i}\times k$
\end_inset
的逆变换为
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; 变换
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\end_inset
的逆变换为
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或
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\end_inset
.
\end_layout
\end_deeper
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\begin_layout Frame
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status open
\begin_layout Plain Layout
矩阵之间的等价关系
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
若矩阵
\begin_inset Formula $A$
\end_inset
经过有限次初等变换变成矩阵
\begin_inset Formula $B$
\end_inset
, 则称
\series bold
矩阵
\begin_inset Formula $A$
\end_inset
与
\begin_inset Formula $B$
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等价
\series default
, 记为
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, (或
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).
\end_layout
\begin_layout Remark*
在理论表述或证明中, 常用记号 “
\begin_inset Formula $\sim$
\end_inset
”, 在对矩阵作初等变换运算的过程中常用记号 “
\begin_inset Formula $\rightarrow$
\end_inset
”.
\end_layout
\begin_layout Standard
矩阵之间的等价关系具有下列基本性质:
\end_layout
\begin_layout Standard
(1) 反身性:
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\end_inset
;
\end_layout
\begin_layout Standard
(2) 对称性: 若
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\end_inset
, 则
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\end_inset
;
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\begin_layout Standard
(3) 传递性: 若
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,
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\end_deeper
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\end_layout
\begin_layout Standard
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\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
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\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
行阶梯形矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
一般地, 称满足下列条件的矩阵为
\series bold
行阶梯形矩阵
\series default
:
\end_layout
\begin_layout Standard
(1).
零行 (元素全为零的行) 位于矩阵的下方;
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\begin_layout Standard
(2).
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\end_layout
\begin_layout Standard
例如:
\end_layout
\begin_layout Standard
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\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
行最简形矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
一般地, 称满足下列条件的阶梯形矩阵为
\series bold
行最简形矩阵
\series default
:
\end_layout
\begin_layout Standard
(1).
各非零行的首非零元都是
\begin_inset Formula $1$
\end_inset
;
\end_layout
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(2).
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\begin_layout Standard
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\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
矩阵的标准形
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
一般地,
\series bold
矩阵
\begin_inset Formula $A$
\end_inset
的标准形
\begin_inset Formula $D$
\end_inset
\series default
具有如下特点:
\end_layout
\begin_layout Standard
\begin_inset Formula $D$
\end_inset
的左上角是一个单位矩阵, 其余元素全为
\begin_inset Formula $0$
\end_inset
.
\end_layout
\begin_layout Standard
例如:
\begin_inset ERT
status open
\begin_layout Plain Layout
$$
\end_layout
\begin_layout Plain Layout
\backslash
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\end_layout
\begin_layout Plain Layout
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\end_layout
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\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
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\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
阶梯型矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
已知矩阵
\begin_inset Formula $A=\begin{bmatrix}3 & 2 & 9 & 6\\
-1 & -3 & 4 & -17\\
1 & 4 & -7 & 3\\
-1 & -4 & 7 & -3
\end{bmatrix}$
\end_inset
, 对其作初等行变换, 先化为行阶梯形矩阵.
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\begin{aligned}A= & \begin{bmatrix}3 & 2 & 9 & 6\\
-1 & -3 & 4 & -17\\
1 & 4 & -7 & 3\\
-1 & -4 & 7 & -3
\end{bmatrix}\xrightarrow{r_{1}\leftrightarrow r_{3}}\begin{bmatrix}1 & 4 & -7 & 3\\
-1 & -3 & 4 & -17\\
3 & 2 & 9 & 6\\
-1 & -4 & 7 & -3
\end{bmatrix}\\
& \xrightarrow[r_{4}+r_{1}]{{r_{2}+r_{1}\atop r_{3}-3r_{1}}}\begin{bmatrix}1 & 4 & -7 & 3\\
0 & 1 & -3 & -14\\
0 & -10 & 30 & -3\\
0 & 0 & 0 & 0
\end{bmatrix}\xrightarrow{r_{3}+10r_{2}}\begin{bmatrix}1 & 4 & -7 & 3\\
0 & 1 & -3 & -14\\
0 & 0 & 0 & -143\\
0 & 0 & 0 & 0
\end{bmatrix}\xlongequal{\text{ 记作 }}B.
\end{aligned}
\]
\end_inset
\end_layout
\begin_layout Standard
这里的矩阵
\begin_inset Formula $B$
\end_inset
依其形状的特征称为
\series bold
行阶梯形矩阵
\series default
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
矩阵的标准形
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none
用初等变换化矩阵
\begin_inset Formula $\begin{bmatrix}0 & 2 & -4\\
-1 & -4 & 5\\
3 & 1 & 7\\
0 & 5 & -10\\
2 & 3 & 0
\end{bmatrix}$
\end_inset
为标准形.
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\begin{aligned}\begin{bmatrix}0 & 2 & -4\\
-1 & -4 & 5\\
3 & 1 & 7\\
0 & 5 & -10\\
2 & 3 & 0
\end{bmatrix} & \xrightarrow{r_{1}\leftrightarrow r_{2}}\begin{bmatrix}-1 & -4 & 5\\
0 & 2 & -4\\
3 & 1 & 7\\
0 & 5 & -10\\
2 & 3 & 0
\end{bmatrix}\xrightarrow[r_{5}+2r_{1}]{r_{3}+3r_{1}}\begin{bmatrix}-1 & -4 & 5\\
0 & 2 & -4\\
0 & -11 & 22\\
0 & 5 & -10\\
0 & -5 & 10
\end{bmatrix}\\
& \hspace{-8em}\xrightarrow[c_{3}+5c_{1}]{c_{2}-4c_{1}}\begin{bmatrix}-1 & 0 & 0\\
0 & 2 & -4\\
0 & -11 & 22\\
0 & 5 & -10\\
0 & -5 & 10
\end{bmatrix}\xrightarrow{c_{3}+2c_{2}}\begin{bmatrix}1 & 0 & 0\\
0 & 2 & 0\\
0 & -11 & 0\\
0 & 5 & 0\\
0 & -5 & 0
\end{bmatrix}\xrightarrow[{c_{4}-5/2c_{2}\atop c_{5}+5/2c_{2}}]{c_{3}+11/2c_{2}}\begin{bmatrix}1 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}\xrightarrow{c_{2}/2}\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}.
\end{aligned}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
矩阵的标准形
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
将矩阵
\begin_inset Formula $A=\begin{bmatrix}2 & 1 & 2 & 3\\
4 & 1 & 3 & 5\\
2 & 0 & 1 & 2
\end{bmatrix}$
\end_inset
化为标准形.
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\begin{aligned}A & =\begin{bmatrix}2 & 1 & 2 & 3\\
4 & 1 & 3 & 5\\
2 & 0 & 1 & 2
\end{bmatrix}\longrightarrow\begin{bmatrix}2 & 1 & 2 & 3\\
0 & -1 & -1 & -1\\
0 & -1 & -1 & -1
\end{bmatrix}\\
& \longrightarrow\begin{bmatrix}2 & 0 & 0 & 0\\
0 & -1 & -1 & -1\\
0 & -1 & -1 & -1
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 0 & 0 & 0\\
0 & -1 & -1 & -1\\
0 & 0 & 0 & 0
\end{bmatrix}\\
& \longrightarrow\begin{bmatrix}1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & 0 & 0
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{aligned}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
化为标准型矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem
\begin_inset CommandInset label
LatexCommand label
name "thm:2.5-1"
\end_inset
任意一个矩阵
\begin_inset Formula $A=\left(a_{ij}\right)_{m\times n}$
\end_inset
经过有限次初等变换, 可以化为下列标准形矩阵
\end_layout
\begin_layout Theorem
\begin_inset ERT
status open
\begin_layout Plain Layout
$$
\end_layout
\begin_layout Plain Layout
A=
\backslash
begin{bNiceMatrix}[last-row,last-col,xdots/line-style={dashed,blue}]
\end_layout
\begin_layout Plain Layout
1 & & & & & &
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&
\backslash
ddots &
\backslash
Vdots & & & &
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&
\backslash
Cdots & 1 &
\backslash
Cdots & & &
\backslash
leftarrow r
\backslash
text{行}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& &
\backslash
Vdots & 0 & & &
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& & & &
\backslash
ddots & &
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& & & & & 0 &
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& &
\backslash
overset{
\backslash
uparrow}{r
\backslash
text{列}} & & & &
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}=
\backslash
begin{bmatrix}
\end_layout
\begin_layout Plain Layout
E_r & O_{r
\backslash
times(n-r)}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
O_{(m-r)
\backslash
times r} & O_{(m-r)
\backslash
times(n-r)}
\end_layout
\begin_layout Plain Layout
\backslash
end{bmatrix}.
\end_layout
\begin_layout Plain Layout
$$
\end_layout
\end_inset
\end_layout
\begin_layout Remark*
定理
\begin_inset CommandInset ref
LatexCommand ref
reference "thm:2.5-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
的证明也实质上给出了下列结论:
\end_layout
\begin_layout Theorem*
\begin_inset Argument 1
status open
\begin_layout Plain Layout
\begin_inset CommandInset ref
LatexCommand ref
reference "thm:2.5-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
'
\end_layout
\end_inset
任一矩阵
\begin_inset Formula $A$
\end_inset
总可以经过有限次初等行变换化为行阶梯形矩阵, 并进而化为行最简形矩阵.
\end_layout
\begin_layout Standard
根据定理
\begin_inset CommandInset ref
LatexCommand ref
reference "thm:2.5-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
的证明及初等变换的可逆性, 有
\end_layout
\begin_layout Corollary
\begin_inset CommandInset label
LatexCommand label
name "cor:5.7"
\end_inset
如果
\begin_inset Formula $A$
\end_inset
为
\begin_inset Formula $n$
\end_inset
阶可逆矩阵, 则矩阵
\begin_inset Formula $A$
\end_inset
经过有限次初等变换可化为单位矩阵
\begin_inset Formula $E$
\end_inset
, 即
\begin_inset Formula $A\sim E$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Subsection
初等矩阵
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
初等矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
对单位矩阵
\begin_inset Formula $E$
\end_inset
施以一次初等变换得到矩阵称为
\series bold
初等矩阵
\series default
.
\end_layout
\begin_layout Standard
三种初等变换分别对应着三种初等矩阵.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
初等矩阵-1
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
(1).
\begin_inset Formula $E$
\end_inset
的第
\begin_inset Formula $i,j$
\end_inset
行 (列) 互换得到的矩阵
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-3mm}
\end_layout
\begin_layout Plain Layout
$$
\end_layout
\begin_layout Plain Layout
\backslash
setcounter{MaxMatrixCols}{12}
\end_layout
\begin_layout Plain Layout
\backslash
newcommand{
\backslash
blue}{
\backslash
color{blue}}
\end_layout
\begin_layout Plain Layout
E(i,j)=
\backslash
begin{bNiceMatrix}[last-row,last-col,nullify-dots,xdots/line-style={dashed,blue}
]
\end_layout
\begin_layout Plain Layout
1& & &
\backslash
Vdots & & & &
\backslash
Vdots
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&
\backslash
Ddots[line-style=standard]
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& & 1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
Cdots[color=blue,line-style=dashed]& & &
\backslash
blue 0 &
\backslash
Cdots & & &
\backslash
blue 1 & & &
\backslash
Cdots &
\backslash
blue
\backslash
leftarrow i
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& & & & 1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& & &
\backslash
Vdots & &
\backslash
Ddots[line-style=standard] & &
\backslash
Vdots
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& & & & & & 1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
Cdots & & &
\backslash
blue 1 &
\backslash
Cdots & &
\backslash
Cdots &
\backslash
blue 0 & & &
\backslash
Cdots &
\backslash
blue
\backslash
leftarrow j
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& & & & & & & & 1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& & & & & & & & &
\backslash
Ddots[line-style=standard]
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& & &
\backslash
Vdots & & & &
\backslash
Vdots & & & 1
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
& & &
\backslash
blue
\backslash
overset{
\backslash
uparrow}{i} & & & &
\backslash
blue
\backslash
overset{
\backslash
uparrow}{j}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}
\backslash
quad.
\end_layout
\begin_layout Plain Layout
$$
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
初等矩阵-2
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
(2).
\begin_inset Formula $E$
\end_inset
的第
\begin_inset Formula $i$
\end_inset
行 (列) 乘以非零数
\begin_inset Formula $k$
\end_inset
得到的矩阵
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
$$E
\backslash
left(i(k)
\backslash
right)=
\backslash
begin{bNiceMatrix}[last-row,last-col,xdots/line-style={dashed,blue}]
\end_layout
\begin_layout Plain Layout
1&&&&&&&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&
\backslash
ddots&&&&&&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&&1&
\backslash
Vdots&&&&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&&
\backslash
Cdots&k&
\backslash
Cdots&&&i
\backslash
text{ 行}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&&&
\backslash
Vdots&1&&&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&&&&&
\backslash
ddots&&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&&&&&&1&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&&&i
\backslash
text{ 列}&&&&
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}.
\end_layout
\begin_layout Plain Layout
$$
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
初等矩阵-3
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
(3).
\begin_inset Formula $E$
\end_inset
的第
\begin_inset Formula $j$
\end_inset
行乘以数
\begin_inset Formula $k$
\end_inset
加到第
\begin_inset Formula $i$
\end_inset
行上, 或
\begin_inset Formula $E$
\end_inset
的第
\begin_inset Formula $i$
\end_inset
列乘以数
\begin_inset Formula $k$
\end_inset
加到第
\begin_inset Formula $j$
\end_inset
列上得到的矩阵
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
$$E
\backslash
left(i+j(k)
\backslash
right)=
\backslash
begin{bNiceMatrix}[last-row,last-col,xdots/line-style={dashed,blue}]
\end_layout
\begin_layout Plain Layout
1&&&&&&&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&
\backslash
ddots&
\backslash
Vdots&&
\backslash
Vdots&&&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&
\backslash
Cdots&1&
\backslash
Cdots&k&
\backslash
Cdots&&i
\backslash
text{ 行}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&&
\backslash
Vdots&
\backslash
ddots&
\backslash
Vdots&&&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&&&&1&
\backslash
Cdots&&j
\backslash
text{ 行}
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&&&&
\backslash
Vdots&
\backslash
ddots&&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&&&&&&1&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&&&&&&&
\backslash
\backslash
\end_layout
\begin_layout Plain Layout
&&i
\backslash
text{ 列}&&j
\backslash
text{ 列}&&&
\end_layout
\begin_layout Plain Layout
\backslash
end{bNiceMatrix}
\end_layout
\begin_layout Plain Layout
$$
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
初等矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Proposition
关于初等矩阵有下列性质:
\end_layout
\begin_layout Proposition
(1).
\begin_inset Formula $E(i,j)^{-1}=E(i,j)$
\end_inset
;
\begin_inset Formula $E(i(k))^{-1}=E\left(i\left(k^{-1}\right)\right)$
\end_inset
;
\begin_inset Formula $E(i+j(k))^{-1}=E(i+j(-k))$
\end_inset
;
\end_layout
\begin_layout Proposition
(2).
\begin_inset Formula $|E(i,j)|=-1$
\end_inset
;
\begin_inset Formula $|E(i(k))|=k$
\end_inset
;
\begin_inset Formula $|E(i+j(k))|=1$
\end_inset
.
\end_layout
\begin_layout Theorem
设
\begin_inset Formula $A$
\end_inset
是一个
\begin_inset Formula $m\times n$
\end_inset
矩阵, 对
\begin_inset Formula $A$
\end_inset
施行一次初等行 (列) 变换, 相当于用同种的
\begin_inset Formula $m(n)$
\end_inset
阶初等矩阵左 (右) 乘
\begin_inset Formula $A$
\end_inset
.
\end_layout
\begin_layout Itemize
\begin_inset Formula $E(i,j)A$
\end_inset
: 交换矩阵
\begin_inset Formula $A$
\end_inset
的第
\begin_inset Formula $i,j$
\end_inset
两行;
\end_layout
\begin_layout Itemize
\begin_inset Formula $E(i(k))A$
\end_inset
: 对矩阵
\begin_inset Formula $A$
\end_inset
的第
\begin_inset Formula $i$
\end_inset
行乘以
\begin_inset Formula $k$
\end_inset
;
\end_layout
\begin_layout Itemize
\begin_inset Formula $E(i+j(k))A$
\end_inset
: 对矩阵
\begin_inset Formula $A$
\end_inset
的第
\begin_inset Formula $j$
\end_inset
行乘以
\begin_inset Formula $k$
\end_inset
加到第
\begin_inset Formula $i$
\end_inset
行上.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
初等矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设有矩阵
\begin_inset Formula $A=\begin{bmatrix}3 & 0 & 1\\
1 & -1 & 2\\
0 & 1 & 1
\end{bmatrix}$
\end_inset
, 而
\begin_inset Formula $E_{3}(1,2)=\begin{bmatrix}0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 1
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $E_{3}(3+1(2))=\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
2 & 0 & 1
\end{bmatrix}$
\end_inset
,
\end_layout
\begin_layout Standard
则
\begin_inset Formula $E_{3}(1,2)A=\begin{bmatrix}0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}3 & 0 & 1\\
1 & -1 & 2\\
0 & 1 & 1
\end{bmatrix}=\begin{bmatrix}1 & -1 & 2\\
3 & 0 & 1\\
0 & 1 & 1
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Standard
即用
\begin_inset Formula $E_{3}(1,2)$
\end_inset
左乘
\begin_inset Formula $A$
\end_inset
, 相当于交换矩阵
\begin_inset Formula $A$
\end_inset
的第
\begin_inset Formula $1$
\end_inset
行与第
\begin_inset Formula $2$
\end_inset
行.
\end_layout
\begin_layout Standard
又
\begin_inset Formula $AE_{3}(31(2))=\begin{bmatrix}3 & 0 & 1\\
1 & -1 & 2\\
0 & 1 & 1
\end{bmatrix}\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
2 & 0 & 1
\end{bmatrix}=\begin{bmatrix}5 & 0 & 1\\
5 & -1 & 2\\
2 & 1 & 1
\end{bmatrix}$
\end_inset
,
\end_layout
\begin_layout Standard
即用
\begin_inset Formula $E_{3}(31(2))$
\end_inset
右乘
\begin_inset Formula $A$
\end_inset
, 相当于将矩阵
\begin_inset Formula $A$
\end_inset
的第
\begin_inset Formula $3$
\end_inset
列乘
\begin_inset Formula $2$
\end_inset
加到第
\begin_inset Formula $1$
\end_inset
列.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
求逆矩阵的初等变换法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
在第二章第三节中, 给出了矩阵
\begin_inset Formula $A$
\end_inset
可逆的充要条件, 也给出了利用伴随矩阵求逆矩阵
\begin_inset Formula $A^{-1}$
\end_inset
的方法, 即
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
A^{-1}=\frac{1}{|A|}A^{*}.
\]
\end_inset
\end_layout
\begin_layout Standard
该方法称为
\series bold
伴随矩阵法
\series default
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
求逆矩阵的初等变换法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
对于较高阶的矩阵, 用伴随矩阵法求逆矩阵计算量太大, 下面介绍一种较为简便的方法:
\series bold
初等变换法
\series default
.
\end_layout
\begin_layout Theorem
\begin_inset Formula $n$
\end_inset
阶矩阵
\begin_inset Formula $A$
\end_inset
可逆的充分必要条件是
\begin_inset Formula $A$
\end_inset
可以表示为若干初等矩阵的乘积.
\end_layout
\begin_layout Proof
\begin_inset Argument 1
status open
\begin_layout Plain Layout
Hint
\end_layout
\end_inset
由推论
\begin_inset CommandInset ref
LatexCommand ref
reference "cor:5.7"
plural "false"
caps "false"
noprefix "false"
\end_inset
, 当
\begin_inset Formula $A$
\end_inset
可逆时, 矩阵
\begin_inset Formula $A$
\end_inset
可以经过有限次初等变换得到
\begin_inset Formula $n$
\end_inset
阶单位阵
\begin_inset Formula $E$
\end_inset
, 即
\begin_inset Formula
\[
P_{s}P_{s-1}\cdots P_{2}P_{1}A=E,
\]
\end_inset
其中
\begin_inset Formula $P_{1},\cdots,P_{s}$
\end_inset
表示对矩阵
\begin_inset Formula $A$
\end_inset
的初等变换矩阵.
故
\begin_inset Formula $A=P_{1}^{-1}P_{2}^{-1}\cdots P_{s}^{-1}$
\end_inset
为
\begin_inset Formula $s$
\end_inset
个初等矩阵的乘积.
\end_layout
\begin_layout Standard
因此, 求矩阵
\begin_inset Formula $A$
\end_inset
的逆矩阵
\begin_inset Formula $A^{-1}$
\end_inset
时, 可构造
\begin_inset Formula $n\times2n$
\end_inset
阶矩阵
\begin_inset Formula
\[
\begin{bmatrix}A & E\end{bmatrix},
\]
\end_inset
\end_layout
\begin_layout Standard
然后对其施以初等行变换将矩阵
\begin_inset Formula $A$
\end_inset
化为单位矩阵
\begin_inset Formula $E$
\end_inset
, 则上述初等变换同时也将其中的单位矩阵
\begin_inset Formula $E$
\end_inset
化为
\begin_inset Formula $A^{-1}$
\end_inset
, 即
\begin_inset Formula
\[
\begin{bmatrix}A & E\end{bmatrix}\xrightarrow{\text{ 初等行变换 }}\begin{bmatrix}E & A^{-1}\end{bmatrix},
\]
\end_inset
\end_layout
\begin_layout Standard
这就是求逆矩阵的
\series bold
初等变换法
\series default
.
再详细点来说, 就是
\begin_inset Formula
\begin{align*}
\begin{bmatrix}A & E\end{bmatrix} & \xrightarrow{P_{1}}\begin{bmatrix}P_{1}A & P_{1}E\end{bmatrix}\\
& \xrightarrow{P_{2}}\begin{bmatrix}P_{2}P_{1}A & P_{2}P_{1}E\end{bmatrix}\\
& \xrightarrow{P_{3}}\cdots\xrightarrow{P_{s}}\begin{bmatrix}P_{s}\cdots P_{2}P_{1}A & P_{s}\cdots P_{2}P_{1}E\end{bmatrix}\\
& =\begin{bmatrix}E & P_{s}\cdots P_{2}P_{1}\end{bmatrix}=\begin{bmatrix}E & A^{-1}\end{bmatrix}.
\end{align*}
\end_inset
再或者, 用矩阵乘法的语言来说
\begin_inset Formula
\[
P_{s}P_{s-1}\cdots P_{1}\begin{bmatrix}A & E\end{bmatrix}=\begin{bmatrix}P_{s}P_{s-1}\cdots P_{1}A & P_{s}P_{s-1}\cdots P_{1}E\end{bmatrix}=\begin{bmatrix}E & A^{-1}\end{bmatrix}.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
求逆矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $A=\begin{bmatrix}1 & 2 & 3\\
2 & 2 & 1\\
3 & 4 & 3
\end{bmatrix}$
\end_inset
, 求
\begin_inset Formula $A^{-1}$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset Formula
\begin{align*}
\begin{bmatrix}A & E\end{bmatrix} & =\begin{bmatrix}1 & 2 & 3 & 1 & 0 & 0\\
2 & 2 & 1 & 0 & 1 & 0\\
3 & 4 & 3 & 0 & 0 & 1
\end{bmatrix}\xrightarrow[r_{3}-3r_{1}]{r_{2}-2r_{1}}\begin{bmatrix}1 & 2 & 3 & 1 & 0 & 0\\
0 & -2 & -5 & -2 & 1 & 0\\
0 & -2 & -6 & -3 & 0 & 1
\end{bmatrix}\\
& \xrightarrow[r_{3}-r_{2}]{r_{1}+r_{2}}\begin{bmatrix}1 & 0 & -2 & -1 & 1 & 0\\
0 & -2 & -5 & -2 & 1 & 0\\
0 & 0 & -1 & -1 & -1 & 1
\end{bmatrix}\xrightarrow[r_{2}-5r_{3}]{r_{1}-2r_{3}}\begin{bmatrix}1 & 0 & 0 & 1 & 3 & -2\\
0 & -2 & 0 & 3 & 6 & -5\\
0 & 0 & -1 & -1 & -1 & 1
\end{bmatrix}\\
& \xrightarrow[r_{3}\div(-1)]{r_{2}\div(-2)}\begin{bmatrix}1 & 0 & 0 & 1 & 3 & -2\\
0 & 1 & 0 & -3/2 & -3 & 5/2\\
0 & 0 & 1 & 1 & 1 & -1
\end{bmatrix},
\end{align*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula
\begin{align*}
\begin{bmatrix}A & E\end{bmatrix} & \xrightarrow{}\begin{bmatrix}1 & 0 & 0 & 1 & 3 & -2\\
0 & 1 & 0 & -3/2 & -3 & 5/2\\
0 & 0 & 1 & 1 & 1 & -1
\end{bmatrix},
\end{align*}
\end_inset
所以
\begin_inset Formula
\[
A^{-1}=\begin{bmatrix}1 & 3 & -2\\
-3/2 & -3 & 5/2\\
1 & 1 & -1
\end{bmatrix}.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
求逆矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
已知矩阵
\begin_inset Formula $A=\begin{bmatrix}1 & 0 & 1\\
2 & 1 & 0\\
-3 & 2 & -5
\end{bmatrix}$
\end_inset
, 求
\begin_inset Formula $(E-A)^{-1}$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset Formula $A=\begin{bmatrix}1 & 0 & 1\\
2 & 1 & 0\\
-3 & 2 & -5
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $E-A=\begin{bmatrix}0 & 0 & -1\\
-2 & 0 & 0\\
3 & -2 & 6
\end{bmatrix}$
\end_inset
,
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-3mm}
\end_layout
\end_inset
\begin_inset Formula
\[
\begin{aligned}\begin{bmatrix}E-A & E\end{bmatrix} & =\begin{bmatrix}0 & 0 & -1 & 1 & 0 & 0\\
-2 & 0 & 0 & 0 & 1 & 0\\
3 & -2 & 6 & 0 & 0 & 1
\end{bmatrix}\longrightarrow\begin{bmatrix}-2 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & -1 & 1 & 0 & 0\\
3 & -2 & 6 & 0 & 0 & 1
\end{bmatrix}\\
& \longrightarrow\begin{bmatrix}-2 & 0 & 0 & 0 & 1 & 0\\
3 & -2 & 6 & 0 & 0 & 1\\
0 & 0 & -1 & 1 & 0 & 0
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 0 & 0 & 0 & -1/2 & 0\\
3 & -2 & 6 & 0 & 0 & 1\\
0 & 0 & -1 & 1 & 0 & 0
\end{bmatrix}\\
& \longrightarrow\begin{bmatrix}1 & 0 & 0 & 0 & -1/2 & 0\\
0 & 1 & -3 & 0 & -3/4 & -1/2\\
0 & 0 & 1 & -1 & 0 & 0
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 0 & 0 & 0 & -1/2 & 0\\
0 & 1 & 0 & -3 & -3/4 & -1/2\\
0 & 0 & 1 & -1 & 0 & 0
\end{bmatrix},
\end{aligned}
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\begin{aligned}\begin{bmatrix}E-A & E\end{bmatrix} & \longrightarrow\begin{bmatrix}1 & 0 & 0 & 0 & -1/2 & 0\\
0 & 1 & 0 & -3 & -3/4 & -1/2\\
0 & 0 & 1 & -1 & 0 & 0
\end{bmatrix},\end{aligned}
\]
\end_inset
所以
\begin_inset Formula
\[
(E-A)^{-1}=\begin{bmatrix}0 & -1/2 & 0\\
-3 & -3/4 & -1/2\\
-1 & 0 & 0
\end{bmatrix}.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
求逆矩阵
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求下列
\begin_inset Formula $n$
\end_inset
阶方阵的逆矩阵:
\begin_inset Formula
\[
A=\begin{bmatrix} & & & a_{1}\\
& & a_{2}\\
& \iddots\\
a_{n}
\end{bmatrix},\ a_{i}\neq0,\ (i=1,2,\cdots,n),
\]
\end_inset
\begin_inset Formula $A$
\end_inset
中空白处表示为零.
\end_layout
\begin_layout Solution*
\begin_inset Formula
\begin{align*}
\begin{bmatrix} & & & a_{1} & 1\\
& & a_{2} & & & 1\\
& \iddots & & & & & \ddots\\
a_{n} & & & & & & & 1
\end{bmatrix} & \rightarrow\begin{bmatrix}a_{n} & & & & & & & 1\\
& a_{n-1} & & & & & \iddots\\
& & \ddots & & & 1\\
& & & a_{1} & 1
\end{bmatrix}\\
& \rightarrow\begin{bmatrix}1 & & & & & & & 1/a_{n}\\
& 1 & & & & & \iddots\\
& & \ddots & & & 1/a_{2}\\
& & & 1 & 1/a_{1}
\end{bmatrix},
\end{align*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula
\begin{align*}
\begin{bmatrix}A & E\end{bmatrix}\rightarrow\begin{bmatrix} & & & a_{1} & 1\\
& & a_{2} & & & 1\\
& \iddots & & & & & \ddots\\
a_{n} & & & & & & & 1
\end{bmatrix} & \rightarrow\begin{bmatrix}1 & & & & & & & 1/a_{n}\\
& 1 & & & & & \iddots\\
& & \ddots & & & 1/a_{2}\\
& & & 1 & 1/a_{1}
\end{bmatrix},
\end{align*}
\end_inset
所以
\begin_inset Formula $A^{-1}=\begin{bmatrix} & & & 1/a_{n}\\
& & \iddots\\
& 1/a_{2}\\
1/a_{1}
\end{bmatrix}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
求逆矩阵的初等变换法
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
把可逆矩阵
\begin_inset Formula $A=\begin{bmatrix}1 & 2 & 0\\
-1 & 1 & 1\\
3 & -2 & 0
\end{bmatrix}$
\end_inset
分解为初等矩阵的乘积.
\end_layout
\begin_layout Solution*
对
\begin_inset Formula $A$
\end_inset
进行如下初等变换:
\begin_inset Formula
\begin{align*}
\begin{bmatrix}1 & 2 & 0\\
-1 & 1 & 1\\
3 & -2 & 0
\end{bmatrix} & \xrightarrow{c_{2}-2c_{1}}\begin{bmatrix}1 & 0 & 0\\
-1 & 3 & 1\\
3 & -8 & 0
\end{bmatrix}\xrightarrow{r_{2}+r_{1}}\begin{bmatrix}1 & 0 & 0\\
0 & 3 & 1\\
3 & -8 & 0
\end{bmatrix}\xrightarrow{r_{3}-3r_{1}}\begin{bmatrix}1 & 0 & 0\\
0 & 3 & 1\\
0 & -8 & 0
\end{bmatrix}\\
& \xrightarrow{c_{3}\leftrightarrow c_{2}}\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 3\\
0 & 0 & -8
\end{bmatrix}\xrightarrow{c_{3}-3c_{2}}\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -8
\end{bmatrix}\xrightarrow{\left(-\frac{1}{8}\right)c_{3}}\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}.
\end{align*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
与每次初等交换对应的矩阵分别为:
\begin_inset Formula
\[
\begin{aligned}P_{1}=\begin{bmatrix}1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}, & P_{2}=\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
-3 & 0 & 1
\end{bmatrix}, & P_{3}=\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -1/8
\end{bmatrix},\\
Q_{1}=\begin{bmatrix}1 & -2 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}, & Q_{2}=\begin{bmatrix}1 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0
\end{bmatrix}, & Q_{3}=\begin{bmatrix}1 & 0 & 0\\
0 & 1 & -3\\
0 & 0 & 1
\end{bmatrix},
\end{aligned}
\]
\end_inset
其中
\begin_inset Formula $P_{i}$
\end_inset
为行变换的初等矩阵,
\begin_inset Formula $Q_{i}$
\end_inset
为列变换的初等矩阵, 其逆矩阵分别为:
\begin_inset Formula
\[
\begin{aligned}P_{1}^{-1}=\begin{bmatrix}1 & 0 & 0\\
-1 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}, & P_{2}^{-1}=\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
3 & 0 & 1
\end{bmatrix}, & P_{3}^{-1}=\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -8
\end{bmatrix},\\
Q_{1}^{-1}=\begin{bmatrix}1 & 2 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}, & Q_{2}^{-1}=\begin{bmatrix}1 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0
\end{bmatrix}, & Q_{3}^{-1}=\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 3\\
0 & 0 & 1
\end{bmatrix},
\end{aligned}
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
于是
\begin_inset Formula
\begin{align*}
A & =P_{1}^{-1}P_{2}^{-1}P_{3}^{-1}Q_{3}^{-1}Q_{2}^{-1}Q_{1}^{-1}\\
& =\begin{bmatrix}1 & 0 & 0\\
-1 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
3 & 0 & 1
\end{bmatrix}\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -8
\end{bmatrix}\cdot\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 3\\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}1 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0
\end{bmatrix}\begin{bmatrix}1 & 2 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}.
\end{align*}
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
用初等变换法求解矩阵方程
\begin_inset Formula $AX=B$
\end_inset
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
设矩阵
\begin_inset Formula $A$
\end_inset
可逆, 则求解矩阵方程
\begin_inset Formula $AX=B$
\end_inset
等价于求矩阵
\begin_inset Formula
\[
X=A^{-1}B,
\]
\end_inset
\end_layout
\begin_layout Standard
为此, 可采用类似初等行变换求矩阵的逆的方法, 构造矩阵
\begin_inset Formula $\begin{pmatrix}A & B\end{pmatrix}$
\end_inset
, 对其施以
\series bold
初等行变换
\series default
\begin_inset Foot
status open
\begin_layout Plain Layout
且只能做初等行变换, 而不能做初等列变换
\end_layout
\end_inset
将矩阵
\begin_inset Formula $A$
\end_inset
化为单位矩阵
\begin_inset Formula $E$
\end_inset
, 则上述初等行变换同时也将其中的单位矩阵
\begin_inset Formula $B$
\end_inset
化为
\begin_inset Formula $A^{-1}B$
\end_inset
, 即
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\begin{pmatrix}A & B\end{pmatrix}\xrightarrow{\text{ 初等行变换 }}\begin{pmatrix}E & A^{-1}B\end{pmatrix}.
\]
\end_inset
\end_layout
\begin_layout Standard
这样就给出了用初等行变换求解矩阵方程
\begin_inset Formula $AX=B$
\end_inset
的方法.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
用初等变换法求解矩阵方程
\begin_inset Formula $XA=B$
\end_inset
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
同理, 求解矩阵方程
\begin_inset Formula $XA=B$
\end_inset
, 等价于计算矩阵
\begin_inset Formula $BA^{-1}$
\end_inset
, 亦可利用
\series bold
初等列变换
\series default
\begin_inset Foot
status open
\begin_layout Plain Layout
且只能做初等列变换, 而不能做初等行变换
\end_layout
\end_inset
求矩阵
\begin_inset Formula $BA^{-1}$
\end_inset
.
即
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\begin{pmatrix}A\\
B
\end{pmatrix}\xrightarrow{\text{ 初等列变换 }}\begin{pmatrix}E\\
BA^{-1}
\end{pmatrix}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Box Boxed
position "t"
hor_pos "c"
has_inner_box 1
inner_pos "t"
use_parbox 0
use_makebox 0
width "100col%"
special "none"
height "1in"
height_special "totalheight"
thickness "0.4pt"
separation "3pt"
shadowsize "4pt"
framecolor "black"
backgroundcolor "none"
status open
\begin_layout Plain Layout
\color red
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-5mm}
\backslash
begin{center}
\end_layout
\begin_layout Plain Layout
左乘变行, 右乘变列.
\end_layout
\begin_layout Plain Layout
\backslash
end{center}
\end_layout
\end_inset
\end_layout
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
解矩阵方程
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求矩阵
\begin_inset Formula $X$
\end_inset
, 使
\begin_inset Formula $AX=B$
\end_inset
, 其中
\begin_inset Formula $A=\begin{bmatrix}1 & 2 & 3\\
2 & 2 & 1\\
3 & 4 & 3
\end{bmatrix}$
\end_inset
,
\begin_inset Formula $B=\begin{bmatrix}2 & 5\\
3 & 1\\
4 & 3
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
若
\begin_inset Formula $A$
\end_inset
可逆, 则
\begin_inset Formula $X=A^{-1}B$
\end_inset
.
\begin_inset Formula
\begin{align*}
\begin{bmatrix}A & B\end{bmatrix} & =\begin{bmatrix}1 & 2 & 3 & 2 & 5\\
2 & 2 & 1 & 3 & 1\\
3 & 4 & 3 & 4 & 3
\end{bmatrix}\xrightarrow[r_{3}-3r_{1}]{r_{2}-2r_{1}}\begin{bmatrix}1 & 2 & 3 & 2 & 5\\
0 & -2 & -5 & -1 & -9\\
0 & -2 & -6 & -2 & -12
\end{bmatrix}\\
& \xrightarrow[r_{3}-r_{2}]{r_{1}+r_{2}}\begin{bmatrix}1 & 0 & -2 & 1 & -4\\
0 & -2 & -5 & -1 & -9\\
0 & 0 & -1 & -1 & -3
\end{bmatrix}\xrightarrow[r_{2}-5r_{3}]{r_{1}-2r_{3}}\begin{bmatrix}1 & 0 & 0 & 3 & 2\\
0 & -2 & 0 & 4 & 6\\
0 & 0 & -1 & -1 & -3
\end{bmatrix}\\
& \xrightarrow[r_{3}\div(-1)]{r_{2}\div(-2)}\begin{bmatrix}1 & 0 & 0 & 3 & 2\\
0 & 1 & 0 & -2 & -3\\
0 & 0 & 1 & 1 & 3
\end{bmatrix},
\end{align*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula
\begin{align*}
\begin{bmatrix}A & B\end{bmatrix} & \xrightarrow{}\begin{bmatrix}1 & 0 & 0 & 3 & 2\\
0 & 1 & 0 & -2 & -3\\
0 & 0 & 1 & 1 & 3
\end{bmatrix}\leftrightarrow\begin{bmatrix}E & A^{-1}B\end{bmatrix},
\end{align*}
\end_inset
所以
\begin_inset Formula $X=\begin{bmatrix}3 & 2\\
-2 & -3\\
1 & 3
\end{bmatrix}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
解矩阵方程
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求解矩阵方程
\begin_inset Formula $AX=A+X$
\end_inset
, 其中
\begin_inset Formula $A=\begin{bmatrix}2 & 2 & 0\\
2 & 1 & 3\\
0 & 1 & 0
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
把所给方程变形为
\begin_inset Formula $(A-E)X=A$
\end_inset
, 则
\begin_inset Formula $X=(A-E)^{-1}A$
\end_inset
.
\begin_inset Formula
\begin{align*}
\begin{bmatrix}A-E & A\end{bmatrix} & =\begin{bmatrix}1 & 2 & 0 & 2 & 2 & 0\\
2 & 0 & 3 & 2 & 1 & 3\\
0 & 1 & -1 & 0 & 1 & 0
\end{bmatrix}\xrightarrow[r_{2}\leftrightarrow r_{3}]{r_{2}-2r_{1}}\begin{bmatrix}1 & 2 & 0 & 2 & 2 & 0\\
0 & 1 & -1 & 0 & 1 & 0\\
0 & -4 & 3 & -2 & -3 & 3
\end{bmatrix}\\
& \xrightarrow[r_{3}\div(-1)]{r_{3}+4r_{2}}\begin{bmatrix}1 & 2 & 0 & 2 & 2 & 0\\
0 & 1 & -1 & 0 & 1 & 0\\
0 & 0 & 0 & 2 & 1 & -3
\end{bmatrix}\xrightarrow{r_{2}+r_{3}}\begin{bmatrix}1 & 2 & 0 & 2 & 2 & 0\\
0 & 1 & 0 & 2 & 0 & -3\\
0 & 0 & 1 & 2 & -1 & -3
\end{bmatrix}\\
& \xrightarrow{r_{1}-2r_{2}}\begin{bmatrix}1 & 2 & 0 & 2 & 2 & 0\\
0 & 1 & 0 & 2 & 0 & -3\\
0 & 0 & 1 & 2 & -1 & -3
\end{bmatrix}
\end{align*}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula
\begin{align*}
\begin{bmatrix}A-E & A\end{bmatrix} & \xrightarrow{}\begin{bmatrix}1 & 2 & 0 & 2 & 2 & 0\\
0 & 1 & 0 & 2 & 0 & -3\\
0 & 0 & 1 & 2 & -1 & -3
\end{bmatrix}
\end{align*}
\end_inset
即得
\begin_inset Formula $X=\begin{bmatrix}-2 & 2 & 6\\
2 & 0 & -3\\
2 & -1 & -3
\end{bmatrix}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
解矩阵方程
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求解矩阵方程
\begin_inset Formula $XA=A+2X$
\end_inset
, 其中
\begin_inset Formula $A=\begin{bmatrix}4 & 2 & 3\\
1 & 1 & 0\\
-1 & 2 & 3
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
先将原方程作恒等变形:
\begin_inset Formula
\[
XA=A+2X\Longleftrightarrow XA-2X=A\Longleftrightarrow X(A-2E)=A,
\]
\end_inset
由于
\begin_inset Formula $A-2E=\begin{bmatrix}2 & 2 & 3\\
1 & -1 & 0\\
-1 & 2 & 1
\end{bmatrix}$
\end_inset
, 而
\begin_inset Formula $|A-2E|=-1\neq0$
\end_inset
, 故
\begin_inset Formula $A-2E$
\end_inset
可逆.
从而
\begin_inset Formula $X=A(A-2E)^{-1}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula
\begin{align*}
\begin{bmatrix}A-2E\\
A
\end{bmatrix} & =\begin{bmatrix}2 & 2 & 3\\
1 & -1 & 0\\
-1 & 2 & 1\\
4 & 2 & 3\\
1 & 1 & 0\\
-1 & 2 & 3
\end{bmatrix}\longrightarrow\begin{bmatrix}-1 & 2 & 3\\
1 & -1 & 0\\
-2 & 2 & 1\\
1 & 2 & 3\\
1 & 1 & 0\\
-4 & 2 & 3
\end{bmatrix}\longrightarrow\begin{bmatrix}-1 & 0 & 0\\
1 & 1 & 3\\
-2 & -2 & -5\\
1 & 4 & 6\\
1 & 3 & 3\\
-4 & -6 & -9
\end{bmatrix}\\
& \longrightarrow\begin{bmatrix}1 & 0 & 0\\
-1 & 1 & 0\\
2 & -2 & 1\\
-1 & 4 & -6\\
-1 & 3 & -6\\
4 & -6 & 9
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 0 & 0\\
-1 & 1 & 0\\
0 & 0 & 1\\
11 & -8 & -6\\
11 & -9 & -6\\
-14 & 12 & 9
\end{bmatrix}\longrightarrow\begin{bmatrix}1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
3 & -8 & -6\\
2 & -9 & -6\\
-2 & 12 & 9
\end{bmatrix},
\end{align*}
\end_inset
即
\begin_inset Formula $X=\begin{bmatrix}3 & -8 & -6\\
2 & -9 & -6\\
-2 & 12 & 9
\end{bmatrix}$
\end_inset
.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Subsection
作业
\end_layout
\begin_layout Frame
\begin_inset Argument 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
作业
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Problem
化矩阵
\begin_inset Formula $A=\begin{bmatrix}1 & 0 & 1\\
2 & 1 & 0\\
-3 & 2 & -5
\end{bmatrix}$
\end_inset
为矩阵的标准形式.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
求下面矩阵的逆矩阵: (1).
\begin_inset Formula $\begin{bmatrix}2 & 0 & 0\\
0 & 3 & 4\\
0 & 1 & 1
\end{bmatrix}$
\end_inset
; (2).
\begin_inset Formula $A=\begin{bmatrix}1 & 0 & 1\\
2 & 1 & 0\\
-3 & 2 & -5
\end{bmatrix}$
\end_inset
;
\end_layout
\begin_layout Problem
(3).
\begin_inset Formula $\begin{bmatrix}0 & a_{1} & 0 & \cdots & 0\\
0 & 0 & a_{2} & \cdots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & \cdots & a_{n-1}\\
a_{n} & 0 & 0 & \cdots & 0
\end{bmatrix}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
已知
\begin_inset Formula $n$
\end_inset
方阵
\begin_inset Formula $A=\begin{bmatrix}2 & 2 & 2 & \cdots & 2\\
0 & 1 & 1 & \cdots & 1\\
0 & 0 & 1 & \cdots & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & \cdots & 1
\end{bmatrix}$
\end_inset
, 求
\begin_inset Formula $A$
\end_inset
中所有元素的代数余子式之和
\begin_inset Formula $\sum_{i,j=1}^{n}A_{ij}$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Standard
\series bold
Hint
\series default
: 要求所有代数余子式的和, 这相当于求
\begin_inset Formula
\[
\begin{bmatrix}1 & 1 & \cdots & 1\end{bmatrix}A^{*}\begin{bmatrix}1\\
1\\
\vdots\\
1
\end{bmatrix},
\]
\end_inset
并注意使用
\begin_inset Formula $A^{*}A=\left|A\right|E$
\end_inset
消去不易计算的伴随矩阵.
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
设
\begin_inset Formula $A,B$
\end_inset
和
\begin_inset Formula $C$
\end_inset
是可逆方阵, 证明: 方阵
\begin_inset Formula $X=\begin{bmatrix}O & O & A\\
O & B & O\\
C & O & O
\end{bmatrix}$
\end_inset
也可逆, 并求
\begin_inset Formula $X^{-1}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
设
\begin_inset Formula $M=\begin{bmatrix}A & B\\
C & D
\end{bmatrix}$
\end_inset
, 其中
\begin_inset Formula $A$
\end_inset
和
\begin_inset Formula $D$
\end_inset
是方阵, 证明:
\end_layout
\begin_layout Problem
(1).
当
\begin_inset Formula $A$
\end_inset
可逆时,
\begin_inset Formula $M$
\end_inset
可逆当且仅当
\begin_inset Formula $D-CA^{-1}B$
\end_inset
可逆;
\end_layout
\begin_layout Problem
(2).
当
\begin_inset Formula $D$
\end_inset
可逆时,
\begin_inset Formula $M$
\end_inset
可逆当且仅当
\begin_inset Formula $A-BD^{-1}C$
\end_inset
可逆;
\end_layout
\begin_layout Problem
(3).
当
\begin_inset Formula $A$
\end_inset
可逆时, 行列式的计算有如下降阶法:
\begin_inset Formula
\[
\left|M\right|=\left|A\right|\cdot\left|D-CA^{-1}B\right|.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\begin_layout Frame
\end_layout
\begin_layout Frame
\end_layout
\end_body
\end_document
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