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\end_header
\begin_body
\begin_layout Section
\begin_inset CommandInset label
LatexCommand label
name "sec:4"
\end_inset
行列式按行 (列) 展开
\end_layout
\begin_layout Subsection
行列式按一行 (列) 展开
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
余子式与代数余子式
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Definition
在
\begin_inset Formula $n$
\end_inset
阶行列式
\begin_inset Formula $D$
\end_inset
中, 去掉元素
\begin_inset Formula $a_{ij}$
\end_inset
所在的第
\begin_inset Formula $i$
\end_inset
行和第
\begin_inset Formula $j$
\end_inset
列后, 余下的
\begin_inset Formula $n-1$
\end_inset
阶行列式, 称为
\begin_inset Formula $D$
\end_inset
中
\series bold
元素
\begin_inset Formula $a_{ij}$
\end_inset
的余子式
\series default
, 记为
\begin_inset Formula $M_{ij}$
\end_inset
, 再记
\begin_inset Formula
\[
A_{ij}=(-1)^{i+j}M_{ij}
\]
\end_inset
称
\begin_inset Formula $A_{ij}$
\end_inset
为
\series bold
元素
\begin_inset Formula $a_{ij}$
\end_inset
的代数余子式
\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
行列式的计算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\begin_inset CommandInset label
LatexCommand label
name "exa:1.4-1"
\end_inset
设有
\begin_inset Formula $5$
\end_inset
阶行列式:
\begin_inset Formula $D=\begin{vmatrix}1 & 0 & -1 & 3 & 1\\
0 & 2 & -5 & 4 & 1\\
3 & -2 & -1 & 1 & 0\\
0 & 0 & 2 & 1 & 3\\
1 & 3 & -1 & 5 & 1
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Example
(1).
\begin_inset Formula $a_{11}=1$
\end_inset
, 其余子式
\begin_inset Formula $M_{11}=\begin{vmatrix}2 & -5 & 4 & 1\\
-2 & -1 & 1 & 0\\
0 & 2 & 1 & 3\\
3 & -1 & 5 & 1
\end{vmatrix}$
\end_inset
, 其代数余子式
\begin_inset Formula
\[
A_{11}=(-1)^{1+1}M_{11}=(-1)^{2}M_{11}=M_{11}\text{. }
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Example*
\begin_inset Argument 1
status open
\begin_layout Plain Layout
\begin_inset CommandInset ref
LatexCommand ref
reference "exa:1.4-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
\end_layout
\end_inset
设有
\begin_inset Formula $5$
\end_inset
阶行列式:
\begin_inset Formula $D=\begin{vmatrix}1 & 0 & -1 & 3 & 1\\
0 & 2 & -5 & 4 & 1\\
3 & -2 & -1 & 1 & 0\\
0 & 0 & 2 & 1 & 3\\
1 & 3 & -1 & 5 & 1
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Example*
(2).
\begin_inset Formula $a_{34}=1$
\end_inset
, 其余子式
\begin_inset Formula $M_{34}=\begin{vmatrix}1 & 0 & -1 & 1\\
0 & 2 & -5 & 1\\
0 & 0 & 2 & 3\\
1 & 3 & -1 & 1
\end{vmatrix}$
\end_inset
, 其代数余子式
\begin_inset Formula
\[
A_{34}=(-1)^{3+4}M_{34}=(-1)^{7}M_{34}=-M_{34}.
\]
\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
行列式按一行 (列) 展开
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Lemma
一个
\begin_inset Formula $n$
\end_inset
阶行列式
\begin_inset Formula $D$
\end_inset
, 若其中第
\begin_inset Formula $i$
\end_inset
行所有元素除
\begin_inset Formula $a_{ij}$
\end_inset
外都为零, 则该行列式等于
\begin_inset Formula $a_{ij}$
\end_inset
与它的代数余子式的乘积, 即
\begin_inset Formula
\[
D=a_{ij}A_{ij}.
\]
\end_inset
\end_layout
\begin_layout Theorem
行列式
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\[
D=\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1j} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2j} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots & \ddots & \vdots\\
a_{i1} & a_{i2} & \cdots & a_{ij} & \cdots & a_{in}\\
\vdots & \vdots & \ddots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nj} & \cdots & a_{nn}
\end{vmatrix}
\]
\end_inset
等于它的任一行 (列) 的各元素与其对应的代数余子式乘积之和, 即
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-3mm}
\end_layout
\end_inset
\begin_inset Formula
\[
D=a_{i1}A_{i1}+a_{i2}A_{i2}+\cdots+a_{in}A_{in},\qquad(i=1,2,\cdots,n),
\]
\end_inset
或
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-3mm}
\end_layout
\end_inset
\begin_inset Formula
\[
D=a_{1j}A_{1j}+a_{2j}A_{2j}+\cdots+a_{nj}A_{nj},\qquad(j=1,2,\cdots,n).
\]
\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
不同行 (列) 的元素与代数余子式的线性组合
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Corollary
\begin_inset CommandInset label
LatexCommand label
name "cor:1.4-1"
\end_inset
行列式某一行 (列) 的元素与另一行 (列) 的对应元素的代数余子式乘积之和等于零, 即:
\begin_inset Formula
\[
a_{i1}A_{j1}+a_{i2}A_{j2}+\cdots+a_{in}A_{jn}=0,\qquad i\neq j,
\]
\end_inset
或
\begin_inset Formula
\[
a_{1i}A_{1j}+a_{2i}A_{2j}+\cdots+a_{ni}A_{nj}=0,\qquad i\neq j.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Corollary*
\begin_inset Argument 1
status open
\begin_layout Plain Layout
\begin_inset CommandInset ref
LatexCommand ref
reference "cor:1.4-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
\end_layout
\end_inset
综上所述, 可得到有关代数余子式的一个重要性质:
\begin_inset Formula
\[
\sum_{k=1}^{n}a_{ki}A_{kj}=D\delta_{ij}=\begin{cases}
D, & \text{ 当 }i=j,\\
0, & \text{ 当 }i\neq j;
\end{cases}
\]
\end_inset
或
\begin_inset Formula
\[
\sum_{k=1}^{n}a_{ik}A_{jk}=D\delta_{ij}=\begin{cases}
D, & \text{ 当 }i=j,\\
0, & \text{ 当 }i\neq j.
\end{cases}
\]
\end_inset
其中,
\begin_inset Formula $\delta_{ij}=\begin{cases}
1, & i=j\\
0, & i\neq j
\end{cases}$
\end_inset
为 Kronecker 符号.
\end_layout
\end_deeper
\begin_layout Subsection
行列式的计算
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
行列式的计算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
直接应用按行 (列) 展开法则计算行列式, 运算量较大, 尤其是高阶行列式.
因此, 计算行列式时, 一般可先用行列式的性质将行列式中某一行 (列) 化为仅含有一个非零元素, 再按此行 (列) 展开, 化为低一阶的行列式,
如此继续下去直到化为三阶或二阶行列式.
\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
行列式的计算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求行列式的值:
\begin_inset Formula $\begin{vmatrix}2 & -1 & 3\\
-1 & 2 & 1\\
4 & 1 & 2
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
按第一列展开行列式
\begin_inset Formula
\begin{align*}
\begin{vmatrix}2 & -1 & 3\\
-1 & 2 & 1\\
4 & 1 & 2
\end{vmatrix} & =2\times\begin{vmatrix}2 & 1\\
1 & 2
\end{vmatrix}-(-1)\times\begin{vmatrix}-1 & 3\\
1 & 2
\end{vmatrix}+4\times\begin{vmatrix}-1 & 3\\
2 & 1
\end{vmatrix}\\
& =2(4-1)+(-2-3)+4(-1-6)=6-5-28=-27.
\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
行列式的计算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求行列式的值:
\begin_inset Formula $\begin{vmatrix}3 & 2 & 7\\
0 & 5 & 2\\
0 & 2 & 1
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Pause
\end_layout
\end_deeper
\begin_layout Solution*
按第一列展开行列式
\begin_inset Formula
\[
\begin{vmatrix}3 & 2 & 7\\
0 & 5 & 2\\
0 & 2 & 1
\end{vmatrix}=3\times\begin{vmatrix}5 & 2\\
2 & 1
\end{vmatrix}=3(5-4)=3.
\]
\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
行列式的计算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
试按第三列展开计算行列式
\begin_inset Formula $D=\begin{vmatrix}1 & 2 & 3 & 4\\
1 & 0 & 1 & 2\\
3 & -1 & -1 & 0\\
1 & 2 & 0 & -5
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_deeper
\begin_layout Pause
\end_layout
\end_deeper
\begin_layout Solution*
将
\begin_inset Formula $D$
\end_inset
按第三列展开, 则有
\begin_inset Formula $D=a_{13}A_{13}+a_{23}A_{23}+a_{33}A_{33}+a_{43}A_{43}$
\end_inset
, 其中
\begin_inset Formula $a_{13}=3$
\end_inset
,
\begin_inset Formula $a_{23}=1$
\end_inset
,
\begin_inset Formula $a_{33}=-1$
\end_inset
,
\begin_inset Formula $a_{43}=0$
\end_inset
,
\begin_inset Formula
\[
\begin{aligned}A_{13}=(-1)^{1+3}\begin{vmatrix}1 & 0 & 2\\
3 & -1 & 0\\
1 & 2 & -5
\end{vmatrix}=19, & A_{23}=(-1)^{2+3}\begin{vmatrix}1 & 2 & 4\\
3 & -1 & 0\\
1 & 2 & -5
\end{vmatrix}=-63\text{, }\\
A_{33}=(-1)^{3+3}\begin{vmatrix}1 & 2 & 4\\
1 & 0 & 2\\
1 & 2 & -5
\end{vmatrix}=18, & A_{43}=(-1)^{4+3}\begin{vmatrix}1 & 2 & 4\\
1 & 0 & 2\\
3 & -1 & 0
\end{vmatrix}=-10,
\end{aligned}
\]
\end_inset
所以
\begin_inset Formula $D=3\times19+1\times(-63)+(-1)\times18+0\times(-10)=-24$
\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
代数余子式的线性组合
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
已知
\begin_inset Formula $4$
\end_inset
阶行列式
\begin_inset Formula
\[
D=\begin{vmatrix}1 & 1 & 1 & 3\\
2 & 1 & 3 & 4\\
3 & 0 & -1 & 0\\
4 & 6 & 3 & 8
\end{vmatrix},
\]
\end_inset
试求其第二行元素的代数余子式之和, 即求
\begin_inset Formula
\[
S=A_{21}+A_{22}+A_{23}+A_{24}.
\]
\end_inset
\end_layout
\begin_deeper
\begin_layout Pause
\end_layout
\end_deeper
\begin_layout Solution*
\begin_inset Argument 1
status open
\begin_layout Plain Layout
解法一
\end_layout
\end_inset
由于
\begin_inset Formula
\[
D=\begin{vmatrix}{\color{magenta}1} & {\color{magenta}1} & {\color{magenta}1} & {\color{magenta}3}\\
2 & 1 & 3 & 4\\
3 & 0 & -1 & 0\\
4 & 6 & 3 & 8
\end{vmatrix},
\]
\end_inset
用行列式的
\uwave on
第一行元素
\uwave default
与
\uwave on
第二行元素的代数余子式
\uwave default
作乘积, 其和为零
\begin_inset Formula
\[
{\color{magenta}1}\cdot A_{21}+{\color{magenta}1}\cdot A_{22}+{\color{magenta}1}\cdot A_{23}+{\color{magenta}3}\cdot A_{24}=0.
\]
\end_inset
所以
\begin_inset Formula
\[
\begin{aligned}S & =A_{21}+A_{22}+A_{23}+A_{24}=-2\cdot A_{24}\\
& =-2\cdot(-1)^{2+4}\begin{vmatrix}1 & 1 & 1\\
3 & 0 & -1\\
4 & 6 & 3
\end{vmatrix}=-22.
\end{aligned}
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Argument 1
status open
\begin_layout Plain Layout
解法二
\end_layout
\end_inset
由于
\begin_inset Formula
\[
\begin{aligned}D & =\begin{vmatrix}1 & 1 & 1 & 3\\
{\color{magenta}2} & {\color{magenta}1} & {\color{magenta}3} & {\color{magenta}4}\\
2 & 0 & -1 & 0\\
4 & 6 & 3 & 8
\end{vmatrix},\end{aligned}
\]
\end_inset
所以
\begin_inset Formula
\[
S=A_{21}+A_{22}+A_{23}+A_{24}=\begin{vmatrix}1 & 1 & 1 & 3\\
{\color{magenta}1} & {\color{magenta}1} & {\color{magenta}1} & {\color{magenta}1}\\
3 & 0 & -1 & 0\\
4 & 6 & 3 & 8
\end{vmatrix}.
\]
\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
(代数) 余子式的线性组合
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
设
\begin_inset Formula $D=\begin{vmatrix}3 & -5 & 2 & 1\\
1 & 1 & 0 & -5\\
-1 & 3 & 1 & 3\\
2 & -4 & -1 & -3
\end{vmatrix}$
\end_inset
,
\begin_inset Formula $D$
\end_inset
中元素
\begin_inset Formula $a_{ij}$
\end_inset
的余子式和代数余子式依次记作
\begin_inset Formula $M_{ij}$
\end_inset
和
\begin_inset Formula $A_{ij}$
\end_inset
, 求
\begin_inset Formula $A_{11}+A_{12}+A_{13}+A_{14}$
\end_inset
及
\begin_inset Formula $M_{11}+M_{21}+M_{31}+M_{41}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
注意到
\begin_inset Formula $A_{11}+A_{12}+A_{13}+A_{14}$
\end_inset
等于用
\begin_inset Formula $1,1,1,1$
\end_inset
代替
\begin_inset Formula $D$
\end_inset
的第
\begin_inset Formula $1$
\end_inset
行所得的行列式, 即
\begin_inset Formula
\[
\begin{aligned}A_{11}+A_{12}+A_{13}+A_{14} & =\begin{vmatrix}1 & 1 & 1 & 1\\
1 & 1 & 0 & -5\\
-1 & 3 & 1 & 3\\
2 & -4 & -1 & -3
\end{vmatrix}\xlongequal[r_{3}-r_{1}]{r_{4}+r_{3}}\begin{vmatrix}{\color{blue}1} & {\color{blue}1} & \boxed{{\color{blue}1}} & {\color{blue}1}\\
1 & 1 & {\color{blue}0} & -5\\
-2 & 2 & {\color{blue}0} & 2\\
1 & -1 & {\color{blue}0} & 0
\end{vmatrix}\\
& =\begin{vmatrix}1 & 1 & -5\\
-2 & 2 & 2\\
1 & -1 & 0
\end{vmatrix}\xlongequal{c_{2}+c_{1}}\begin{vmatrix}1 & 2 & -5\\
-2 & 0 & 2\\
\boxed{1} & 0 & 0
\end{vmatrix}=\begin{vmatrix}2 & -5\\
0 & 2
\end{vmatrix}=4.
\end{aligned}
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
又按定义知,
\begin_inset Formula
\[
\begin{aligned}M_{11}+M_{21}+M_{31}+M_{41} & =A_{11}-A_{21}+A_{31}-A_{41}=\begin{vmatrix}1 & -5 & 2 & 1\\
-1 & 1 & 0 & -5\\
1 & 3 & 1 & 3\\
-1 & -4 & -1 & -3
\end{vmatrix}\\
& \xlongequal{r_{4}+r_{3}}\begin{vmatrix}1 & {\color{blue}-5} & 2 & 1\\
-1 & {\color{blue}1} & 0 & -5\\
1 & {\color{blue}3} & 1 & 3\\
{\color{blue}0} & {\color{blue}\boxed{-1}} & {\color{blue}0} & {\color{blue}0}
\end{vmatrix}=(-1)\begin{vmatrix}1 & 2 & 1\\
-1 & 0 & -5\\
1 & 1 & 3
\end{vmatrix}\\
& \xlongequal{r_{1}-2r_{3}}-\begin{vmatrix}-1 & 0 & -5\\
-1 & 0 & -5\\
1 & 1 & 3
\end{vmatrix}=0.
\end{aligned}
\]
\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
行列式的综合计算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
计算行列式
\begin_inset Formula $D=\begin{vmatrix}1 & 2 & 3 & 4\\
1 & 0 & 1 & 2\\
3 & -1 & -1 & 0\\
1 & 2 & 0 & -5
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\begin{aligned}D & =\begin{vmatrix}1 & 2 & 3 & 4\\
1 & 0 & 1 & 2\\
3 & -1 & -1 & 0\\
1 & 2 & 0 & -5
\end{vmatrix}\xlongequal[r_{4}+2r_{3}]{r_{1}+2r_{3}}\begin{vmatrix}7 & 0 & 1 & 4\\
1 & 0 & 1 & 2\\
3 & -1 & -1 & 0\\
7 & 0 & -2 & -5
\end{vmatrix}\\
& =(-1)\times(-1)^{3+2}\begin{vmatrix}7 & 1 & 4\\
1 & 1 & 2\\
7 & -2 & -5
\end{vmatrix}\xlongequal[r_{3}+2r_{2}]{r_{1}-r_{2}}\begin{vmatrix}6 & 0 & 2\\
1 & 1 & 2\\
9 & 0 & -1
\end{vmatrix}\\
& =1\times(-1)^{2+2}\begin{vmatrix}6 & 2\\
9 & -1
\end{vmatrix}=-6-18=-24.
\end{aligned}
\]
\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
行列式的综合计算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
计算行列式
\begin_inset Formula $D=\begin{vmatrix}5 & 3 & -1 & 2 & 0\\
1 & 7 & 2 & 5 & 2\\
0 & -2 & 3 & 1 & 0\\
0 & -4 & -1 & 4 & 0\\
0 & 2 & 3 & 5 & 0
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\begin{aligned}D & =\begin{vmatrix}5 & 3 & -1 & 2 & 0\\
1 & 7 & 2 & 5 & 2\\
0 & -2 & 3 & 1 & 0\\
0 & -4 & -1 & 4 & 0\\
0 & 2 & 3 & 5 & 0
\end{vmatrix}=(-1)^{2+5}2\begin{vmatrix}5 & 3 & -1 & 2\\
0 & -2 & 3 & 1\\
0 & -4 & -1 & 4\\
0 & 2 & 3 & 5
\end{vmatrix}\\
& =-2\cdot5\begin{vmatrix}-2 & 3 & 1\\
-4 & -1 & 4\\
2 & 3 & 5
\end{vmatrix}\xlongequal[r_{3}+r_{1}]{r_{2}+(-2)r_{1}}-10\begin{vmatrix}-2 & 3 & 1\\
0 & -7 & 2\\
0 & 6 & 6
\end{vmatrix}\\
& =-10\cdot(-2)\begin{vmatrix}-7 & 2\\
6 & 6
\end{vmatrix}=20(-42-12)=-1080.
\end{aligned}
\]
\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
行列式的综合计算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求证:
\begin_inset Formula $\begin{vmatrix}1 & 2 & 3 & 4 & \cdots & n-1 & n\\
1 & 1 & 2 & 3 & \cdots & n-2 & n-1\\
1 & x & 1 & 2 & \cdots & n-3 & n-2\\
1 & x & x & 1 & \cdots & n-4 & n-3\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
1 & x & x & x & \cdots & 1 & 2\\
1 & x & x & x & \cdots & x & 1
\end{vmatrix}=(-1)^{n+1}x^{n-2}$
\end_inset
.
\end_layout
\begin_layout Proof
\begin_inset Formula
\begin{align*}
D & \xlongequal[\substack{r_{3}-r_{4}\\
\vdots\\
r_{n-1}-r_{n}
}
]{\substack{r_{1}-r_{2}\\
r_{2}-r_{3}
}
}\left|\begin{array}{cccccc}
0 & 1 & 1 & \cdots & 1 & 1\\
0 & 1-x & 1 & \cdots & 1 & 1\\
0 & 0 & 1-x & \cdots & 1 & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 1-x & 1\\
1 & x & x & \cdots & x & 1
\end{array}\right|_{\xcancel{n\times n}}\\
& \xlongequal[\substack{r_{3}-r_{4}\\
\vdots\\
r_{n-2}-r_{n-1}
}
]{\substack{r_{1}-r_{2}\\
r_{2}-r_{3}
}
}(-1)^{n+1}\left|\begin{array}{cccccc}
x & 0 & 0 & \cdots & 0 & 0\\
1-x & x & 0 & \cdots & 0 & 0\\
0 & 1-x & x & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & x & 0\\
0 & 0 & 0 & \cdots & 1-x & 1
\end{array}\right|_{\xcancel{(n-1)\times(n-1)}}\hspace{-3em}=(-1)^{n+1}x^{n-2}.
\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 3
status open
\begin_layout Plain Layout
allowframebreaks
\end_layout
\end_inset
\begin_inset Argument 4
status open
\begin_layout Plain Layout
Vandermonde 行列式
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
证明范德蒙德 (Vandermonde) 行列式
\begin_inset Formula
\begin{equation}
D_{n}=\begin{vmatrix}1 & 1 & \cdots & 1\\
x_{1} & x_{2} & \cdots & x_{n}\\
x_{1}^{2} & x_{2}^{2} & \cdots & x_{n}^{2}\\
\vdots & \vdots & \ddots & \vdots\\
x_{1}^{n-1} & x_{2}^{n-1} & \cdots & x_{n}^{n-1}
\end{vmatrix}=\prod_{n\geq i>j\geq1}\left(x_{i}-x_{j}\right),\label{eq:4.2-1}
\end{equation}
\end_inset
其中记号 ``
\begin_inset Formula $\prod$
\end_inset
'' 表示全体同类因子的乘积.
\end_layout
\begin_layout Proof
用数学归纳法.
\begin_inset Formula $D_{2}=\begin{vmatrix}1 & 1\\
x_{1} & x_{2}
\end{vmatrix}=x_{2}-x_{1}=\prod_{2\geq i>j\geq1}\left(x_{i}-x_{j}\right)$
\end_inset
, 所以当
\begin_inset Formula $n=2$
\end_inset
时 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:4.2-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 式成立.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Proof
假设 (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:4.2-1"
plural "false"
caps "false"
noprefix "false"
\end_inset
) 式对于
\begin_inset Formula $n-1$
\end_inset
时成立, 则
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\[
\begin{aligned}D_{n} & =\begin{vmatrix}1 & 1 & \cdots & 1\\
x_{1} & x_{2} & \cdots & x_{n}\\
x_{1}^{2} & x_{2}^{2} & \cdots & x_{n}^{2}\\
\vdots & \vdots & \ddots & \vdots\\
x_{1}^{n-1} & x_{2}^{n-1} & \cdots & x_{n}^{n-1}
\end{vmatrix}\\
& \xlongequal[\substack{r_{4}-x_{1}r_{3}\\
\vdots\\
r_{n}-x_{1}r_{n-1}
}
]{\substack{r_{2}-x_{1}r_{1}\\
r_{3}-x_{1}r_{2}
}
}\begin{vmatrix}1 & 1 & 1 & \cdots & 1\\
0 & x_{2}-x_{1} & x_{3}-x_{1} & \cdots & x_{n}-x_{1}\\
0 & x_{2}\left(x_{2}-x_{1}\right) & x_{3}\left(x_{3}-x_{1}\right) & \cdots & x_{n}\left(x_{n}-x_{1}\right)\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & x_{2}^{n-2}\left(x_{2}-x_{1}\right) & x_{3}^{n-2}\left(x_{3}-x_{1}\right) & \cdots & x_{n}^{n-2}\left(x_{n}-x_{1}\right)
\end{vmatrix}.
\end{aligned}
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Proof
因此
\end_layout
\begin_layout Proof
\begin_inset Formula
\[
\scriptsize D_{n}=\begin{vmatrix}1 & 1 & \cdots & 1\\
x_{1} & x_{2} & \cdots & x_{n}\\
x_{1}^{2} & x_{2}^{2} & \cdots & x_{n}^{2}\\
\vdots & \vdots & \ddots & \vdots\\
x_{1}^{n-1} & x_{2}^{n-1} & \cdots & x_{n}^{n-1}
\end{vmatrix}=\left(x_{2}-x_{1}\right)\left(x_{3}-x_{1}\right)\cdots\left(x_{n}-x_{1}\right)\begin{vmatrix}1 & 1 & \cdots & 1\\
x_{2} & x_{3} & \cdots & x_{n}\\
\vdots & \vdots & \ddots & \vdots\\
x_{2}^{n-2} & x_{3}^{n-2} & \cdots & x_{n}^{n-2}
\end{vmatrix},
\]
\end_inset
由归纳假设便得
\begin_inset Formula
\[
D_{n}=\left(x_{2}-x_{1}\right)\left(x_{3}-x_{1}\right)\cdots\left(x_{n}-x_{1}\right)\prod_{n\geq i>j\geq2}\left(x_{i}-x_{j}\right)=\prod_{n\geq i>j\geq1}\left(x_{i}-x_{j}\right).
\]
\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
行列式的计算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
计算
\begin_inset Formula $2n$
\end_inset
阶行列式
\begin_inset Formula $D_{2n}=\begin{vmatrix}a & & & & & b\\
& \ddots & & & \iddots\\
& & a & b\\
& & c & d\\
& \iddots & & & \ddots\\
c & & & & & d
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Solution*
把
\begin_inset Formula $D_{2n}$
\end_inset
中的第
\begin_inset Formula $2n$
\end_inset
行依次与第
\begin_inset Formula $2n-1$
\end_inset
行,
\begin_inset Formula $\cdots$
\end_inset
, 第
\begin_inset Formula $2$
\end_inset
行对调 (作
\begin_inset Formula $2n-2$
\end_inset
次相邻对换), 再把第
\begin_inset Formula $2n$
\end_inset
列依次与第
\begin_inset Formula $2n-1$
\end_inset
列,
\begin_inset Formula $\cdots$
\end_inset
, 第
\begin_inset Formula $2$
\end_inset
列对调, 得 (注意使用例
\begin_inset CommandInset ref
LatexCommand ref
reference "exa:laplace"
plural "false"
caps "false"
noprefix "false"
\end_inset
的结论)
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{-4mm}
\end_layout
\end_inset
\begin_inset Formula
\[
D_{2n}=(-1)^{2(2n-2)}\left|\begin{array}{llllllll}
a & b & 0 & \cdots & \cdots & \cdots & \cdots & 0\\
c & d & 0 & \cdots & \cdots & \cdots & \cdots & 0\\
0 & 0 & a & \cdots & \cdots & \cdots & \cdots & b\\
\vdots & \vdots & \vdots & \ddots & 0 & 0 & \iddots & \vdots\\
\vdots & \vdots & \vdots & 0 & a & b & 0 & \vdots\\
\vdots & \vdots & \vdots & 0 & c & d & 0 & \vdots\\
\vdots & \vdots & \vdots & \iddots & 0 & 0 & \mathrm{\ddots} & \vdots\\
0 & 0 & c & \cdots & \cdots & \cdots & \cdots & d
\end{array}\right|.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
D_{2n}=D_{1}D_{2(n-1)}=(ad-bc)D_{2(n-1)}.
\]
\end_inset
以此作递推公式, 得
\begin_inset Formula
\[
D_{2n}=(ad-bc)D_{2(n-1)}=\cdots=(ad-bc)^{n-1}D_{2}=(ad-bc)^{n}.
\]
\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
行列式的计算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
\begin_inset Argument 1
status open
\begin_layout Plain Layout
*
\end_layout
\end_inset
用拉普拉斯定理求行列式
\begin_inset Formula $\begin{vmatrix}2 & 3 & 0 & 0\\
1 & 2 & 3 & 0\\
0 & 1 & 2 & 3\\
0 & 0 & 1 & 2
\end{vmatrix}$
\end_inset
的值.
\end_layout
\begin_layout Solution*
按第一行和第二行展开
\begin_inset Formula
\begin{align*}
\begin{vmatrix}2 & 3 & 0 & 0\\
1 & 2 & 3 & 0\\
0 & 1 & 2 & 3\\
0 & 0 & 1 & 2
\end{vmatrix} & =\begin{vmatrix}2 & 3\\
1 & 2
\end{vmatrix}\times(-1)^{1+2+1+2}\begin{vmatrix}2 & 3\\
1 & 2
\end{vmatrix}+\begin{vmatrix}2 & 0\\
1 & 3
\end{vmatrix}\times(-1)^{1+2+1+3}\begin{vmatrix}1 & 3\\
0 & 2
\end{vmatrix}\\
& \qquad+\begin{vmatrix}3 & 0\\
2 & 3
\end{vmatrix}\times(-1)^{1+2+2+3}\begin{vmatrix}0 & 3\\
0 & 2
\end{vmatrix}\\
& =1-12+0=-11.
\end{align*}
\end_inset
\end_layout
\end_deeper
\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 Exercise
计算行列式
\begin_inset Formula $D=\begin{vmatrix}3 & 1 & -1 & 2\\
-5 & 1 & 3 & -4\\
2 & 0 & 1 & -1\\
1 & -5 & 3 & -3
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise
按第三行展开, 计算行列式
\begin_inset Formula
\[
\begin{vmatrix}2 & -3 & 4 & 1\\
4 & 2 & 3 & 2\\
a & b & c & d\\
3 & -1 & 4 & 3
\end{vmatrix}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise
按第二列展开, 计算行列式
\begin_inset Formula
\[
\begin{vmatrix}5 & a & 2 & -1\\
4 & b & 4 & -3\\
2 & c & 3 & -2\\
4 & d & 5 & -4
\end{vmatrix}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise
计算行列式: (1).
\begin_inset Formula $\begin{vmatrix}a & 3 & 0 & 5\\
0 & b & 0 & 2\\
1 & 2 & c & 3\\
0 & 0 & 0 & d
\end{vmatrix}$
\end_inset
.
(2).
\begin_inset Formula $\begin{vmatrix}1 & 0 & 2 & a\\
2 & 0 & b & 0\\
3 & c & 4 & 5\\
d & 0 & 0 & 0
\end{vmatrix}$
\end_inset
.
(3).
\begin_inset Formula $\begin{vmatrix}x & a & b & 0 & c\\
0 & y & 0 & 0 & d\\
0 & e & z & 0 & f\\
g & h & k & u & l\\
0 & 0 & 0 & 0 & v
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise
讨论当
\begin_inset Formula $k$
\end_inset
为何值时
\begin_inset Formula $\begin{vmatrix}1 & 1 & 0 & 0\\
1 & k & 1 & 0\\
0 & 0 & k & 2\\
0 & 0 & 2 & k
\end{vmatrix}\neq0$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise
\begin_inset Argument 1
status open
\begin_layout Plain Layout
*
\end_layout
\end_inset
设
\begin_inset Formula $n$
\end_inset
阶行列式
\begin_inset Formula $D_{n}=\begin{vmatrix}1 & 2 & 3 & \cdots & n\\
1 & 2 & 0 & \cdots & 0\\
1 & 0 & 3 & \cdots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 0 & 0 & \cdots & n
\end{vmatrix}$
\end_inset
, 求第一行各元素的代数余子式之和
\begin_inset Formula
\[
A_{11}+A_{12}+\cdots+A_{1n}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise
\begin_inset Argument 1
status open
\begin_layout Plain Layout
*
\end_layout
\end_inset
设
\begin_inset Formula $A=\left(\alpha_{ij}\right)$
\end_inset
,
\begin_inset Formula $A_{ij}$
\end_inset
是
\begin_inset Formula $\alpha_{ij}$
\end_inset
在
\begin_inset Formula $\mathrm{det}A$
\end_inset
中的代数余子式, 在求行列式的各种方法中, 我们有时会通过
\begin_inset Quotes eld
\end_inset
升阶
\begin_inset Quotes erd
\end_inset
来简化计算, 比如在一些问题中计算
\begin_inset Formula $n$
\end_inset
阶行列式的值比较困难, 我们考虑一个
\begin_inset Formula $n+1$
\end_inset
阶的
\begin_inset Quotes eld
\end_inset
升阶
\begin_inset Quotes erd
\end_inset
行列式来简化求解次数, 进而降低计算上的错误概率.
\end_layout
\begin_layout Standard
(1).
求证
\begin_inset Formula
\[
(-1)^{n}\begin{vmatrix}1 & 1 & \cdots & 1 & 0 & 1\\
\alpha_{11} & \alpha_{12} & \cdots & \alpha_{1,n-1} & 1 & \alpha_{1n}\\
\alpha_{21} & \alpha_{22} & \cdots & \alpha_{2,n-1} & 1 & \alpha_{2n}\\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
\alpha_{n1} & \alpha_{n2} & \cdots & \alpha_{n,n-1} & 1 & \alpha_{nn}
\end{vmatrix}_{(n+1)\times(n+1)}=\begin{vmatrix}\alpha_{11}-\alpha_{12} & \alpha_{12}-\alpha_{13} & \cdots & \alpha_{1,n-1}-\alpha_{1n} & 1\\
\alpha_{21}-\alpha_{22} & \alpha_{22}-\alpha_{23} & \cdots & \alpha_{2,n-1}-\alpha_{2n} & 1\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
\alpha_{n1}-\alpha_{n2} & \alpha_{n2}-\alpha_{n3} & \cdots & \alpha_{n,n-1}-\alpha_{nn} & 1
\end{vmatrix}_{n\times n}.
\]
\end_inset
\end_layout
\begin_layout Standard
(2).
证明:
\begin_inset Formula
\[
\sum_{i,j=1}^{n}A_{ij}=\begin{vmatrix}\alpha_{11}-\alpha_{12} & \alpha_{12}-\alpha_{13} & \cdots & \alpha_{1,n-1}-\alpha_{1n} & 1\\
\alpha_{21}-\alpha_{22} & \alpha_{22}-\alpha_{23} & \cdots & \alpha_{2,n-1}-\alpha_{2n} & 1\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
\alpha_{n1}-\alpha_{n2} & \alpha_{n2}-\alpha_{n3} & \cdots & \alpha_{n,n-1}-\alpha_{nn} & 1
\end{vmatrix}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise
\begin_inset Argument 1
status open
\begin_layout Plain Layout
*
\end_layout
\end_inset
证明: 行列式
\begin_inset Formula
\[
\begin{vmatrix}a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\cdots & \cdots & \cdots & \cdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix}
\]
\end_inset
所有元素的代数余子式的和等于行列式
\begin_inset Formula
\[
\begin{vmatrix}1 & 1 & \cdots & 1\\
a_{21}-a_{11} & a_{22}-a_{12} & \cdots & a_{2n}-a_{1n}\\
a_{31}-a_{11} & a_{32}-a_{12} & \cdots & a_{3n}-a_{1n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1}-a_{11} & a_{n2}-a_{12} & \cdots & a_{nn}-a_{1n}
\end{vmatrix}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise
\begin_inset Argument 1
status open
\begin_layout Plain Layout
**
\end_layout
\end_inset
证明: 如果行列式的某一行 (或列) 的所有元素等于
\begin_inset Formula $1$
\end_inset
, 则这行列式所有元素的代数余子式之和等于该行列式自己.
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\end_body
\end_document
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