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\begin_body
\begin_layout Section
二阶与三阶行列式
\end_layout
\begin_layout Subsection
二阶行列式的定义与应用
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
解二元一次方程组
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Itemize
二阶线性方程组
\begin_inset Formula
\[
\begin{cases}
a_{11}x_{1}+a_{12}x_{2}=b_{1}\\
a_{21}x_{1}+a_{22}x_{2}=b_{2}
\end{cases}
\]
\end_inset
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
pause{}
\end_layout
\end_inset
\end_layout
\begin_layout Itemize
将第一个方程两边同时乘以
\begin_inset Formula $a_{21}$
\end_inset
, 第二个方程两边同时乘以
\begin_inset Formula $a_{11}$
\end_inset
, 使得两个方程的未知数
\begin_inset Formula $x_{1}$
\end_inset
的系数相同
\begin_inset Foot
status collapsed
\begin_layout Plain Layout
为什么我们不分别除以
\begin_inset Formula $a_{11}$
\end_inset
,
\begin_inset Formula $a_{21}$
\end_inset
使未知数
\begin_inset Formula $x_{1}$
\end_inset
的系数都为
\begin_inset Formula $1$
\end_inset
?
\end_layout
\end_inset
, 得到
\begin_inset Formula
\[
\begin{cases}
a_{11}a_{21}x_{1}+a_{12}a_{21}x_{2}=b_{1}a_{21},\\
a_{11}a_{21}x_{1}+a_{11}a_{22}x_{2}=b_{2}a_{11}.
\end{cases}
\]
\end_inset
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
pause{}
\end_layout
\end_inset
\end_layout
\begin_layout Itemize
两式相减得到
\begin_inset Formula
\[
\left(a_{11}a_{22}-a_{12}a_{21}\right)x_{2}=a_{11}b_{2}-a_{21}b_{1}.
\]
\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
解二元一次方程组
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Itemize
当
\begin_inset Formula $a_{11}a_{22}-a_{12}a_{21}\ne0$
\end_inset
时, 解得
\begin_inset Formula
\[
x_{2}=\frac{{\color{blue}a_{11}}b_{2}-{\color{green}a_{21}}b_{1}}{{\color{blue}a_{11}}a_{22}-a_{12}{\color{green}a_{21}}};
\]
\end_inset
\end_layout
\begin_layout Itemize
同理可得
\begin_inset Formula
\[
x_{1}=\frac{{\color{magenta}a_{22}}b_{1}-{\color{teal}a_{12}}b_{2}}{a_{11}{\color{magenta}a_{22}}-{\color{teal}a_{12}}a_{21}}.
\]
\end_inset
\end_layout
\begin_layout Itemize
二阶行列式
\begin_inset Note Note
status open
\begin_layout Plain Layout
对角线方式
\end_layout
\end_inset
\begin_inset Formula
\[
\begin{vmatrix}a_{11} & a_{12}\\
a_{21} & a_{22}
\end{vmatrix}=a_{11}a_{22}-a_{12}a_{21}.
\]
\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
二阶行列式的定义与应用
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Standard
解方程组
\begin_inset Formula
\[
\begin{cases}
a_{11}x_{1}+a_{12}x_{2}=b_{1}\\
a_{21}x_{1}+a_{22}x_{2}=b_{2}
\end{cases}
\]
\end_inset
当系数行列式
\begin_inset Formula $\begin{vmatrix}a_{11} & a_{12}\\
a_{21} & a_{22}
\end{vmatrix}\ne0$
\end_inset
时, 方程的解为
\begin_inset Formula
\[
x_{1}=\frac{b_{1}a_{22}-b_{2}a_{12}}{a_{11}a_{22}-a_{12}a_{21}}=\frac{\begin{vmatrix}b_{1} & a_{12}\\
b_{2} & a_{22}
\end{vmatrix}}{\begin{vmatrix}a_{11} & a_{12}\\
a_{21} & a_{22}
\end{vmatrix}}\eqqcolon\frac{D_{1}}{D},
\]
\end_inset
\begin_inset Formula
\[
x_{2}=\frac{b_{2}a_{11}-b_{1}a_{21}}{a_{11}a_{22}-a_{12}a_{21}}=\frac{\begin{vmatrix}a_{11} & b_{1}\\
a_{21} & b_{2}
\end{vmatrix}}{\begin{vmatrix}a_{11} & a_{12}\\
a_{21} & a_{22}
\end{vmatrix}}\eqqcolon\frac{D_{2}}{D}.
\]
\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
二阶行列式的应用
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
解方程组
\begin_inset Formula
\[
\begin{cases}
2x_{1}+3x_{2}=8\\
x_{1}-2x_{2}=-3
\end{cases}
\]
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
\begin{aligned}D & =\begin{vmatrix}2 & 3\\
1 & -2
\end{vmatrix}=2\times(-2)-3\times1=-7,\\
D_{1} & =\begin{vmatrix}8 & 3\\
-3 & -2
\end{vmatrix}=8\times(-2)-3\times(-3)=-7,\\
D_{2} & =\begin{vmatrix}2 & 8\\
1 & -3
\end{vmatrix}=2\times(-3)-8\times1=-14.
\end{aligned}
\]
\end_inset
因
\begin_inset Formula $D=-7\neq0$
\end_inset
, 故所给方程组有唯一解
\begin_inset Formula
\[
x_{1}=\frac{D_{1}}{D}=\frac{-7}{-7}=1,x_{2}=\frac{D_{2}}{D}=\frac{-14}{-7}=2.
\]
\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
二阶行列式的应用
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
解线性方程组
\begin_inset Formula
\[
\begin{cases}
x_{1}+2x_{2}=0\\
3x_{1}+4x_{2}=1
\end{cases}
\]
\end_inset
\end_layout
\begin_layout Solution*
由于方程组的系数行列式
\begin_inset Formula
\[
D=\begin{vmatrix}1 & 2\\
3 & 4
\end{vmatrix}=4-6=-2\neq0,
\]
\end_inset
又
\begin_inset Formula
\[
D_{1}=\begin{vmatrix}0 & 2\\
1 & 4
\end{vmatrix}=-2,\quad D_{2}=\begin{vmatrix}1 & 0\\
3 & 1
\end{vmatrix}=1.
\]
\end_inset
所以方程组的解为
\begin_inset Formula
\[
x_{1}=\frac{D_{1}}{D}=1,\quad x_{2}=\frac{D_{2}}{D}=-\frac{1}{2}.
\]
\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
二阶行列式的应用
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem*
\begin_inset Argument 1
status open
\begin_layout Plain Layout
\begin_inset CommandInset href
LatexCommand href
name "知乎: 鞋带定理"
target "https://zhuanlan.zhihu.com/p/110025234"
literal "false"
\end_inset
,
\begin_inset CommandInset href
LatexCommand href
name "AOPS: Shoelace Theorem"
target "https://artofproblemsolving.com/wiki/index.php?title=Shoelace_Theorem"
literal "false"
\end_inset
,
\begin_inset CommandInset href
LatexCommand href
name "Wolfram: Shoelace Theorem"
target "https://mathworld.wolfram.com/ShoelaceFormula.html"
literal "false"
\end_inset
, 线积分
\end_layout
\end_inset
假设平面上点
\begin_inset Formula $P_{1}(x_{1},y_{1})$
\end_inset
,
\begin_inset Formula $P_{2}(x_{2},y_{2})$
\end_inset
,
\begin_inset Formula $\cdots$
\end_inset
,
\begin_inset Formula $P_{n}(x_{n},y_{n})$
\end_inset
围成的平面多边形为
\begin_inset Formula $P_{1}P_{2}\cdots P_{n}$
\end_inset
, 则此平面多边形的 (符号) 面积为
\begin_inset Formula
\begin{align*}
S & =\frac{1}{2}\left(\begin{vmatrix}x_{1} & y_{1}\\
x_{2} & y_{2}
\end{vmatrix}+\begin{vmatrix}x_{2} & y_{2}\\
x_{3} & y_{3}
\end{vmatrix}+\begin{vmatrix}x_{3} & y_{3}\\
x_{4} & y_{4}
\end{vmatrix}+\cdots+\begin{vmatrix}x_{n-1} & y_{n-1}\\
x_{n} & y_{n}
\end{vmatrix}+\begin{vmatrix}x_{n} & y_{n}\\
x_{1} & y_{1}
\end{vmatrix}\right)\\
& \xcancel{=\frac{1}{2}\begin{vmatrix}x_{1} & x_{2} & \cdots & x_{n} & x_{1}\\
y_{1} & y_{2} & \cdots & y_{n} & y_{1}
\end{vmatrix}}.
\end{align*}
\end_inset
这里的 (符号) 面积在点
\begin_inset Formula $P_{1},P_{2},\cdots,P_{n}$
\end_inset
按逆时针排列时取正面积; 在点
\begin_inset Formula $P_{1},P_{2},\cdots,P_{n}$
\end_inset
按顺时针排列时取负面积.
\end_layout
\begin_layout Pause
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Theorem*
设平面上点
\begin_inset Formula $A,B$
\end_inset
的坐标分别为
\begin_inset Formula $(x_{1},y_{1})$
\end_inset
,
\begin_inset Formula $(x_{2},y_{2})$
\end_inset
, 点
\begin_inset Formula $O$
\end_inset
为坐标原点
\begin_inset Formula $(0,0)$
\end_inset
, 则由向量
\begin_inset Formula $\overrightarrow{OA},\overrightarrow{OB}$
\end_inset
组成的平行四边形的面积
\begin_inset Formula $S$
\end_inset
为
\begin_inset Formula
\[
S=\left|\begin{vmatrix}x_{1} & y_{1}\\
x_{2} & y_{2}
\end{vmatrix}\right|=\left|\overrightarrow{OA}\times\overrightarrow{OB}\right|.
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Subsection
三阶行列式的定义与应用
\end_layout
\begin_layout Frame
\begin_inset Argument 4
status open
\begin_layout Plain Layout
三元一次 (线性) 方程组
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Itemize
类似于二元线性方程组的讨论, 对三元线性方程组
\begin_inset Formula
\[
\begin{cases}
a_{11}x_{1}+a_{12}x_{2}+a_{13}x_{3}=b_{1}\\
a_{21}x_{1}+a_{22}x_{2}+a_{23}x_{3}=b_{2}\\
a_{31}x_{1}+a_{32}x_{2}+a_{33}x_{3}=b_{3}
\end{cases}
\]
\end_inset
\end_layout
\begin_layout Itemize
记
\begin_inset Formula
\[
\begin{aligned}D=\begin{vmatrix}a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{vmatrix}, & D_{1}=\begin{vmatrix}{\color{purple}b_{1}} & a_{12} & a_{13}\\
{\color{purple}b_{2}} & a_{22} & a_{23}\\
{\color{purple}b_{3}} & a_{32} & a_{33}
\end{vmatrix},\\
D_{2}=\begin{vmatrix}a_{11} & {\color{purple}b_{1}} & a_{13}\\
a_{21} & {\color{purple}b_{2}} & a_{23}\\
a_{31} & {\color{purple}b_{3}} & a_{33}
\end{vmatrix}, & D_{3}=\begin{vmatrix}a_{11} & a_{12} & {\color{purple}b_{1}}\\
a_{21} & a_{22} & {\color{purple}b_{2}}\\
a_{31} & a_{32} & {\color{purple}b_{3}}
\end{vmatrix}.
\end{aligned}
\]
\end_inset
\end_layout
\begin_layout Itemize
若
\series bold
系数行列式
\begin_inset Formula $D\neq0$
\end_inset
\series default
, 则该方程组存在唯一解:
\begin_inset Formula
\[
x_{1}=\frac{D_{1}}{D},\quad x_{2}=\frac{D_{2}}{D},\quad x_{3}=\frac{D_{3}}{D}.
\]
\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
三阶行列式
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Itemize
三阶行列式
\begin_inset Note Note
status open
\begin_layout Plain Layout
沙路法则
\end_layout
\end_inset
\begin_inset Formula
\begin{align*}
\begin{vmatrix}a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{vmatrix} & =a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}\\
& \quad-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}.
\end{align*}
\end_inset
\end_layout
\begin_layout Itemize
三阶行列式有
\begin_inset Formula $6$
\end_inset
项, 每一项均为不同行不同列的三个元素之积再冠于正负号, 其运算的规律性可用``对角线法则''或``沙路法则''来表述.
\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
三阶行列式的计算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
计算三阶行列式
\begin_inset Formula $\begin{vmatrix}1 & 2 & 3\\
4 & 0 & 5\\
-1 & 0 & 6
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
解:
\begin_inset Formula
\[
\begin{aligned}\begin{vmatrix}1 & 2 & 3\\
4 & 0 & 5\\
-1 & 0 & 6
\end{vmatrix} & =1\times0\times6+2\times5(-1)+3\times4\times0\\
& \qquad-3\times0\times(-1)-1\times5\times0-2\times4\times6\\
& =-10-48=-58.
\end{aligned}
\]
\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
三阶行列式的计算
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
求解方程
\begin_inset Formula $D=\begin{vmatrix}1 & 1 & 1\\
2 & 3 & x\\
4 & 9 & x^{2}
\end{vmatrix}=0$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
pause{}
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
方程左端
\begin_inset Formula $D=3x^{2}+4x+18-12-9x-2x^{2}=x^{2}-5x+6$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
vspace{2mm}
\end_layout
\end_inset
由
\begin_inset Formula $x^{2}-5x+6=0$
\end_inset
解得
\begin_inset Formula $x=2$
\end_inset
或
\begin_inset Formula $x=3$
\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
用三阶行列式解方程
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
解三元线性方程组
\begin_inset Formula $\begin{cases}
x_{1}-2x_{2}+x_{3}=-2\\
2x_{1}+x_{2}-3x_{3}=1\\
-x_{1}+x_{2}-x_{3}=0
\end{cases}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
由于方程组
\begin_inset Formula
\[
\begin{cases}
x_{1}-2x_{2}+x_{3}=-2\\
2x_{1}+x_{2}-3x_{3}=1\\
-x_{1}+x_{2}-x_{3}=0
\end{cases}
\]
\end_inset
的系数行列式
\begin_inset Formula
\[
\begin{aligned}D & =\begin{vmatrix}1 & -2 & 1\\
2 & 1 & -3\\
-1 & 1 & -1
\end{vmatrix}=1\times1\times(-1)+(-2)\times(-3)\times(-1)+1\times2\times1\\
& \quad-(-1)\times1\times1-1\times(-3)\times1-(-2)\times2\times(-1)\\
& =-5\neq0,
\end{aligned}
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula
\[
D_{1}=\begin{vmatrix}-2 & -2 & 1\\
1 & 1 & -3\\
0 & 1 & -1
\end{vmatrix}=-5,D_{2}=\begin{vmatrix}1 & -2 & 1\\
2 & 1 & -3\\
-1 & 0 & -1
\end{vmatrix}=-10,D_{3}=\begin{vmatrix}1 & -2 & -2\\
2 & 1 & 1\\
-1 & 1 & 0
\end{vmatrix}=-5,
\]
\end_inset
故所求方程组的解为:
\begin_inset Formula
\[
x_{1}=\frac{D_{1}}{D}=1,\quad x_{2}=\frac{D_{2}}{D}=2,\quad x_{3}=\frac{D_{3}}{D}=1.
\]
\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
用三阶行列式解方程
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Example
解线性方程组
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
systeme{-x_{1}+2x_{2}+3x_{3}=-1,3x_{1}-4x_{2}+x_{3}=2,x_{1}+3x_{2}-x_{3}=1}
\end_layout
\end_inset
.
\end_layout
\begin_layout Overprint
\begin_inset Argument item:1
status open
\begin_layout Plain Layout
2
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Solution*
系数行列式
\begin_inset Formula
\[
\begin{aligned}D & =\begin{vmatrix}-1 & 2 & 3\\
3 & -4 & 1\\
1 & 3 & -1
\end{vmatrix}\\
& =-1\times(-4)\times(-1)+2\times1\times1+3\times3\times3\\
& -3\times(-4)\times1-2\times3\times(-1)-1\times3\times(-1)\\
& =46.
\end{aligned}
\]
\end_inset
\end_layout
\end_deeper
\begin_layout Overprint
\begin_inset Argument item:1
status open
\begin_layout Plain Layout
3
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Solution*
同理
\begin_inset Formula
\[
\scriptsize D_{1}=\begin{vmatrix}-1 & 2 & 3\\
2 & -4 & 1\\
1 & 3 & -1
\end{vmatrix}=35,\ D_{2}=\begin{vmatrix}-1 & -1 & 3\\
3 & 2 & 1\\
1 & 1 & -1
\end{vmatrix}=2,\ D_{3}=\begin{vmatrix}-1 & 2 & -1\\
3 & -4 & 2\\
1 & 3 & 1
\end{vmatrix}=-5,
\]
\end_inset
于是方程组的解为
\begin_inset Formula
\[
x_{1}=\frac{D_{1}}{D}=\frac{35}{46},\quad x_{2}=\frac{D_{2}}{D}=\frac{1}{23},\quad x_{3}=\frac{D_{3}}{D}=-\frac{5}{46}.
\]
\end_inset
\end_layout
\end_deeper
\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
另一个面积公式
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem*
已知平面上三点
\begin_inset Formula $A,B,C$
\end_inset
的坐标分别为
\begin_inset Formula $(x_{1},y_{1})$
\end_inset
,
\begin_inset Formula $(x_{2},y_{2})$
\end_inset
,
\begin_inset Formula $(x_{3},y_{3})$
\end_inset
, 则三角形
\begin_inset Formula $ABC$
\end_inset
的 (符号) 面积
\begin_inset Formula $S$
\end_inset
为
\begin_inset Formula
\[
S=\frac{1}{2}\begin{vmatrix}x_{1} & y_{1} & 1\\
x_{2} & y_{2} & 1\\
x_{3} & y_{3} & 1
\end{vmatrix}.
\]
\end_inset
\end_layout
\begin_layout Itemize
当三角形的三个点
\begin_inset Formula $A,B,C$
\end_inset
按逆时针顺序排列时, 上式算得的值就是通常的面积
\begin_inset Formula $S_{\triangle ABC}$
\end_inset
;
\end_layout
\begin_layout Itemize
当三角形的三个点
\begin_inset Formula $A,B,C$
\end_inset
按顺时针顺序排列时, 上式算得的值是通常的面积
\begin_inset Formula $S_{\triangle ABC}$
\end_inset
的相反数.
\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
还可以算体积公式
\end_layout
\end_inset
\end_layout
\begin_deeper
\begin_layout Theorem*
\begin_inset Argument 1
status open
\begin_layout Plain Layout
空间解析几何
\end_layout
\end_inset
已知三维坐标系中的三点
\begin_inset Formula $A,B,C$
\end_inset
的坐标分别为
\begin_inset Formula $(x_{1},y_{1},z_{1})$
\end_inset
,
\begin_inset Formula $(x_{2},y_{2},z_{2})$
\end_inset
,
\begin_inset Formula $(x_{3},y_{3},z_{3})$
\end_inset
, 点
\begin_inset Formula $O$
\end_inset
为原点, 则由向量
\begin_inset Formula $\overrightarrow{OA},\overrightarrow{OB},\overrightarrow{OC}$
\end_inset
构成的平行六面体的体积
\begin_inset Formula $V$
\end_inset
为
\begin_inset Formula
\[
V=\begin{vmatrix}x_{1} & y_{1} & z_{1}\\
x_{2} & y_{2} & z_{2}\\
x_{3} & y_{3} & z_{3}
\end{vmatrix}=\left(\overrightarrow{OA},\overrightarrow{OB},\overrightarrow{OC}\right),
\]
\end_inset
其中
\begin_inset Formula $\left(\overrightarrow{OA},\overrightarrow{OB},\overrightarrow{OC}\right)$
\end_inset
为矢量的混合积, 定义为
\begin_inset Formula $\overrightarrow{OA}\cdot\left(\overrightarrow{OB}\times\overrightarrow{OC}\right)$
\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 Problem
设
\begin_inset Formula $D=\begin{vmatrix}a & 1 & 0\\
1 & a & 0\\
4 & 0 & 1
\end{vmatrix}$
\end_inset
, 试给出
\begin_inset Formula $D>0$
\end_inset
的充分必要条件.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
\begin_inset CommandInset label
LatexCommand label
name "prob:inter"
\end_inset
\begin_inset Argument 1
status open
\begin_layout Plain Layout
插值公式
\begin_inset ERT
status open
\begin_layout Plain Layout
$^{
\backslash
star}$
\end_layout
\end_inset
\end_layout
\end_inset
求一个二次多项式
\begin_inset Formula $f(x)$
\end_inset
, 使
\begin_inset Formula
\[
f(1)=0,\quad f(2)=3,\quad f(-3)=28.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
求行列式
\begin_inset Formula $\begin{vmatrix}\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{vmatrix}$
\end_inset
的值.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
求行列式
\begin_inset Formula $\begin{vmatrix}a+b & a-b\\
a-b & a+b
\end{vmatrix}$
\end_inset
的值.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
求行列式
\begin_inset Formula $\begin{vmatrix}\cos\alpha & \sin\alpha\\
\sin\beta & \cos\beta
\end{vmatrix}$
\end_inset
的值.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
证明:
\begin_inset Formula $\begin{vmatrix}1 & \log_{b}a\\
\log_{a}b & 1
\end{vmatrix}=0$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
证明:
\begin_inset Formula $\begin{vmatrix}\cos\alpha+\ui\sin\alpha & 1\\
1 & \cos\alpha-\ui\sin\alpha
\end{vmatrix}=0$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
利用行列式求解以下方程:
\end_layout
\begin_layout Problem
(1).
\begin_inset Formula $\begin{cases}
2x+5y=1\\
3x+7y=2
\end{cases}$
\end_inset
\end_layout
\begin_layout Problem
(2).
\begin_inset Formula $\begin{cases}
2x-3y=4\\
4x-5y=10
\end{cases}$
\end_inset
\end_layout
\begin_layout Problem
(3).
\begin_inset Formula $\begin{cases}
5x-7y=1\\
x-2y=0
\end{cases}$
\end_inset
\end_layout
\begin_layout Problem
(4).
\begin_inset Formula $\begin{cases}
4x+7y+13=0\\
5x+8y+14=0
\end{cases}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
证明: 对于实数
\begin_inset Formula $a,b,c$
\end_inset
, 方程
\begin_inset Formula $\begin{vmatrix}a-x & b\\
b & c-x
\end{vmatrix}=0$
\end_inset
的根是实数.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
证明: 复系数二次三项式
\begin_inset Formula $ax^{2}+bx+c$
\end_inset
, 当且仅当
\begin_inset Formula
\[
\begin{vmatrix}a & b\\
b & c
\end{vmatrix}=0
\]
\end_inset
时, 是完全平方.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
计算下列三阶行列式:
\end_layout
\begin_layout Problem
(1).
\begin_inset Formula $\begin{vmatrix}2 & 1 & 3\\
5 & 3 & 2\\
1 & 4 & 3
\end{vmatrix}$
\end_inset
; (2).
\begin_inset Formula $\begin{vmatrix}3 & 2 & 1\\
2 & 5 & 3\\
3 & 4 & 2
\end{vmatrix}$
\end_inset
; (3).
\begin_inset Formula $\begin{vmatrix}4 & -3 & 5\\
3 & -2 & 8\\
1 & -7 & -5
\end{vmatrix}$
\end_inset
; (4).
\begin_inset Formula $\begin{vmatrix}3 & 2 & -4\\
4 & 1 & -2\\
5 & 2 & -3
\end{vmatrix}$
\end_inset
;
\end_layout
\begin_layout Problem
(5).
\begin_inset Formula $\begin{vmatrix}3 & 4 & -5\\
8 & 7 & -2\\
2 & -1 & 8
\end{vmatrix}$
\end_inset
; (6).
\begin_inset Formula $\begin{vmatrix}1 & 1 & 1\\
1 & 2 & 3\\
1 & 3 & 6
\end{vmatrix}$
\end_inset
; (7).
\begin_inset Formula $\begin{vmatrix}0 & 1 & 1\\
1 & 0 & 1\\
1 & 1 & 0
\end{vmatrix}$
\end_inset
; (8).
\begin_inset Formula $\begin{vmatrix}1 & 1 & 1\\
4 & 5 & 9\\
16 & 25 & 81
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
计算行列式
\begin_inset Formula $\begin{vmatrix}a & b & c\\
c & a & b\\
b & c & a
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
计算如下行列式:
\end_layout
\begin_layout Problem
(1).
\begin_inset Formula $\begin{vmatrix}a+x & x & x\\
x & b+x & x\\
x & x & c+x
\end{vmatrix}$
\end_inset
; (2).
\begin_inset Formula $\begin{vmatrix}\alpha^{2}+1 & \alpha\beta & \alpha\gamma\\
\alpha\beta & \beta^{2}+1 & \beta\gamma\\
\alpha\gamma & \beta\gamma & \gamma^{2}+1
\end{vmatrix}$
\end_inset
;
\end_layout
\begin_layout Problem
(3).
\begin_inset Formula $\begin{vmatrix}\cos\alpha & \sin\alpha\cos\beta & \sin\alpha\sin\beta\\
-\sin\alpha & \cos\alpha\cos\beta & \cos\alpha\sin\beta\\
0 & -\sin\beta & \cos\beta
\end{vmatrix}$
\end_inset
; (4).
\begin_inset Formula $\begin{vmatrix}\sin\alpha & \cos\alpha & 1\\
\sin\beta & \cos\beta & 1\\
\sin\gamma & \cos\gamma & 1
\end{vmatrix}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
在怎样的条件下下列等式成立:
\begin_inset Formula
\[
\begin{vmatrix}1 & \cos\alpha & \cos\beta\\
\cos\alpha & 1 & \cos\gamma\\
\cos\beta & \cos\gamma & 1
\end{vmatrix}=\begin{vmatrix}0 & \cos\alpha & \cos\beta\\
\cos\alpha & 0 & \cos\gamma\\
\cos\beta & \cos\gamma & 0
\end{vmatrix}.
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
已知三阶行列式由
\begin_inset Formula $9$
\end_inset
个元素, 如果这
\begin_inset Formula $9$
\end_inset
个元素的取值为
\begin_inset Formula $\pm1$
\end_inset
两种数值中的某一个, 求能够取到最大值的那个三阶行列式.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem
利用行列式求解下列方程组:
\end_layout
\begin_layout Problem
(1).
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
systeme{2x+3y+5z=10,3x+7y+4z=3,x+2y+2z=3}
\end_layout
\end_inset
; (2).
\begin_inset ERT
status open
\begin_layout Plain Layout
\backslash
systeme{5x-6y+4z=3,3x-3y+2z=1,4x-5y+2z=1}
\end_layout
\end_inset
.
\end_layout
\end_deeper
\begin_layout Frame
\end_layout
\end_body
\end_document
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