线性代数简明教程笔记_1
线性代数第一章知识点总结篇
行列式定义
逆序数
1.n!个n排列中，唯有123...(n–1)n是按顺序排列的，称该排列为标准排列
2.任意排列经过一次变换必将改变奇偶性
3.一段排列的逆序总数称为逆序数，例如:15243的逆序数为4，记τ(15243)=4
(为了方便，以下用\(\tau\)标识)
4.逆序数为偶数的排列称为偶排列，同样，逆序数为奇数的排列称为奇排列
二阶行列式计算
\[\left|
\begin{matrix}
a & b \\
c & d
\end{matrix}
\right|
\]
\[=\sum_{j_1j_2}(-1)^{\tau(j_1j_2)}a_{1j1}a_{2j2}= a*d-b*c
\]
三阶行列式计算
\[\left|
\begin{matrix}
a & b & c\\
d & e & f\\
g & h & i\\
\end{matrix}
\right|
\]
\[=\sum_{j_1j_2j_3}-1^{\tau(j_1j_2j_3)}a_{1j_1}a_{2j_2}a_{3j_3}
= a*e*i+b*f*g+c*d*h-c*e*g-f*h*a-i*b*d
\]
对角线法则
上下三角形
\[\left|
\begin{matrix}
a & * & *\\
0 & e & *\\
0 & 0 & i\\
\end{matrix}
\right|=上三角
\]
\[\left|
\begin{matrix}
a & 0 & 0\\
x & e & 0\\
x & x & i\\
\end{matrix}
\right| = 下三角
\]
注意点(杂项)
1.行列式中，行列地位平等
(证明请看参考下篇的证明习题篇)
2.一阶行列式 |a|=a
3.代数和中每一项的正负号的决定方法：当行指标取成标准排列时，由列指标组成的排列的奇偶性确定，偶者为正，奇者为负
见上面的\(\tau(j_1j_2j_3)\)
行列式性质
1.D=DT(行列式行列互换，值不变)
证明：
\[D=
\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
b_{21} & b_{22} & ... & b_{2n}\\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right|
\]
\[= \sum_{j_1j_2j_3}(-1)^{\tau(j_1j_2j_3)}b_{1j_1}b_{2j_2}b_{3j_3}…b_{nj_n}
\]
\[= \sum_{j_1j_2j_3}(-1)^{\tau(j_1j_2j_3)}b_{j_11}b_{j_22}b_{j_33}…b_{nj_n}
\]
\[=
\left|
\begin{matrix}
b_{11} & b_{21} & ... & b_{n1}\\
b_{12} & b_{22} & ... & b_{n2}\\
... & ... & & ...\\
b_{1n} & b_{2n} & ... & b_{nn}\\
\end{matrix}
\right|
\]
2.行列式中两行(列)交换一次，行列式的值增加一个负号。
i行和j行交换:
\[\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
... & ... & & ...\\
—&—&—&— \\
|b_{i1} & b_{i2} & ... & b_{in}|\\
|... & ... & & ...|\\
|b_{j1} & b_{j2} & ... & b_{jn}|\\
—&—&—&— \\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right| =
\]
\[-
\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
... & ... & & ...\\
—&—&—&— \\
|b_{j1} & b_{j2} & ... & b_{jn}|\\
|... & ... & & ...|\\
|b_{i1} & b_{i2} & ... & b_{in}|\\
—&—&—&— \\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right|
\]
推1.行列式两行(列)对应元素全相等，则行列式为零。
\[\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
... & ... & & ...\\
—&—&—&— \\
|b_{j1} & b_{j2} & ... & b_{jn}|\\
|... & ... & & ...|\\
| b_{j1} & b_{j2} & ... & b_{jn}|\\
—&—&—&— \\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right| = 0
\]
3.K倍行列式=行列式某一行(列)的K倍
\[\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
kb_{21} & kb_{22} & ... & kb_{2n}\\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right| =
\]
\[k
\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
b_{21} & b_{22} & ... & b_{2n}\\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right|
\]
推2.行列式中某一行元素全为0则行列式的值为0。
\[\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
0 & 0 & ... & 0\\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right| = 0
\]
推3.行列式中某一行(列)元素与另一行(列)元素成比例，则行列式的值为0。
4.行列式中第i行(列)的k倍加(减)第j行(列)的m倍，行列式的值不变。
\[\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
... & ... & & ...\\
—&—&—&— \\
|b_{i1} & b_{i2} & ... & b_{in}|\\
|... & ... & & ...|\\
| b_{j1} & b_{j2} & ... & b_{jn}|\\
—&—&—&— \\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right| =
\]
\[\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
... & ... & & ...\\
mb_{i1}+nb_{j1} & mb_{i2}+nb_{j2} & ... & mb_{in}+nb_{jn}\\
... & ... & & ...\\
b_{j1} & b_{j2} & ... & b_{jn}\\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right|
\]
5.行列式分行相加性
\[\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
... & ... & & ...\\
b_{i1}+c_{i1} & b_{i2}+c_{i2} & ... & b_{in}+c_{in}\\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right| =
\]
\[\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
... & ... & & ...\\
b_{i1} & b_{i2} & ... & b_{in}\\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right| +
\]
\[\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
... & ... & & ...\\
c_{i1} & c_{i2} & ... & c_{in}\\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right|
\]
推广到n：
\[\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
... & ... & & ...\\
b_{i1}+c_{i1}+...+k_{i1} & b_{i2}+c_{i2}+...+k_{i2} & ... & b_{in}+c_{in}+...+k_{in}\\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right| =
\]
\[\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
... & ... & & ...\\
b_{i1} & b_{i2} & ... & b_{in}\\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right| +
\]
\[\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
... & ... & & ...\\
c_{i1} & c_{i2} & ... & c_{in}\\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right|+...
\]
\[+
\left|
\begin{matrix}
b_{11} & b_{12} & ... & b_{1n}\\
... & ... & & ...\\
k_{i1} & k_{i2} & ... & k_{in}\\
... & ... & & ...\\
b_{n1} & b_{n2} & ... & b_{nn}\\
\end{matrix}
\right|
\]
行列式按行或列展开(降阶)
1.余子式：去掉元素\(a_{ij}\)所在的第i行第j列的元素，留下的元素按原来位置构成的n-1阶行列式为元素\(a_{ij}\)的余子式，记：\(M_{ij}\)
2.代数余子式：\(A_{ij} = (-1)^{i+j}M_{ij}\)
\[D=
\left|
\begin{matrix}
b_{11} & b_{12} & b_{13}\\
b_{21} & b_{22} & b_{23}\\
b_{31} & b_{32}& b_{33}\\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[M_{32}=
\left|
\begin{matrix}
b_{11} & b_{13}\\
b_{21} & b_{23}\\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[A_{32} = (-1)^{3+2}M_{32}
\]
<Empty \space Math \space Block>
\[=(-1)^5M_{32}=-M_{32}
\]
3.n阶行列式D = |\(a_{ij}\)|等于它的任意一行（列）的各元素与其对应的代数余子式（\(A_{ij}\)）乘积之和
规则:利用行列式的性质，使某一行（列）尽可能地有较多的0，再按行列展开！
例：
\[D=
\left|
\begin{matrix}
-1 & 1 & -1 & 2\\
1 & 0 & 1 & -1\\
2 & 4 & 3 & 1\\
-1 & 1 & 2 & -2\\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[ =
\left|
\begin{matrix}
1 & 1 & 1 & 2\\
0 & 0 & 0 & -1\\
3 & 4 & 4 & 1\\
-3 & 1 & 0 & -2\\
\end{matrix}
\right| =
(-1)^{2+4}(-1)*
\left|
\begin{matrix}
1 & 1 & 1 \\
3 & 4 & 4 \\
-3 & 1 & 0\\
\end{matrix}
\right| = 1
\]
克拉默法则
线性方程组的系数 矩阵 行列式\(D≠0\),则行列式有唯一解，且解为：
\[x_1 = \frac {D_1}{D},x_2 = \frac {D_2}{D},x_3 = \frac {D_3}{D}
\]
八个基本型
上三角:
\[\left|
\begin{matrix}
b_{11} & 0 & 0 & 0\\
* & b_{22} & 0 & 0\\
* & *& b_{33} & 0\\
... & ... & & ...\\
*& * &*& b_{nn}\\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[
= b_{11}*b_{22}*b_{33}*...*b_{nn}
\]
<Empty \space Math \space Block>
\[=
\left|
\begin{matrix}
b_{11} & * & * & *\\
0 & b_{22} & * & *\\
0 & 0& b_{33} & *\\
... & ... & & ...\\
0& 0 &0& b_{nn}\\
\end{matrix}
\right|
\]
下三角:
\[\left|
\begin{matrix}
0& 0 & 0 & b_{1n}\\
0 & 0 & b_{2(n-1)} & *\\
0 & b_{3(n-2)} & *& *\\\\
... & ... & .... & ...\\
b_{n1}& * &*&*\\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[=
\left|
\begin{matrix}
*& * & * & b_{1n}\\
* & * & b_{2(n-1)} & 0\\
* & b_{3(n-2)} & 0& 0\\\\
... & ... & .... & ...\\
b_{n1}& 0 &0&0\\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[
= (-1)^{\frac {n(n-1)}{2}}b_{1(n-1)}*b_{2(n-2)}*b_{3(n-3)}*...*b_{1n}
\]
组合三角:
\[\left|
\begin{matrix}
b_{11} & b_{12} & 0 & 0\\
b_{21} & b_{22} & 0 & 0\\
* & *&c_{11} & c_{12}\\
*& * &c_{21}& c_{22}\\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[
=
\left|
\begin{matrix}
b_{11} & b_{12} & * & *\\
b_{21} & b_{22} & * & *\\
0 & 0& c_{11} & c_{12}\\
0& 0 &c_{21}& c_{22}\\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[=
\left|
\begin{matrix}
b_{11} & b_{12} \\
b_{21} & b_{22} \\
\end{matrix}
\right|*
\left|
\begin{matrix}
c_{11} & c_{12}\\
c_{21}& c_{22}\\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[\left|
\begin{matrix}
0 & 0&b_{11} & b_{12} \\
0 & 0&b_{21} & b_{22}\\
c_{11} & c_{12}&* & *\\
c_{21}& c_{22}&*& * \\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[=
\left|
\begin{matrix}
* & * &b_{11} & b_{12}\\
* & *& b_{21} & b_{22}\\
c_{11} & c_{12}&0 & 0 \\
c_{21}& c_{22}&0& 0 \\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[= (-1)^{2*2}
\left|
\begin{matrix}
b_{11} & b_{12} \\
b_{21} & b_{22} \\
\end{matrix}
\right|*
\]
<Empty \space Math \space Block>
\[\left|
\begin{matrix}
c_{11} & c_{12}\\
c_{21}& c_{22}\\
\end{matrix}
\right|
\]
范德蒙行列式
\[D=(a_1,a_2,...,a_n)
\]
<Empty \space Math \space Block>
\[=
\left|
\begin{matrix}
1& 1&... & 1\\
a_1& a_2&... & a_n\\
a_1^2& a_2^2&... & a_n^2\\
... & ... & ... & ...\\
a_1^{(n-1)}& a_2^{(n-1)}&... & a_n^{(n-1)}\\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[=
\prod_{1≤i<j≤n}(a_j-a_i)
\]
重要行列式解法
type1:
\[f(x)=
\left|
\begin{matrix}
x& a_1&a_2&...&a_{n-1} & 1\\
a_1&x& a_2&...&a_{n-1} & 1\\
a_1& a_2&x&...&a_{n-1} & 1\\
... & ... & ... & ...& ... & ...\\
a_1& a_2&a_3&... & x&1\\
a_1& a_2&a_3&... & a_n&1\\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[=
\left|
\begin{matrix}
x-a_1& a_1-x&0&...&0 & 0\\
0&x-a_2& a_2-x&...&0 & 0\\
0& 0&x-a_3&...&0& 0\\
... & ... & ... & ...& ... & ...\\
0& 0&0&... & x-a_n&0\\
a_1& a_2&a_3&... & a_n&1\\
\end{matrix}
\right| (列展开)
\]
<Empty \space Math \space Block>
\[=\left|
\begin{matrix}
x-a_1& a_1-x&0&...&0 \\
0&x-a_2& a_2-x&...&0 \\
0& 0&x-a_3&...&0\\
... & ... & ... & ...& ... \\
0& 0&0&... & x-a_n\\
\end{matrix}
\right| = (x-a_1)(x-a_2)...(x-a_n)
\]
type2:
\[f(x)=
\left|
\begin{matrix}
a& b& b&...& b\\
b&a& b&...&b \\
b&b& a&...&b \\
... & ... & ... & ...& ...\\
b&b& b&...&a \\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[=
\left|
\begin{matrix}
a+(n-1)b& b& b&...& b\\
a+(n-1)b&a& b&...&b \\
a+(n-1)b&b& a&...&b \\
... & ... & ... & ...& ...\\
a+(n-1)b&b& b&...&a \\
\end{matrix}
\right| (提取a+(n-1)b)
\]
<Empty \space Math \space Block>
\[=[a+(n-1)b]
\left|
\begin{matrix}
1 & b& b&...& b\\
1&a& b&...&b \\
1&b& a&...&b \\
... & ... & ... & ...& ...\\
1&b& b&...&a \\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[=[a+(n-1)b]
\left|
\begin{matrix}
1 & b& b&...& b\\
0&a-b& 0&...&0 \\
0&0& a-b&...&0 \\
... & ... & ... & ...& ...\\
0&0& 0&...&a-b \\
\end{matrix}
\right|
\]
<Empty \space Math \space Block>
\[ = [a+(n-1)b](a-b)_{n-1}
\]
浙公网安备 33010602011771号