2. N阶行列式
数域
K
\textbf{K}
K上的二元方程组
{
a
11
x
1
+
a
12
x
2
=
b
1
a
21
x
1
+
a
22
x
2
=
b
2
\left\{\begin{array}{l} a_{11}x_{1}+a_{12}x_{2}=b_{1}\\ a_{21}x_{1}+a_{22}x_{2}=b_{2} \end{array}\right.
{a11x1+a12x2=b1a21x1+a22x2=b2
写其系数矩阵：
(
a
11
a
12
a
21
a
22
)
\begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{pmatrix}
(a11a21a12a22)
上节课发现若
a
11
a
22
−
a
12
a
21
=
0
a_{11}a_{22}-a_{12}a_{21}=0
a11a22−a12a21=0，则此方程组无解或有无穷多个解；若
a
11
a
22
−
a
12
a
21
≠
0
a_{11}a_{22}-a_{12}a_{21}\ne 0
a11a22−a12a21=0，则此方程组有唯一解。该表达式称为二阶行列式，记为：
∣
a
11
a
12
a
21
a
22
∣
=
a
11
a
22
−
a
12
a
21
\begin{vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{vmatrix}=a_{11}a_{22}-a_{12}a_{21}
a11a21a12a22
=a11a22−a12a21
它也称为矩阵
A
\textbf{A}
A的行列式，记作
∣
A
∣
|\textbf{A}|
∣A∣，或
det
A
\det \textbf{A}
detA
- 数域 K \textbf{K} K上系数矩阵为 A \textbf{A} A的二元一次方程组有唯一解 ⇔ ∣ A ∣ ≠ 0 \Leftrightarrow|\textbf{A}|\ne 0 ⇔∣A∣=0
- 数域
K
\textbf{K}
K上系数矩阵为
A
\textbf{A}
A的二元一次方程组有无穷多个解或无解
⇔
∣
A
∣
=
0
\Leftrightarrow|\textbf{A}|=0
⇔∣A∣=0
对推广到 n n n元线性方程组，我们要研究 n n n阶行列式。
观察二阶行列式，我们发现，其行列式展开后的式子的每一项是取自不同行不同列的两个元素的乘积，每一项按照行指标按自然序（从小到大）排好位置，列指标所成的排列有 2 ! 2! 2!项的代数和，当列指标形成的排列时，该项带正号，当列指标形成的排列时，该项带负号
2.1 n n n元排列
- 1 , 2 , . . . , n 1,2,...,n 1,2,...,n或 n n n个不同的正整数的全排列就称为一个 n n n元排列。
- 从而
1
,
2
,
.
.
.
,
n
1,2,...,n
1,2,...,n或
n
n
n个不同的正整数形成的
n
n
n元排列有
n
!
n!
n!个
【例】 3 3 3元排列有 3 ! = 6 3!=6 3!=6个， 123 , 132 , 213 , 231 , 312 , 321 123,132,213,231,312,321 123,132,213,231,312,321
【例】 4 4 4元排列： 2431 2431 2431从左到右，顺序（从小到大）的数对有： 24 , 23 24,23 24,23；逆序（从大到小）的数对有 21 , 43 , 41 , 31 21,43,41,31 21,43,41,31
- 排列中逆序的数对的数目称为这个排列的逆序数，比如上面例子中的逆序数为4，逆序数记作 τ \tau τ，比如上面例子记作 τ ( 2431 ) = 4 \tau(2431)=4 τ(2431)=4
2.2 排列的奇偶性
把逆序数是偶数的排列称为偶排列，逆序数是奇数的排列称为奇排列，比如刚才例子中的排列2431就是偶排列。
我们将偶排列2431的4和1交换位置，其余的数不动，我们称这样的变换为一个对换，记作 ( 4 , 1 ) (4,1) (4,1)，对换后变成排列2134， τ ( 2134 ) = 1 \tau(2134)=1 τ(2134)=1，所以2134是一个奇排列。
【定理1】对换会改变排列的奇偶性。
【证】先看对换的两个数相邻的情形
.
.
.
p
.
.
.
i
j
.
.
.
q
.
.
.
...p...ij...q...
...p...ij...q...（排列1）
对换
(
i
,
j
)
(i,j)
(i,j)后
.
.
.
p
.
.
.
j
i
.
.
.
q
.
.
.
...p...ji...q...
...p...ji...q...（排列2）
排列（1）与（2）的逆序数就相差1，所以排列（1）和（2）的奇偶性相反，
一般情形，
.
.
.
i
k
1
.
.
.
k
s
j
.
.
.
...ik_{1}...k_{s}j...
...ik1...ksj...（排列3）
对换
(
i
,
j
)
(i,j)
(i,j)后
.
.
.
j
k
1
.
.
.
k
s
i
.
.
.
...jk_{1}...k_{s}i...
...jk1...ksi...（排列4）
排列（3）到排列（4）相当于做若干次对换
(
i
,
k
1
)
,
.
.
.
,
(
i
,
k
s
)
,
(
i
,
j
)
(i,k_{1}),...,(i,k_{s}),(i,j)
(i,k1),...,(i,ks),(i,j)变成
.
.
.
k
1
.
.
.
k
s
j
i
.
.
.
...k_{1}...k_{s}ji...
...k1...ksji...，
再经过对换
(
j
,
k
s
)
,
.
.
.
,
(
j
,
k
1
)
(j,k_{s}),...,(j,k_{1})
(j,ks),...,(j,k1)变成
.
.
.
j
k
1
.
.
.
k
s
i
.
.
.
...jk_{1}...k_{s}i...
...jk1...ksi...，排列（3）变成排列（4）经过
s
+
1
+
s
=
2
s
+
1
,
s
∈
N
+
s+1+s=2s+1,s\in\mathbb{N}^{+}
s+1+s=2s+1,s∈N+次变换，经过奇数次相邻两数对换，从而排列（3）与（4）奇偶性相反。
12345是偶排列，将25143（ τ ( 25143 ) = 5 \tau(25143)=5 τ(25143)=5，奇排列）通过对换变成12345，25143(5,3)→23145(3,1)→21345(2,1)→12345
【定理2】任一
n
n
n元排列
j
1
,
j
2
,
.
.
.
,
j
n
j_{1},j_{2},...,j_{n}
j1,j2,...,jn与
1
,
2
,
.
.
.
,
n
1,2,...,n
1,2,...,n可以经过一系列的对换互变，且所做对换的次数与
j
1
,
j
2
,
.
.
.
,
j
n
j_{1},j_{2},...,j_{n}
j1,j2,...,jn有相同的奇偶性。
【证】
j
1
,
j
2
,
.
.
.
,
j
n
j_{1},j_{2},...,j_{n}
j1,j2,...,jn经过
s
s
s次对换变成
1
,
2
,
.
.
.
,
n
1,2,...,n
1,2,...,n（偶排列），设
j
1
,
j
2
,
.
.
.
,
j
n
j_{1},j_{2},...,j_{n}
j1,j2,...,jn是奇排列，则
s
s
s必为奇数，设
j
1
,
j
2
,
.
.
.
,
j
n
j_{1},j_{2},...,j_{n}
j1,j2,...,jn是偶排列，则
s
s
s必为偶数。
2.3 n n n阶行列式的定义
【定义1】
n
n
n阶行列式：
∣
a
11
a
12
.
.
.
a
1
n
a
21
a
22
.
.
.
a
2
n
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.
.
.
.
.
.
a
n
1
a
n
2
.
.
.
a
n
n
∣
\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}
a11a21...an1a12a22...an2............a1na2n...ann
n
n
n阶行列是
n
!
n!
n!项的代数和，其中每一项是不同行和不同列的
n
n
n个元素的乘积，每一项按行指标成自然序排好位置，当列指标形成的排列是偶排列时，该项带正号，当列指标形成的排列是奇排列的时候，该项带负号。
比如二阶行列式 ∣ a 11 a 12 a 21 a 22 ∣ = a 11 a 22 − a 12 a 21 = ∑ j 1 j 2 ( − 1 ) τ ( j 1 j 2 ) a 1 j 1 a 2 j 2 \begin{vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{vmatrix}=a_{11}a_{22}-a_{12}a_{21}=\sum\limits_{j_{1}j_{2}}(-1)^{\tau(j_{1}j_{2})}a_{1j_{1}}a_{2j_{2}} a11a21a12a22 =a11a22−a12a21=j1j2∑(−1)τ(j1j2)a1j1a2j2
则
∣
a
11
a
12
.
.
.
a
1
n
a
21
a
22
.
.
.
a
2
n
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.
a
n
1
a
n
2
.
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.
a
n
n
∣
=
∑
j
1
j
2
.
.
.
j
n
a
1
j
1
(
−
1
)
τ
(
j
1
j
2
.
.
.
j
n
)
a
2
j
2
.
.
.
a
n
j
n
\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}=\sum\limits_{j_{1}j_{2}...j_{n}}a_{1j_{1}}(-1)^{\tau(j_{1}j_{2}...j_{n})}a_{2j_{2}}...a_{nj_{n}}
a11a21...an1a12a22...an2............a1na2n...ann
=j1j2...jn∑a1j1(−1)τ(j1j2...jn)a2j2...anjn（行指标按自然序排好）
设矩阵
A
=
(
a
11
a
12
.
.
.
a
1
n
a
21
a
22
.
.
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a
2
n
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a
n
1
a
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a
n
n
)
\textbf{A}=\begin{pmatrix} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn} \end{pmatrix}
A=
a11a21...an1a12a22...an2............a1na2n...ann
，该矩阵可简记为简记成
A
=
(
a
i
j
)
\textbf{A}=(a_{ij})
A=(aij)，其中
a
i
j
a_{ij}
aij是第
i
i
i行第
j
j
j列的交叉位置元素即
A
\textbf{A}
A的
(
i
,
j
)
(i,j)
(i,j)元，则上述行列式也可以记作
n
n
n阶矩阵的阶行列式，可记为
∣
A
∣
|\textbf{A}|
∣A∣或
det
A
\det\textbf{A}
detA
- 1阶行列式： ∣ a ∣ = a |a|=a ∣a∣=a
- 2阶行列式： ∣ a 11 a 12 a 21 a 22 ∣ = a 11 a 22 − a 12 a 21 = ∑ j 1 j 2 ( − 1 ) τ ( j 1 j 2 ) a 1 j 1 a 2 j 2 \begin{vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{vmatrix}=a_{11}a_{22}-a_{12}a_{21}=\sum\limits_{j_{1}j_{2}}(-1)^{\tau(j_{1}j_{2})}a_{1j_{1}}a_{2j_{2}} a11a21a12a22 =a11a22−a12a21=j1j2∑(−1)τ(j1j2)a1j1a2j2
- 3阶行列式：3元排列
偶排列：123,231,312
奇排列：132,213,321
∣ a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ∣ = a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 − a 11 a 23 a 32 − a 12 a 21 a 33 − a 13 a 22 a 31 \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}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31} a11a21a31a12a22a32a13a23a33 =a11a22a33+a12a23a31+a13a21a32−a11a23a32−a12a21a33−a13a22a31
2.4 上三角形行列式
∣
a
11
a
12
a
13
.
.
.
a
1
,
n
−
1
a
1
n
0
a
22
a
23
.
.
.
a
2
,
n
−
1
a
2
n
0
0
a
33
.
.
.
a
3
,
n
−
1
a
3
n
.
.
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.
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.
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.
.
.
.
.
.
.
0
0
0
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.
.
a
n
−
1
,
n
−
1
a
n
−
1
,
n
0
0
0
.
.
.
0
a
n
n
∣
=
a
11
a
22
a
33
.
.
.
a
n
−
1
,
n
−
1
a
n
n
\begin{vmatrix} a_{11} & a_{12} &a_{13}& ... & a_{1,n-1}&a_{1n}\\ 0 & a_{22} &a_{23}& ... & a_{2,n-1}&a_{2n}\\ 0 & 0 &a_{33}& ... & a_{3,n-1}&a_{3n}\\ ... & ... &...& ... & ...&...\\ 0 & 0 &0& ... & a_{n-1,n-1}&a_{n-1,n}\\ 0 & 0 &0& ... & 0&a_{nn} \end{vmatrix}=a_{11}a_{22}a_{33}...a_{n-1,n-1}a_{nn}
a1100...00a12a220...00a13a23a33...00..................a1,n−1a2,n−1a3,n−1...an−1,n−10a1na2na3n...an−1,nann
=a11a22a33...an−1,n−1ann（因为取自不同行不同列，且列指标按自然序排列，是偶排列，所以带正号）
主对角线下方元素全为0这样的
n
n
n阶行列式，我们称为上三角形行列式。
【命题】
n
n
n阶上三角形行列式的值等于它的主对角线上
n
n
n个元素的乘积。
2.5 行列式的转置
n
n
n阶行列式的一项为
(
−
1
)
τ
(
j
1
j
2
.
.
.
j
n
)
a
1
j
1
a
2
j
2
.
.
.
a
n
j
n
(-1)^{\tau(j_{1}j_{2}...j_{n})}a_{1j_{1}}a_{2j_{2}}...a_{nj_{n}}
(−1)τ(j1j2...jn)a1j1a2j2...anjn经过
s
s
s次两个元素互换位置，相应的行指标所成的排列
12...
n
12...n
12...n经过
s
s
s次对换变为
i
1
,
i
2
,
.
.
.
,
i
n
i_{1},i_{2},...,i_{n}
i1,i2,...,in，列指标排列
j
1
,
j
2
,
.
.
.
,
j
n
j_{1},j_{2},...,j_{n}
j1,j2,...,jn经过
s
s
s次对换变为
k
1
,
k
2
,
.
.
.
,
k
n
k_{1},k_{2},...,k_{n}
k1,k2,...,kn，
(
−
1
)
τ
(
i
1
,
i
2
,
.
.
.
,
i
n
)
=
(
−
1
)
s
(-1)^{\tau(i_{1},i_{2},...,i_{n})}=(-1)^{s}
(−1)τ(i1,i2,...,in)=(−1)s（1次对换改变奇偶性，
s
s
s次，就改变
s
s
s次奇偶性）
(
−
1
)
τ
(
k
1
,
k
2
,
.
.
.
,
k
n
)
=
(
−
1
)
s
(
−
1
)
τ
(
j
1
,
j
2
,
.
.
.
,
j
n
)
(-1)^{\tau(k_{1},k_{2},...,k_{n})}=(-1)^{s}(-1)^{\tau(j_{1},j_{2},...,j_{n})}
(−1)τ(k1,k2,...,kn)=(−1)s(−1)τ(j1,j2,...,jn)（列指标原来的排列
j
1
,
j
2
,
.
.
.
,
j
n
j_{1},j_{2},...,j_{n}
j1,j2,...,jn的正负号是由
(
−
1
)
τ
(
j
1
,
j
2
,
.
.
.
,
j
n
)
(-1)^{\tau(j_{1},j_{2},...,j_{n})}
(−1)τ(j1,j2,...,jn)所决定，所以要乘一个
(
−
1
)
τ
(
j
1
,
j
2
,
.
.
.
,
j
n
)
(-1)^{\tau(j_{1},j_{2},...,j_{n})}
(−1)τ(j1,j2,...,jn)），这两个式子相乘得
(
−
1
)
τ
(
k
1
,
k
2
,
.
.
.
,
k
n
)
+
τ
(
j
1
,
j
2
,
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,
j
n
)
=
(
−
1
)
2
s
+
τ
(
j
1
,
j
2
,
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.
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,
j
n
)
=
1
⋅
(
−
1
)
τ
(
j
1
,
j
2
,
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,
j
n
)
=
(
−
1
)
τ
(
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2
,
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,
j
n
)
(-1)^{\tau(k_{1},k_{2},...,k_{n})+\tau(j_{1},j_{2},...,j_{n})}=(-1)^{2s+\tau(j_{1},j_{2},...,j_{n})}=1\cdot(-1)^{\tau(j_{1},j_{2},...,j_{n})}=(-1)^{\tau(j_{1},j_{2},...,j_{n})}
(−1)τ(k1,k2,...,kn)+τ(j1,j2,...,jn)=(−1)2s+τ(j1,j2,...,jn)=1⋅(−1)τ(j1,j2,...,jn)=(−1)τ(j1,j2,...,jn)，于是我们等量代替，
(
−
1
)
τ
(
j
1
j
2
.
.
.
j
n
)
a
1
j
1
a
2
j
2
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a
n
j
n
=
(
−
1
)
τ
(
k
1
,
k
2
,
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,
k
n
)
+
τ
(
j
1
,
j
2
,
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,
j
n
)
a
2
j
2
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.
a
n
j
n
(-1)^{\tau(j_{1}j_{2}...j_{n})}a_{1j_{1}}a_{2j_{2}}...a_{nj_{n}}=(-1)^{\tau(k_{1},k_{2},...,k_{n})+\tau(j_{1},j_{2},...,j_{n})}a_{2j_{2}}...a_{nj_{n}}
(−1)τ(j1j2...jn)a1j1a2j2...anjn=(−1)τ(k1,k2,...,kn)+τ(j1,j2,...,jn)a2j2...anjn
特别地，按列指标排自然序
A
=
(
a
i
j
)
\textbf{A}=(a_{ij})
A=(aij)的行列式
∣
A
∣
=
∑
i
1
,
i
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|\textbf{A}|=\sum\limits_{i_{1},i_{2},...,i_{n}}(-1)^{\tau(i_{1},i_{2},...,i_{n})}a_{i_{1}1}a_{i_{2}2}...a_{i_{n}n}
∣A∣=i1,i2,...,in∑(−1)τ(i1,i2,...,in)ai11ai22...ainn
设 n n n阶矩阵 A = ( a i j ) = ( a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . . . . . . . . . . a n 1 a n 2 . . . a n n ) \textbf{A}=(a_{ij})=\begin{pmatrix} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn} \end{pmatrix} A=(aij)= a11a21...an1a12a22...an2............a1na2n...ann ，把行列互换得到的矩阵 ( a 11 a 21 . . . a n 1 a 12 a 22 . . . a n 2 . . . . . . . . . . . . a 1 n a 2 n . . . a n n ) \begin{pmatrix} a_{11} & a_{21} & ... & a_{n1}\\ a_{12} & a_{22} & ... & a_{n2}\\ ... & ...&...&...\\ a_{1n} & a_{2n} & ... & a_{nn} \end{pmatrix} a11a12...a1na21a22...a2n............an1an2...ann ，将这个矩阵称为矩阵 A \textbf{A} A的转置，记作 A ′ \textbf{A}' A′或 A T \textbf{A}^{T} AT或 A t \textbf{A}^{t} At（国内外教材写法不同，工科常用 A T \textbf{A}^{T} AT，数学专业常用 A ′ \textbf{A}' A′） ∣ A ′ ∣ ( 列指标呈自然序 ) = ∣ a 11 a 21 . . . a n 1 a 12 a 22 . . . a n 2 . . . . . . . . . . . . a 1 n a 2 n . . . a n n ∣ = ∑ i 1 , i 2 , . . . , i n ( − 1 ) τ ( i 1 , i 2 , . . . , i n ) a i 1 , 1 a i 2 , 2 . . . a i n , n = ( 刚才证明的，行指标呈自然序） ∣ A ∣ |\textbf{A}'|(列指标呈自然序)=\begin{vmatrix} a_{11} & a_{21} & ... & a_{n1}\\ a_{12} & a_{22} & ... & a_{n2}\\ ... & ...&...&...\\ a_{1n} & a_{2n} & ... & a_{nn} \end{vmatrix}=\sum\limits_{i_{1},i_{2},...,i_{n}}(-1)^{\tau(i_{1},i_{2},...,i_{n})}a_{i_{1},1}a_{i_{2},2}...a_{i_{n},n}=(刚才证明的，行指标呈自然序）|\textbf{A}| ∣A′∣(列指标呈自然序)= a11a12...a1na21a22...a2n............an1an2...ann =i1,i2,...,in∑(−1)τ(i1,i2,...,in)ai1,1ai2,2...ain,n=(刚才证明的，行指标呈自然序）∣A∣
2.6 行列式的性质
- 性质1： ∣ A ∣ = ∣ A ′ ∣ |\textbf{A}|=|\textbf{A}'| ∣A∣=∣A′∣
- 性质2：
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|k\textbf{A}|=k|\textbf{A}|
∣kA∣=k∣A∣（
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k=0也成立）
【证】 ∣ k A ∣ = ∣ a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . . . . . . . . . . k a i 1 k a i 2 . . . k a i n . . . . . . . . . . . . a n 1 a n 2 . . . a n n ∣ = ∑ j 1 , j 2 , . . . , j n ( − 1 ) τ ( j 1 , j 2 , . . . , j n ) a 1 j 1 . . . ( k a i j i ) . . . a n j n = k ( − 1 ) τ ( j 1 , j 2 , . . . , j n ) a 1 j 1 . . . a i j i . . . a n j n = k ∣ A ∣ |k\textbf{A}|=\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ ... & ...&...&...\\ ka_{i1} & ka_{i2} & ... & ka_{in}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}=\sum\limits_{j_{1},j_{2},...,j_{n}}(-1)^{\tau(j_{1},j_{2},...,j_{n})}a_{1j_{1}}...(ka_{ij_{i}})...a_{nj_{n}}=k(-1)^{\tau(j_{1},j_{2},...,j_{n})}a_{1j_{1}}...a_{ij_{i}}...a_{nj_{n}}=k|\textbf{A}| ∣kA∣= a11a21...kai1...an1a12a22...kai2...an2..................a1na2n...kain...ann =j1,j2,...,jn∑(−1)τ(j1,j2,...,jn)a1j1...(kaiji)...anjn=k(−1)τ(j1,j2,...,jn)a1j1...aiji...anjn=k∣A∣ - 性质3：
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\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ ... & ...&...&...\\ b_{1}+c_{1}& b_{2}+c_{2} & ... & b_{n}+c_{n}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}=\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ ... & ...&...&...\\ b_{1}& b_{2} & ... & b_{n}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}+\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ ... & ...&...&...\\ c_{1}& c_{2} & ... & c_{n}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}
a11a21...b1+c1...an1a12a22...b2+c2...an2..................a1na2n...bn+cn...ann
=
a11a21...b1...an1a12a22...b2...an2..................a1na2n...bn...ann
+
a11a21...c1...an1a12a22...c2...an2..................a1na2n...cn...ann
【证】 ∣ a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . . . . . . . . . . b 1 + c 1 b 2 + c 2 . . . b n + c n . . . . . . . . . . . . a n 1 a n 2 . . . a n n ∣ = ∑ j 1 , j 2 , . . . , j n ( − 1 ) τ ( j 1 , j 2 , . . . , j n ) a 1 j 1 . . . ( b j i + c j i ) . . . a n j n = ∑ j 1 , j 2 , . . . , j n ( − 1 ) τ ( j 1 , j 2 , . . . , j n ) a 1 j 1 . . . b j i . . . a n j n + ∑ j 1 , j 2 , . . . , j n ( − 1 ) τ ( j 1 , j 2 , . . . , j n ) a 1 j 1 . . . c j i . . . a n j n = ∣ a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . . . . . . . . . . b 1 b 2 . . . b n . . . . . . . . . . . . a n 1 a n 2 . . . a n n ∣ + ∣ a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . . . . . . . . . . c 1 c 2 . . . c n . . . . . . . . . . . . a n 1 a n 2 . . . a n n ∣ \begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ ... & ...&...&...\\ b_{1}+c_{1}& b_{2}+c_{2} & ... & b_{n}+c_{n}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}=\sum\limits_{j_{1},j_{2},...,j_{n}}(-1)^{\tau(j_{1},j_{2},...,j_{n})}a_{1j_{1}}...(b_{j_{i}}+c_{j_{i}})...a_{nj_{n}}=\sum\limits_{j_{1},j_{2},...,j_{n}}(-1)^{\tau(j_{1},j_{2},...,j_{n})}a_{1j_{1}}...b_{j_{i}}...a_{nj_{n}}+\sum\limits_{j_{1},j_{2},...,j_{n}}(-1)^{\tau(j_{1},j_{2},...,j_{n})}a_{1j_{1}}...c_{j_{i}}...a_{nj_{n}}=\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ ... & ...&...&...\\ b_{1}& b_{2} & ... & b_{n}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}+\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ ... & ...&...&...\\ c_{1}& c_{2} & ... & c_{n}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix} a11a21...b1+c1...an1a12a22...b2+c2...an2..................a1na2n...bn+cn...ann =j1,j2,...,jn∑(−1)τ(j1,j2,...,jn)a1j1...(bji+cji)...anjn=j1,j2,...,jn∑(−1)τ(j1,j2,...,jn)a1j1...bji...anjn+j1,j2,...,jn∑(−1)τ(j1,j2,...,jn)a1j1...cji...anjn= a11a21...b1...an1a12a22...b2...an2..................a1na2n...bn...ann + a11a21...c1...an1a12a22...c2...an2..................a1na2n...cn...ann - 性质4：若矩阵
C
\textbf{C}
C是矩阵
A
\textbf{A}
A经过两行互换后的来的矩阵，则
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|\textbf{C}|=-|\textbf{A}|
∣C∣=−∣A∣（两行互换，行列式反号）
【证】假设将第 k k k行和第 i i i行互换， ∣ A ∣ = ∣ a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . . . . . . . . . . a i 1 b i 2 . . . b i n . . . . . . . . . . . . a k 1 b k 2 . . . b k n . . . . . . . . . . . . a n 1 a n 2 . . . a n n ∣ |\textbf{A}|=\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ ... & ...&...&...\\ a_{i1}& b_{i2} & ... & b_{in}\\ ... & ...&...&...\\ a_{k1}& b_{k2} & ... & b_{kn}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix} ∣A∣= a11a21...ai1...ak1...an1a12a22...bi2...bk2...an2........................a1na2n...bin...bkn...ann
∣ C ∣ = ∣ a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . . . . . . . . . . a k 1 b k 2 . . . b k n . . . . . . . . . . . . a i 1 b i 2 . . . b i n . . . . . . . . . . . . a n 1 a n 2 . . . a n n ∣ = ∑ j 1 , j 2 , . . . , j k , . . . , j i , . . . , j n ( − 1 ) τ ( j 1 , j 2 , . . . , j k , . . . , j i , . . . , j n ) a 1 j 1 . . . a k j i . . . a i j k . . . a n j n ( 乘法交换律 ) = ∑ j 1 , j 2 , . . . , j k , . . . , j i , . . . , j n ( − 1 ) τ ( j 1 , j 2 , . . . , j k , . . . , j i , . . . , j n ) a 1 j 1 . . . a i j k . . . a k j i . . . a n j n ( 对换，改变奇偶性，前面乘 − 1 ) = ∑ j 1 , j 2 , . . . , j i , . . . , j k , . . . , j n ( − 1 ) ⋅ ( − 1 ) τ ( j 1 , j 2 , . . . , j i , . . . , j k , . . . , j n ) a 1 j 1 . . . a i j k . . . a k j i . . . a n j n = − ∑ j 1 , j 2 , . . . , j i , . . . , j k , . . . , j n ( − 1 ) τ ( j 1 , j 2 , . . . , j i , . . . , j k , . . . , j n ) a 1 j 1 . . . a i j k . . . a k j i . . . a n j n = − ∣ A ∣ |\textbf{C}|=\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ ... & ...&...&...\\ a_{k1}& b_{k2} & ... & b_{kn}\\ ... & ...&...&...\\ a_{i1}& b_{i2} & ... & b_{in}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}=\sum\limits_{j_{1},j_{2},...,j_{k},...,j_{i},...,j_{n}}(-1)^{\tau(j_{1},j_{2},...,j_{k},...,j_{i},...,j_{n})}a_{1j_{1}}...a_{kj_{i}}...a_{ij_{k}}...a_{nj_{n}}(乘法交换律)=\sum\limits_{j_{1},j_{2},...,j_{k},...,j_{i},...,j_{n}}(-1)^{\tau(j_{1},j_{2},...,j_{k},...,j_{i},...,j_{n})}a_{1j_{1}}...a_{ij_{k}}...a_{kj_{i}}...a_{nj_{n}}(对换，改变奇偶性，前面乘-1)=\sum\limits_{j_{1},j_{2},...,j_{i},...,j_{k},...,j_{n}}(-1)\cdot(-1)^{\tau(j_{1},j_{2},...,j_{i},...,j_{k},...,j_{n})}a_{1j_{1}}...a_{ij_{k}}...a_{kj_{i}}...a_{nj_{n}}=-\sum\limits_{j_{1},j_{2},...,j_{i},...,j_{k},...,j_{n}}(-1)^{\tau(j_{1},j_{2},...,j_{i},...,j_{k},...,j_{n})}a_{1j_{1}}...a_{ij_{k}}...a_{kj_{i}}...a_{nj_{n}}=-|\textbf{A}| ∣C∣= a11a21...ak1...ai1...an1a12a22...bk2...bi2...an2........................a1na2n...bkn...bin...ann =j1,j2,...,jk,...,ji,...,jn∑(−1)τ(j1,j2,...,jk,...,ji,...,jn)a1j1...akji...aijk...anjn(乘法交换律)=j1,j2,...,jk,...,ji,...,jn∑(−1)τ(j1,j2,...,jk,...,ji,...,jn)a1j1...aijk...akji...anjn(对换，改变奇偶性，前面乘−1)=j1,j2,...,ji,...,jk,...,jn∑(−1)⋅(−1)τ(j1,j2,...,ji,...,jk,...,jn)a1j1...aijk...akji...anjn=−j1,j2,...,ji,...,jk,...,jn∑(−1)τ(j1,j2,...,ji,...,jk,...,jn)a1j1...aijk...akji...anjn=−∣A∣ - 性质5：行列式两行相等，行列式的值为0；【证】先看两行相等的情况
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\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ ... & ...&...&...\\ a_{i1} & a_{i2} & ... & a_{in}\\ ... & ...&...&...\\ a_{i1} & a_{i2} & ... & a_{in}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}(换相同的两行)=-\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ ... & ...&...&...\\ a_{i1} & a_{i2} & ... & a_{in}\\ ... & ...&...&...\\ a_{i1} & a_{i2} & ... & a_{in}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}
a11...ai1...ai1...an1a12...ai2...ai2...an2.....................a1n...ain...ain...ann
(换相同的两行)=−
a11...ai1...ai1...an1a12...ai2...ai2...an2.....................a1n...ain...ain...ann
所以 ∣ a 11 a 12 . . . a 1 n . . . . . . . . . . . . a i 1 a i 2 . . . a i n . . . . . . . . . . . . a i 1 a i 2 . . . a i n . . . . . . . . . . . . a n 1 a n 2 . . . a n n ∣ = 0 \begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ ... & ...&...&...\\ a_{i1} & a_{i2} & ... & a_{in}\\ ... & ...&...&...\\ a_{i1} & a_{i2} & ... & a_{in}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}=0 a11...ai1...ai1...an1a12...ai2...ai2...an2.....................a1n...ain...ain...ann =0 - 性质6：行列式两行成比例，行列式的值为0；
【证】现在讨论两行成比例的情况：
∣ a 11 a 12 . . . a 1 n . . . . . . . . . . . . a i 1 a i 2 . . . a i n . . . . . . . . . . . . l a i 1 l a i 2 . . . l a i n . . . . . . . . . . . . a n 1 a n 2 . . . a n n ∣ = l ∣ a 11 a 12 . . . a 1 n . . . . . . . . . . . . a i 1 a i 2 . . . a i n . . . . . . . . . . . . a i 1 a i 2 . . . a i n . . . . . . . . . . . . a n 1 a n 2 . . . a n n ∣ = 0 \begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ ... & ...&...&...\\ a_{i1} & a_{i2} & ... & a_{in}\\ ... & ...&...&...\\ la_{i1} & la_{i2} & ... & la_{in}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}=l\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ ... & ...&...&...\\ a_{i1} & a_{i2} & ... & a_{in}\\ ... & ...&...&...\\ a_{i1} & a_{i2} & ... & a_{in}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}=0 a11...ai1...lai1...an1a12...ai2...lai2...an2.....................a1n...ain...lain...ann =l a11...ai1...ai1...an1a12...ai2...ai2...an2.....................a1n...ain...ain...ann =0 - 性质7：将矩阵
A
\textbf{A}
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∣A∣=∣D∣
【证】 ∣ D ∣ = ∣ a 11 a 12 . . . a 1 n . . . . . . . . . . . . a i 1 a i 2 . . . a i n . . . . . . . . . . . . l a i 1 + a k 1 l a i 2 + a k 2 . . . l a i n + a k n . . . . . . . . . . . . a n 1 a n 2 . . . a n n ∣ = l ∣ a 11 a 12 . . . a 1 n . . . . . . . . . . . . a i 1 a i 2 . . . a i n . . . . . . . . . . . . a i 1 a i 2 . . . a i n . . . . . . . . . . . . a n 1 a n 2 . . . a n n ∣ + ∣ a 11 a 12 . . . a 1 n . . . . . . . . . . . . a i 1 a i 2 . . . a i n . . . . . . . . . . . . a k 1 a k 2 . . . a k n . . . . . . . . . . . . a n 1 a n 2 . . . a n n ∣ = 0 + ∣ A ∣ = ∣ A ∣ |\textbf{D}|=\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ ... & ...&...&...\\ a_{i1} & a_{i2} & ... & a_{in}\\ ... & ...&...&...\\ l a_{i1} +a_{k1} & l a_{i2} +a_{k2} & ... & l a_{in} +a_{kn}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}=l\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ ... & ...&...&...\\ a_{i1} & a_{i2} & ... & a_{in}\\ ... & ...&...&...\\ a_{i1} & a_{i2} & ... & a_{in}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}+\begin{vmatrix} a_{11} & a_{12} & ... & a_{1n}\\ ... & ...&...&...\\ a_{i1} & a_{i2} & ... & a_{in}\\ ... & ...&...&...\\ a_{k1} & a_{k2} & ... & a_{kn}\\ ... & ...&...&...\\ a_{n1} & a_{n2} & ... & a_{nn}\\ \end{vmatrix}=0+|\textbf{A}|=|\textbf{A}| ∣D∣= a11...ai1...lai1+ak1...an1a12...ai2...lai2+ak2...an2.....................a1n...ain...lain+akn...ann =l a11...ai1...ai1...an1a12...ai2...ai2...an2.....................a1n...ain...ain...ann + a11...ai1...ak1...an1a12...ai2...ak2...an2.....................a1n...ain...akn...ann =0+∣A∣=∣A∣
【例1】计算 n ( n ≥ 2 ) n(n\ge 2) n(n≥2)阶行列式 ∣ k λ λ . . . λ λ k λ . . . λ . . . . . . . . . . . . . . . λ λ λ . . . k ∣ \begin{vmatrix} k & \lambda & \lambda&... & \lambda\\ \lambda & k &\lambda& ... & \lambda\\ ... & ...&...&...&...\\ \lambda & \lambda & \lambda & ... & k\\ \end{vmatrix} kλ...λλk...λλλ...λ............λλ...k
【解】原式（将后 2 , 3 , . . . , n 2,3,...,n 2,3,...,n列的1倍依次加到第1列上） = ∣ k + ( n − 1 ) λ λ λ . . . λ k + ( n − 1 ) λ k λ . . . λ . . . . . . . . . . . . . . . k + ( n − 1 ) λ λ λ . . . k ∣ = ( k + ( n − 1 ) λ ) ∣ 1 λ λ . . . λ 1 k λ . . . λ . . . . . . . . . . . . . . . 1 λ λ . . . k ∣ ( 将第 1 行的负 1 倍依次加到第 2 , 3 , . . . , n 行 ) = ( k + ( n − 1 ) ∣ 1 λ λ . . . λ 0 k − λ 0 . . . 0 . . . . . . . . . . . . . . . 0 0 0 . . . k − λ ∣ = ( k + ( n − 1 ) λ ) ( k − λ ) n − 1 =\begin{vmatrix} k+(n-1)\lambda & \lambda & \lambda&... & \lambda\\ k+(n-1)\lambda & k &\lambda& ... & \lambda\\ ... & ...&...&...&...\\ k+(n-1)\lambda & \lambda & \lambda & ... & k\\ \end{vmatrix}=(k+(n-1)\lambda)\begin{vmatrix} 1& \lambda & \lambda&... & \lambda\\ 1 & k &\lambda& ... & \lambda\\ ... & ...&...&...&...\\ 1 & \lambda & \lambda & ... & k\\ \end{vmatrix}(将第1行的负1倍依次加到第2,3,...,n行)=(k+(n-1)\begin{vmatrix} 1& \lambda & \lambda&... & \lambda\\ 0 & k-\lambda &0& ... & 0\\ ... & ...&...&...&...\\ 0 & 0 & 0 & ... & k-\lambda\\ \end{vmatrix}=(k+(n-1)\lambda)(k-\lambda)^{n-1} = k+(n−1)λk+(n−1)λ...k+(n−1)λλk...λλλ...λ............λλ...k =(k+(n−1)λ) 11...1λk...λλλ...λ............λλ...k (将第1行的负1倍依次加到第2,3,...,n行)=(k+(n−1) 10...0λk−λ...0λ0...0............λ0...k−λ =(k+(n−1)λ)(k−λ)n−1
2.5 代数余子式初步
3阶行列式：
∣
A
∣
=
∣
a
11
a
12
a
13
a
21
a
22
a
23
a
31
a
32
a
33
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11
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a
22
a
33
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a
23
a
32
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a
12
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a
23
a
31
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a
21
a
33
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a
13
(
a
21
a
32
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a
22
a
31
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=
a
11
(
a
22
a
33
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a
23
a
32
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−
a
12
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a
21
a
33
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31
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a
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21
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32
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a
31
)
|\textbf{A}|=\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_{23}a_{32})+a_{12}(a_{23}a_{31}-a_{21}a_{33})+a_{13}(a_{21}a_{32}-a_{22}a_{31})=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31})
∣A∣=
a11a21a31a12a22a32a13a23a33
=a11(a22a33−a23a32)+a12(a23a31−a21a33)+a13(a21a32−a22a31)=a11(a22a33−a23a32)−a12(a21a33−a23a31)+a13(a21a32−a22a31)
后面课继续记