二阶行列式推导
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[abcd]=[a0cd]+[0bcd]=[a00d]+[a0c0]+[0bc0]+[0b0d]=[a00d]−[b00c]=ad−bc
[acbd]=[ac0d]+[0cbd]=[a00d]+[ac00]+[0cb0]+[00bd]=[a00d]−[b00c]=ad−bc
三阶行列式推导
[ a b c d e f g h i ] = [ a 0 0 d e f g h i ] + [ 0 b 0 d e f g h i ] + [ 0 0 c d e f g h i ] = [ a 0 0 0 e 0 g h i ] + [ a 0 0 0 0 f g h i ] + [ 0 b 0 d 0 0 g h i ] + [ 0 b 0 0 0 f g h i ] + [ 0 0 c d 0 0 g h i ] + [ 0 0 c 0 e 0 g h i ] = [ a 0 0 0 e 0 0 0 i ] + [ a 0 0 0 0 f 0 h 0 ] + [ 0 b 0 d 0 0 0 0 i ] + [ 0 b 0 0 0 f g 0 0 ] + [ 0 0 c d 0 0 0 h 0 ] + [ 0 0 c 0 e 0 g 0 0 ] = a e i + b f g + c d h − a h f − b d i − c e g [abcdefghi] =[a00defghi]+ [0b0defghi] +[00cdefghi]\\ =[a000e0ghi] +[a0000fghi]+ [0b0d00ghi]+ [0b000fghi]+ [00cd00ghi]+ [00c0e0ghi]\\= [a000e000i]+ [a0000f0h0] +[0b0d0000i]+ [0b000fg00]+ [00cd000h0] +[00c0e0g00] \\= aei+bfg+cdh-ahf-bdi-ceg adgbehcfi = adg0eh0fi + 0dgbeh0fi + 0dg0ehcfi = a0g0eh00i + a0g00h0fi + 0dgb0h00i + 00gb0h0fi + 0dg00hc0i + 00g0ehc0i = a000e000i + a0000h0f0 + 0d0b0000i + 00gb000f0 + 0d000hc00 + 00g0e0c00 =aei+bfg+cdh−ahf−bdi−ceg
行列式公式
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det\ A=\sum_{j_1,j_2,j_3\quad is\ permutaion}\pm a_{1j_1}a_{2j_2}...a_{nj_n}\\ \forall j_{t_1},j_{t_2} \wedge t_1 \ne t_2 \Rightarrow j_{t_1} \ne j_{t_2}
det A=j1,j2,j3is permutaion∑±a1j1a2j2...anjn∀jt1,jt2∧t1=t2⇒jt1=jt2
即选取的列坐标不重复,构成了一个排列。
所以非0项共有
n
!
n!
n!项。
余子式
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:
方阵去掉
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列后的方阵的行列式
M_{ij}:方阵去掉i行j列后的方阵的行列式
Mij:方阵去掉i行j列后的方阵的行列式
代数余子式
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A_{ij}:(-1)^{i+j}M_{ij}
Aij:(−1)i+jMij
方阵行列式:
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det\ A=\sum_{1}^{n}A_{ik}, 1 \le i \le n
det A=1∑nAik,1≤i≤n
[ 1 1 0 0 1 1 1 0 0 1 1 1 0 0 1 1 ] [1100111001110011] 1100111001110011
∣
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|A_1|=1
∣A1∣=1
∣
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|A_2|=0
∣A2∣=0
∣
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|A_3|=--1
∣A3∣=−−1
∣
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|A_4|=-1
∣A4∣=−1
∣
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|A_5|=-0
∣A5∣=−0
∣
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|A_6|=1
∣A6∣=1
周期为6
A n A_n An的意思是以 1 , 1 1,1 1,1为起始点的向右向下扩展 k k k个单位的矩阵。
如
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A_3= [110111011]
A3=
110111011