1.4 行列式按行(列)展开定理
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1.4 行列式按行(列)展开定理
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一、余子式与代数余子式容易验证:
问题:一个高阶行列式是否可以转化为若干个 低阶行列式来计算?
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在 阶行列式中,把元素 所在的第 行和第 列划去后,留下来的 阶行列式叫做元素 的余子式,记作
n ija i j1n
ija.M ij
,记 ijji
ij MA 1 叫做元素 的代数余子式.ija
例如
44434241
34333231
24232221
14131211
αααα
αααα
αααα
αααα
A
444241
343231
141211
23
aaa
aaa
aaa
M
2332
23 1 MA .23M
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例 1 设
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定理 1.2 行列式等于它的任一行(列)的各元素与其对应的代数余子式乘积之和,即
ininiiii AαAαAαA 2211 ni ,,2,1
证
nnnn
inii
n
ααα
ααα
ααα
A
21
21
11211
000000
二、行列式按行(列)展开法则
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nnnn
i
n
aaa
a
aaa
21
1
11211
00
nnnn
i
n
aaa
a
aaa
21
2
11211
00
nnnn
in
n
aaa
a
aaa
21
11211
00 ininiiii AaAaAa 2211
ni ,,2,1
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例 2 计算行列式
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例 3 计算范德蒙德 (Vandermonde) 行列式
112
11
222
21
21
111
nn
nn
n
n
n
ααα
ααα
ααα
A
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)()()(0
)()()(0
0
1111
)2其中(
12
132
3122
2
1133122
11312
11
ααααααααα
ααααααααα
αααααα
niRαR
A
nnn
nn
nn
n
ii
n
就有提出,因子列展开,并把每列的公按第 )(1 1xxi
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)()())((2
11312 jjin
inn ααααααααA
).(1
jjin
i αα
223
22
3211312
111
)())((
nn
nn
nn
ααα
ααααααααα
n-1 阶范德蒙德行列式
递推可得
例 4, 5 略
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定理 1.3 行列式任一行(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
.ji,AaAaAa jninjiji 02211
,
1
1
1
111
11
nnn
jnj
ini
n
jnjnjj
aa
aa
aa
aa
AaAa
证 行展开,有按第把行列式 jA
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,
1
1
1
111
11
nnn
ini
ini
n
jninji
aa
aa
aa
aa
AaAa
可得换成把 ),,,1( nkaa ikjk
行第 j
行第 i
,时当 ji ).(,02211 jiAaAaAa jninjiji
同理 ).(,02211 jiAaAaAa njnijiji
相同
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关于代数余子式的重要性质
;当,0
,当,
11 ji
jiAAαAα
n
kkjki
n
kjkik
7.4
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证明从略
三、拉普拉斯 (Laplace) 定理
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例 6 略
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例 7 计算行列式
nnn
n
n
n
nnn
n
bb
bb
cnc
ccαα
αα
A
1
111
1
111
1
111
0
计算从略 由拉普拉斯定理可得
nnn
n
nnn
n
bb
bb
αα
αα
A
1
111
1
111