行列式5
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1 0 0 " n 求第一行各元素的代数余子式之和 A11 + A12 + " + A1n . 解 第一行各元素的代数余子式之和可以表示成 1 1 1 " 1 1 2 0 " 0 A11 + A12 + " + A1n = 1 0 3 " 0 # # # % # 1 0 0 " n
5
举例
1 1 1 " 1 1 2 0 " 0 =1 0 3 " 0 # # # % # 1 0 0 " n
3
1
3
5
7
= −256 0 − 2
举例
例3
1 1 1 1 2 1 1 3 1 1 4
计算行列式
1 1 4 1 9 1 16
范德蒙行列式
1 1 8 1 1 1 2 3 1 1 1 . = ( − )( − )( − )( − )( − )( − ) = 3 4 6 4 12 1152 2 27 1 1 1 " 1 64
分析 解
每一行/列所有元素的和 = n-1 把第 2, …, n 列都加到第一列,得
7
举例
1 1 1 " 1 1 n −1 1 1 " 1 1 1 0 1 " 1 1 n −1 0 1 " 1 1 1 1 0 " 1 1 n −1 1 0 " 1 1 = ( n − 1) A= " " " " " " " " " " " " 1 1 1 " 0 1 n −1 1 1 " 0 1 1 1 1 " 1 0 n −1 1 1 " 1 0 1 1 1 " 1 1
1 1− ∑ j =2 j 0 0 # 0
n
1 1 1 c1 − c2 − c3 − " − cn 2 3 n
1 1 " 1 2 0 " 0 n 1 0 3 " 0 = n! (1 − ∑ j ) . j =2 # # % # 0 0 " n
=
6
举例
例5
0 1 1 1 0 1 1 1 0 " 1 1 " 1 1 " 1 1 = A= " " " " " " 1 1 1 " 0 1 1 1 1 " 1 0 1 1 3 5
1 1 −3 2 2 −6 0 2 2 −6 = 16 = 16 4 − 4 − 4 = −256 1 − 1 − 1 0 4 −4 −4 −2 −2 −2 1 1 1 0 −2 −2 −2
1 0 1 0 −3 2 = 2048 . 4
9
举例
⎧ x+ y+ z = a+b+c ⎪ 2 2 2 ax + by + cz = a + b + c 方程组 例7 ⎨ ⎪ bcx + acy + abz = 3abc ⎩
在 a , b, c 满足什么条件时有唯一解? 并求这个解。
解
1 1 1 1 0 0 D= a b c = a b−a c−a bc ac ab bc ac − bc ab − bc
2 1 23 6 = −385 . 0
2 3 = −7 10 8
21 13
r3 + r2
2
3
21
−9 −14
−7 10 13 = −7 17 1 1 −1 1 0
2
举例
1 3 计算行列式 例2 5 7 3 5 7 1 5 7 1 3 7 16 3 1 16 5 = 3 16 7 5 16 1
5 7 7 1 1 3 3 5
∴ D ≠ 0 即 a , b, c 互不相等时方程组有唯一解。
a+b+c D1 = a 2 + b 2 + c 2 3abc
= aD ,
1 1 1 1 1 1 1 a b c = a2 b c =a a b c ac ab abc ac ab bc ac ab
同理有 D2 = bD , D3 = cD ,
LOGO
第一章 行列式
习题课
胡倩倩 qianqian_hu@
举例
例1
10 − 13 −3 17
计算行列式
−4 2 1 −3 1 8 −1 −6 −1
−2 −4 −3 −21 2 1 10 0
−7 3 = 0 −5 8 1
−3 −9 −14
− 2 − 3 − 21 13 = − − 7 10 13 0 8 − 9 − 14
x1
2 Dn = x1 # n −1 x1
2 2 = ∏ ( x i − x j ). " xn x2 n ≥ i > j ≥1 # # n −1 n −1 " xn x2
x2
"
xn
4
举例
1 2 3 " n 1 2 0 " 0
例4 设 n 阶行列式 Dn = 1 0 3 " 0 , # # # % #
n≥ i > j ≥1
1
n " n n −1
范德蒙行列式 的转置
= n!
∏ (x
i
− xj)
= n ! ( 2 − 1 )( 3 − 1 ) " ( n − 1 )
⋅ ( 3 − 2 )( 4 − 2 ) " ( n − 2 ) "[n − ( n − 1)]
= n ! ( n − 1 ) ! ( n − 2 ) !" 2 ! 1 ! .
1 1 = (b − a )(c − a ) = (b − a )(c − a ) (c − b ) , −c −b
10
举例
⎧ x+ y+ z = a+b+c ⎪ 2 2 2 ax + by + cz = a + b + c D = (b − a )(c − a ) (c − b ) , ⎨ ⎪ bcx + acy + abz = 3abc ⎩
11
D3 D2 D1 =c. =b, z= 所以解为 x = =a, y= D D D
−1 0 0 −1 = ( n − 1) " " " 0 0 0 0 0 0 0 0
0 " 0 0 " 0 n −1 = ( − 1 ) ( n − 1 ) . " " " " −1 0 " 0 −1
8
举例
例6
1 2
1
"
1
1 1
1 2
"
1
22 " 2n
" 2 n −1
Dn = 3 3 2 " 3 n = n ! 1 3 " 3 n−1 " " " " " " " " n n2 " nn