发布时间 : 星期二 文章行列式的例题更新完毕开始阅读0bc2574be45c3b3567ec8baa
行列式的例题
一.直接用行列式的性质计算行列式 1.试证明
b?cb1?c1b2?c2c?ac1?a1c2?a2a?babb1b2cc1证明:先用行列式的加法性质拆第一列,c2a1?b1?2a1a2?b2a2再用初等变换化简得
b左?b1b2b?b1b2b?b1b2b?b1b2c?ac1?a1c2?a2c?ac1?a1c2?a2cc1c2cc1c2aa?bcc?ac1?a1c2?a2aa1a2aa1a2cc1c2bb1 b2aa?ba1?b1 a2?b2a?ba1?b1a2?b2a1?b1?c1a2?b2acc2
a1?c1a2cc2a1?c1a2ac2bbcc1c2aa1=右 a2a1?b1a2b2a1 ?2b1a2b2
2.计算n阶行列式
a1?b1Dn?a2?b1an?b1a1?b2a2?b2?an?b2????an?bna1?bna2?bn
解:当n=1时,D1=a1+b1 , 当n=2时,D2=(a1+b1)(a2+b2)-(a1+b2)(a2+b1) =(a1-a2)(b1-b2)
当n≥3时,将第一行乘(-1)加到其余各行后,可得这些行对应成比例,即
a1?b1a2?a1Dn?a3?a1?an?a1a1?b2a2?a1a3?a1?an?a1????a1?bna2?a1a3?a1?0 ?an?a1综上所述
a1?b1,??Dn??(a1?a2)(b1?b2),?0,?n?1n?2 。 n?3 3. n阶行列式D中每一个元素aij分别用数bi-j(b≠0)去乘得到另一个行列式D1 ,试证明D1=D 。 证明: 首先将行列式D的每行分别提出b1,b2…,bn,再由每列分别提出b-1,b-2…,b-n可得
a11bD1?a21b?an1bn?11?12?1a12ba22b?an2b1?1?11?22?2??a1nba2nb?annb?2?21?n2?n
n?nn?2?
a11bb?a21bb?an1bbn?12a12bba22bb?an2bba11b?1?1n21??a1nbba2nbb?annbb??n21?n?n
?n?2??2?2a12ba22b?an2ba1nba2nb?annba12a22?an2?n?n ?(bb?b)12na21b?an1b
?n?1?2?a11?(bb?b)(bb12n?1?2??a1na2n?ann?b?n)a21?an1
?a11?a21?an1a12a22?an2??a1na2n?ann?D
?
15252263552543412532,求 134.已知A?324
(1)A51+2A52+3A53+4A54+5A55;
(2)A31+A32+A33及A34+A35 。 解:由行列式的性质可知
1525354353 (1) A51+2A52+3A53+4A54+5A55=32542?02122231415
(2)
152552635225355254311243312531?0 13533?0 135A31+5A32+5A33+3A34+3A35 =52415252262A31+2A32+A33+A34+A35 =224解出A31+A32+A33=0,A34+A35 =0 。
二.利用行列式的性质化为上(下)三角形行列式计算 1. 计算n阶行列式
x?1xDn?anan?1?1??x?a2?1a1?x
解:(解法1) 依次按第i列的x倍加到第i-1列去(i=n,n-1,?,2),再将最后1行依次换到第一行得
?1?1Dn?x?a1xnn?1???1???an?a1?xxn?an?11x???an?a1?x?1?(?1)n?1?
??1= xn+a1xn-1+…+an .
(解法2) 直接按第n行展开
Dn=an(-1)n+1(-1)n-1+an-1(-1)n+2x(-1)n-2+…
+(-1)n+n(a1+x)xn-1
=an+an-1x+…+a1xn-1+xn
(解法3)递推法,按第一列展开得 Dn=xDn-1+an(-1)n+1(-1)n-1=xDn-1+an
= x(Dn-2+an-1)+an=…= xn-1D1+a2xn-2+…+an = xn-1(a1+x)+a2xn-2+…+an = xn+a1xn-1+a2 xn-2+…+an 。
2.计算n阶行列式
123?n234?1Dn?????。
n?1n1?n?2n12?n?1解:依次将第i行乘(-1)加到第i+1行(i=n-1,n-2,?,1),再将第2,3,?,列全加到第1列。
123?n111?1?nDn?????
111?n?111?n1?1n(n?1)223?n011?1?n?????
011?n?101?n1?111?1?n?n(n?1)???211?n?1
1?n1?1n