第三章 行列式习题3.1
3-1-6.用定义计算行列式
a10a20解：设D4?aij4?4（1）D4?0c10c2b10b200d10d2?ai,bi,ci,di?0,i?1,2?
则D4中第1行的非0元为a11?a1,a13?b1，故j1?1,3
同法可求：j2?2,4;j3?1,3;j4?2,4
∵j1,j2,j3,j4可组成四个4元排列 1 2 3 4，1 4 3 2，3 2 1 4，3 4 1 2，
故D4中相应的非0项有4项，分别为a1c1b2d2，?a1d1b2c2,?b1c1a2d2，b1d1a2c2 其代数和即为D4的值，整理后得 D4??a1b2?a2b1??c1d2?c2d1?
010...0002...0(2)Dn?
000...0n00...0解：由行列式的定义Dn?仅当j1,j2,j1j2?(?1)?(j1j2jn)jna1j1a2j2anjn
,jn分别取2，3,…,n-1,n,1 时，对应项不为零，其余各项都为零
jn)Dn?(?1)?(j1j2?(?1)
(n?1)a1j1a2j2anjn?(?1)?(23(n?1)n1)a12a23an?1nan11?2n?(?1)?n!
习题3.2
cos2?3.2-2.证明(1)cos2?cos2?cos2?cos2?sin2?sin2??0 sin2?cos2?证明:
cos2??sin2? 左?cos??sin?22cos2?cos2?sin2?cos2?cos2?cos2?sin2?sin2??0 sin2?c1?c3cos2?cos2??sin2?cos2?sin2?cos2?cos2?sin2?a2abb2(2) 2aa?b2b?(a?b)3
111a2?abab?b2b2证明:左c1?c2a?ba?b2b?a(a?b)b(a?b)?(a?b)2ab
c2?c3001a?ba?b11?(a?b)3?右
x?10000x?100(3)
?xn?a1xn?1?a?22xn??an?1x?an
000x?1anan?1an?2a2a1?x证明: 按最后一行展开，得
?1000x0000x?1000?1000左?an?1n(?1)0x?an?1(?1)n?200?10000?1000x?1000x?1x0000x?1000x0000x?10?a?3n?2(?1)n??a2(?1)2n?1000?10000x000x?10000000?1
x?100x?1 ?(x?a1)(?1)2n0000
000000x?10x?an(?1)2n?an?1x(?1)2n?an?2x2(?1)2n??an?an?1x?an?2x2?3=2-3．计算下列行列式 （1）
?a2xn?2(?1)2n?(x?a1)xn?1(?1)2n
?a2xn?2?a1xn?1?xn?右
xaaxaaaax?(x?(n?1)a)11axaa1ax?(x?(n?1)a)110x?a0010x?a
?(x?a)n?1[x?(n?1)a]
(2)
anDn?1?a1?a?1?n(a?n)n1?(?1)n(n?1)21a?11a?nan?1(a?1)n?1a?11?a?n?n?1aan?1ana?n1?a?1?n?a?1?n?1?a?n?n?a?n?n?1（最后一行（n+1）行依次与第n,n-1,…,2,1行交换，经过n次交换；再将新的行列式的最后一行（即原来的n行）依次换到第二行，经过n-1次交换；。。。。最后一共经过
n?(n?1)?...?2?1?n(n?1)次换行。使原行列式化为范德蒙德行列式） 2n(n?1)2?(?1)(3)
n(n?1)20?i?j?n???a?j???a?i???(?1)0?i?j?n?(i?j)?0?i?j?n?(j?i)