線代第三章習(xí)題解答

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3-1-6.用定義計算行列式

a10a20解：設(shè)D4?aij4?4（1）D4?0c10c2b10b200d10d2?ai,bi,ci,di?0,i?1,2?

則D4中第1行的非0元為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可組成四個4元排列1234，1432，3214，3412，

故D4中相應(yīng)的非0項有4項，分別為a1c1b2d2，?a1d1b2c2,?b1c1a2d2，b1d1a2c2其代數(shù)和即為D4的值，整理后得D4??a1b2?a2b1??c1d2?c2d1?

010...0002...0(2)Dn?

000...0n00...0解：由行列式的定義Dn?僅當(dāng)j1,j2,j1j2?(?1)?(j1j2jn)jna1j1a2j2anjn

,jn分別取2，3,…,n-1,n,1時，對應(yīng)項不為零，其余各項都為零

jn)Dn?(?1)?(j1j2?(?1)

(n?1)a1j1a2j2anjn?(?1)?(23(n?1)n1)a12a23an?1nan11?2n?(?1)?n!

習(xí)題3.2

cos2?3.2-2.證明(1)cos2?cos2?cos2?cos2?sin2?sin2??0sin2?cos2?證明:

cos2??sin2?左?cos??sin?22cos2?cos2?sin2?cos2?cos2?cos2?sin2?sin2??0sin2?c1?c3cos2?cos2??sin2?cos2?sin2?cos2?cos2?sin2?a2abb2(2)2aa?b2b?(a?b)3

111a2?abab?b2b2證明: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證明:按最終一行展開，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．計算以下行列式（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（最終一行（n+1）行依次與第n,n-1,…,2,1行交換，經(jīng)過n次交換；再將新的行列式的最終一行（即原來的n行）依次換到其次行，經(jīng)過n-1次交換；。。。。最終一共經(jīng)過

n?(n?1)?...?2?1?n(n?1)次換行。使原行列式化為范德蒙德行列式）2n(n?1)2?