段正敏主编《线性代数》习题解答(4)

41241202（2）D?105200?1?1?7?9

r2?2r3r1?2r3r1?4r2r3?10r20?72?412020?152?200?1?1?7按第二列展开按第一列展开??1?2?1?72?4?152?20 ?1?1?70?18??170?34?1?1?7???1?3?2?9?18推论4?17?340

31（3）

111311211311113c1?ci,i?2,3,466660100013116013011311113ri?r,i?2,3,416000010001200102010?6?23?48 0261130130234305164009r1?2r2r3?2r2r4?r2r5?3r21（4）D?213

按第一列展开再按第二列展开?31?5204?15?8r4?3r321?4?36?3r1?6r30?7?1302041?15?8 05?14200?36?3?7?135?1420?36?3c2+2c1c3?c1?7?15105?3?40150按第三行展开??3????1?3?1???15?15?4?10?

?555000023000056（5）D?1008?OB4?4A2?2C4?2???1?2?4100800230000045000006700?232300? 5604500067?230100?1+44+4??2?6?3?5???8???1??045?7???1??230??

??006045????3???8?2?4?6?7?1?3?5??837B按第四列展开1?abcda1?bcd（6）方法一：D?ab1?cdabc1?d

1?abcr4?r3?110r3?r20?11r2?r100?1d0 0115

按第一列展开1002?1bcd

?1?a??1010???1????1??11?110

0?11bcddd1?d

?1?a?b?c?d1?aa方法二：aabcddd1?dc1?ci,i?2,3,41?a?b?c?d1?bcb1?cbc1?a?b?c?d1?bc1?a?b?c?db1?c1?a?b?c?dbc1?a?b?c?dbcdri?r1,i?2,3,4010000100001ab（7）bd上三角行列式1?a?b?c?d

ac?cdcfaede?ef每一行提一个公因子性质3bcebfadfb?ce

bc?e1r2?r1每一列提一个公因子性质311111adfbce1?11abcdef0?20 r3?r111?100?2?4abcdef1a2（8）方法一：考虑新的行列式aa3a41bb2b3b41cc2c3c41dd2d3d41x4?5x2，则A45???1?x3x41aa2a41bb2b41cc2c41d，2dd4即为x3 的系数，因为将D按最后一列展开时，A45即为x3 的系数所在项，而由D为范德蒙行列式知：

D??b?a??c?a??d?a??x?a??c?b??d?b??x?b??d?c??x?c??x?d???b?a??c?a??d?a??c?b??d?b??d?c????x?a??x?b??x?c??x?d??? ?A45??b?a??c?a??d?a??c?b??d?b??d?c????1???a?b?c?d?因此有：

1aa2a4

1bb2b41cc2c41d??b?a??c?a??d?a??c?b??d?b??d?c??a?b?c?d? 2dd416

1a方法二：2aa41bb2b41cc2c41dd2d4r4?a2r3r3?ar2r2?ar11001b?ab?b?a?1c?ac?c?a?1d?a

d?d?a?0b2?b2?a2?c2?c2?a2?d2?d2?a2?c?ad?ad?d?a?1c1c?b1d1d?b11?c?b?a?d?d?b?a?

按第一列展开b?ab?b?a?c?c?a?1b2?b2?a2?c2?c2?a2?d2?d2?a2?性质3?b?a??c?a??d?a?b1b2?b?a?c2?c?a?d2?d?a?r3?b?b?a?r2r2?br1?b?a??c?a??d?a?00c?c?b??c?b?a?d?d?b??d?b?a?按第一列展开性质3?b?a??c?a??d?a??c?b??d?b?c??b?a??c?a??d?a??c?b??d?b??d?c??a?b?c?d?注：此方法的因式分解有点难！ 4．计算下列n阶行列式

xaaaxa（1）Dn?aaxaaaaaa； x（2）Dn??(i?j)， （即aij?i?j）；

?i（3）Dn?(aij),其中aij???2a1?（4）

i?j； i?jb1?ancnbndn?d1，其中未写出元素为零；

?c117

a?b1（5）

aba?bab1a?b???ab1a?b，其中未写出元素为零.

xaaaxa解：（1）Dn?aaxaaaaaaxc1?ci,2?i?n性质31aa1xa??x??n?1?a??1ax1aaaaa x1ri?r1,2?i?naa0x?a0a00x?a

0x?a?0?x??n?1?a??000

???x??n?1?a???x?a?01（2）方法一：Dn?2101n?1210n?1n?2n?30n?1000按最后一列展开0ri?ri?1,i?n:212n?1?1?1 ?11?1?111?1111n?1n?2n?3012

ci?c1,i?2:n100120122?n?1???1?1?n2n?2

01方法二：Dn?2101210n?1n?2n?302n?1?1?1?118

c1?cn性质311?n?1?1101210n?1n?2n?3 02n?1?1?2?2n?1n?2n?31ri?ri?1,i?n:21n?2n?31ri?r2,i?3110?1?1?n?1?01?10110?1?1?n?1?00?2000

??n?1???1???2?n?2??n?1???1?n?12n?2

122222（3）Dn?223222222nri?r1,i?2:n122100101100200n?2

按第二行展开

??1?22010020n?2b1???1??2??n?2?!???2??n?2?!

a1ancnc10a2an ?b1a2ana1c20b2a2按最后一行展开b2bndnd20（4）D2n?bndnd1按第一行展开cn

d1b2ancnc2bndnd2bndnd20cnc2c1a2ancnc2bndna1d1

b2??a1d1?b1c1?D2n?2

?b1c1d2???aidi?bici?

i?1n?D2n??a1d1?b1c1?D2n?2??a1d1?b1c1??a2d2?b2c2?D2n?4?19

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