A.
???1??1????1x C. D.dx B.?dxedxsinxdx 2??101xx (三)解答题
1.计算下列不定积分
3x()3x3x3xe?c?x?c (1)?xdx原式=?()dx =
3eee(ln3?1)lne(2)
?(1?x)2x12dx答案:原式=?(x?12?2x?x)dx
325432 =2x?x2?x2?c
351x2?4dx答案:原式=?(x?2)dx?x2?2x?c (3)?x?22(4)
1d(1?2x)11???ln1?2x?c 答案:原式=dx?1?2x2?1?2x223112222(5)?x2?xdx答案:原式=?2?xd(2?x) =(2?x)?c
32(6)
?sinxxdx答案:原式=2?sinxdx??2cosx?c
(7)
xxsin?2dx
答案:∵(+) x sinx 2x (-) 1 ?2cos 2 (+) 0 ?4sinx 2∴原式=?2xcosxx?4sin?c 22(8)
?ln(x?1)dx
1 x x?1x?x?1dx
答案:∵ (+) ln(x?1) 1
(-) ?∴ 原式=xln(x?1)? =xln(x?1)??(1?1)dx x?1 =xln(x?1)?2.计算下列定积分 (1)
x?ln(x?1)?c
?2?11?xdx
2125922?(x?x)?2?? (1?x)dx?(x?1)dx=1??1?12221答案:原式=
1
x
(2)
?
2
1
edx 2x
答案:原式=
?2111e122(?x)d=?ex1?e?e22xx1x
(3)
?e31x1?lnx1dx
3答案:原式=
?e31exd(1?lnx)=21?lnx?2
1x1?lnx?(4)
?20xcos2xdx
答案:∵ (+)x cos2x (-)1 1sin2x 2 (+)0 ?1cos2x 4?112∴ 原式=(xsin2x?cos2x)0
24 =?e111??? 442(5)
?1xlnxdx
答案:∵ (+) lnx x
x21 (-)
2x121eexlnx?xdx ∴ 原式=1?122e212?x =
24(6)
e112?(e?1) 4?(1?xe04?x)dx
答案:∵原式=4??40xe?xdx
?x又∵ (+)x e
(-)1 -e (+)0 e∴
?x?x?404xe?xdx?(?xe?x?e?x)0 =?5e故:原式=5?5e 作业三 (一)填空题
?4?4?1
?104?5???,则的元素a?__________________.答案:3
21.设矩阵A?3?23A23????216?1??2.设A,B均为3阶矩阵,且
A?B??3,则?2ABT=________. 答案:?72
3. 设A,B均为n阶矩阵,则等式(A?答案:
B)2?A2?2AB?B2成立的充分必要条件是 .
AB?BA
?B)可逆,则矩阵A?BX?X的解X?______________.答案:
4. 设A,B均为n阶矩阵,(I(I?B)?1A
?100???,则A?105. 设矩阵A?02????00?3??(二)单项选择题
1. 以下结论或等式正确的是( C ). A.若
??1??__________.答案:A??0??0??0120?0??0? ?1??3??A,B均为零矩阵,则有A?B B.若AB?AC,且A?O,则B?C
O,B?O,则AB?O
C.对角矩阵是对称矩阵 D.若A?2. 设
A为3?4矩阵,B为5?2矩阵,且乘积矩阵ACBT有意义,则CT为( A )矩阵.
D.5?3
A.2?4 B.4?2 C.3?5
3. 设A,B均为n阶可逆矩阵,则下列等式成立的是( C ). `
A.(A?B)?1?A?1?B?1, B.(A?B)?1?A?1?B?1
C.
AB?BA D.AB?BA
4. 下列矩阵可逆的是( A ).
?123???10?1? A.??023? B.?01?
?003???1?????123?? C.??11?00 D.?11? ?????22???25. 矩阵A??22??333?的秩是( B ). ?? ?444??A.0 B.1 C.2 D.3
三、解答题 1.计算 (1)???21??01?=?1?2??53????10????35? ?(2)??02??0?3????11??00??00?????00?
????3?4??0?(3)?125??=?0???1?
?2???2.计算?123???124??245???122??143???610?
?1?32??????????23?1????3?27???解
?123???124??245??719??122??143???61?32?????0?7?????712?1????23?1????3?2???0?4? =?5152??1110?????3?2?14???3.设矩阵A??23?1??111???123?,B?112?,求AB。
?1?????0?1???011??解 因为
AB?AB
7??20???6?7??????345?10??27???
2A?131?111023122?(?1)2?3(?1)1?10?1123011所以
0?10213122?2
12B?112?0-1-1?0
AB?AB?2?0?0
?124???4.设矩阵A?2?1,确定?的值,使r(A)最小。 ????110??24?24??124?②?①?(?2)?1?1??③?①?(?1)??(②,?? ③)??0?1?4解:A?2?1?????0??4?7?????????????110???0?1?4???0??4?7??4??12? ?②?(??4)?③??????0?1?4????009?4???
9时,秩r(A)最小为2。 4?2?5321??5?8543??的秩。 5.求矩阵A???1?7420????4?1123? 所以当???2?532?5?854答案:解:A???1?742??4?1121??1?742?5?8543?(①,③)????????2?5320???3??4?1120?②?①?(?5)③?①?(?2)3??①?(?4)??④???? 1??3?420?420??1?7?1?7?027?15?63??09?③②?(?3)?5?21(②,③)??②?(?3)?????④????????
?09?027?15?63??5?21?????027?15?63027?15?63????2?1?74?09?5?2??0000?00?00所以秩r(A)=2。 6.求下列矩阵的逆矩阵:
0?1?? 0??0??1?32???
01(1)A??3???1?1??1?
电大经济数学基础形成性考核册答案[1]教学内容
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