好文档 - 专业文书写作范文服务资料分享网站

电大秋经济数学基础复习资料-解答

天下 分享 时间: 加入收藏 我要投稿 点赞

1设矩阵

?102?,?21?(2I?AT)B

A????124?,求B???13?????311?????03??解: 因为

2I?AT=

?100?T??102?

2?010????124??001????????311?? =?200??1?13?=?11?3??

?020???021????002????00?1??????241?????2?41??所以 (2I?AT)B=?11?3??21?=?1?5??

?00?1???13??0?3???2?41??????????03????0?11??2设矩阵

A??102?212?,??61?计BAT?C.

??,?1?20??B???010???C???22???002?????42??解:BAT?C=?212???11????61?

?010????0?2?????22??002???20???????42?? =?60????61? =?01?0?2???22??

?0??????20??4????42?????02??3设矩阵A =??13?6?3??,求

A?1

??4?2?1???211???解 因为 (A I )=

??13?6?3100??114107?? ??4?2?1010???001012??211001??????211001????

?114107??1101?4?1

?100?130??100?130?

???001012?? ???12????0?10?271?????0102?7?1??0?1?7?20?13?0010??????0?10?271????001012?????001012??所以 A-

1 =??130??

?2?7?1???012???4设矩阵A =?012??,求逆矩阵A?1

?114??0??2?1??因为(A I ) =

?012100??114010??102?110??1002?11??1002?11???114010???012100?????012100?????0104?21?????0104?21???2?10001??????0?3?80?21????00?23?21?0?23?21????0????001?321?12??A-

1=?2?11??

?4?21??21?12???3??5设矩阵 A =?10?2??,B =?63-1

?1?20????,计算(AB)

?12????41??解 因为AB =?10?2?63???1?20???=??21?

??12?????4?1???41??所以

(AB I ) =??2?4?1?10???2110? ??20?1?1?10????????2??101??0121??012?0121??11? 2?1?? 所以 (AB)1=

-

?1?2?2?1? 2?1??

?2?3???1?. 7解矩阵方程?X??3?2?4?????11解 因为??2?310????34?3401????即 ??2?3??34????111??1111??1043? ???????01??01?3?2??01?3?2?3???1?=?2?

?????2????2???13? 所以,X =?4?4????3???3?2??8解矩阵方程X?1?3?2??1??5???2?1?

0??解:因为??1210??1210??10?52? ??????013?1??3501??0?1?31????1?12???52???即 ? ???35??3?1??1 所以,X =?1?1??12?=?1?1???5?2??0???35???2??0???32?= ??8??10?1???3?10

4???2x3??1?x1?9、设线性方程组 ??x1?x2?3x3?2,求其系数矩阵和增广矩阵的并.

?2x?x?5x?023?1?102?1??102?1?解 因为A???11?32???01?11?

??????2?150????0?112???102?1?

????01?11??3??000?

所以 r(A) = 2,r(A) = 3. 又因为r(A) ? r(A),所以方程组无解. ?2x3?x4?0?x111求下列线性方程组的一般解:???x1?x2?3x3?2x4?0

?2x?x?5x?3x?0234?1解:因为系数矩

2?1??10?102?1??102?1?

????A???11?32???01?11???01?11???????2?15?30?11?10000????????

所以一般解为??x1??2x3?x4 (其中x3,x4是自由未知量)

?x2?x3?x4?2x1?5x2?2x3??3?12.求下列线性方程组的一般解:?x1?2x2?x3?3

??2x?14x?6x?12123?解:因为增广矩阵

?2?52?3??12?13??10?191?

?????A??12?13???0?94?9????01?491????00???214?612???018?818???00?

1?x?x?1所以一般解为 ? (其中x3是自由未知量) ?193??x?4x?123?9??x1?3x2?2x3?0?13、设齐次线性方程组?2x1?5x2?3x3?0问?取何值时方程组有非零解,并求一般解.

?3x?8x??x?023?1解 因为系数矩阵A =?1?32?2??1?3?10?1?

?2?53???01???1???????01?1?????3?8????01??6???00??5??

所以当? = 5时,方程组有非零解. 且一般解为

?x1?x3? (其中x3是自由未知量) x?x3?2?x1?x2?x3?114、当?取何值时,线性方程组??2x1?x2?4x3?? 有解?并求解

??x?5x3?1?1解 因为增广矩阵

11??1111??11?10?5?1?

????A?21?4??0?1?6??2??0162???????62???????1051????01??000?

所以当?=0时,线性方程组有无穷多解,且一般解为:??x1?5x3?1

x??6x?23?2(x3是自由未知量〕

(1)y?x2?2x?log2x?22,求y?答案:y??2x?2xln2?1 xln2a(cx?d)?c(ax?b)ad?bc?(2)y?ax?b,求y? 答案:y??22

(cx?d)(cx?d)cx?d(3)y?1,求

3x?5y? 答案:

y???32(3x?5)3

(4)

y?x?xex,求y? 答案:y??n12x?(ex?xex)=

12x?ex?xex

(5)y?sinx?sinnx,求y?答案:(6)

y??n(sinn?1xcosx?cosnx)

y?ln(x?1?x2),求y?答案:

12y??=

x?1?x11?x?(x?1?x2)?=

1x?1?x2?(1?x1?x2)=

1x?1?x2?1?x2?x1?x2

2 (7)

y?2cot1x?1?3x2?2xx,求y?。

11?1y??2?ln2?(cos)??(x2?x6?2)?x答:

1cos1x111??2?2ln2?sin??xx2x36x5cos1x

(8)sin(x?y)?exy4?yexy?cos(x?y)?4x,求y?答案:y??

xexy?cos(x?y)(1)

?3xdx xe(3x)3x3xe答案:原式=?()dx=?c?x?c

3ee(ln3?1)lne(2)

?(1?x)2dx

x答案:原式=(3)

?(x?12543222?2x?x)dx=2x?x?x?c

353212?12x2?4x?2x?c dx 答案:原式=(x?2)dx?x?22?(4)

1 ?1?2xdx 答案:?1d(1?2x)1??ln1?2x?c ?21?2x231122222(2?x)?c 2?xd(2?x)x2?xdx(5)?答案:原式=?=

32(6)

?sinxxdx 答案:原式=2?sinxdx??2cosx?c

xx(7)xsinxdx 答案:?2xcos?4sin?c

?222(8)ln(x?1)dx=xln(x?1)???x?1dxx=xln(x?1)?(1?1)dx=xln(x?1)?x?ln(x?1)?c

?x?1(1)

1?2?11?xdx

2125922?(x?x)?2?? (x?1)dx原式=??1(1?x)dx??=11222(2)

?221edx 2x1x1x原式=

?1e12(?x)dx2x=?e1x21?e?e12

(3)

e3?e311dxx1?lnx

原式=

?1e3x?2 d(1?lnx) =21?lnx1x1?lnx?20(4)

?xcos2xdx

?111112原式=(xsin2x?cos2x)0=????24442

(5)

?e1xlnxdx

e1原式=

12xlnx21?2?e1e212?xxdx=

24e1?12(e?1) 4(6)

?40(1?xe?x)dx

∵原式=

4??40xe?xdx

?4∵0?44?4xe?xdx?(?xe?x?e?x)0?1故:原式=5?5e=?5e

(四)代数计算题

1.计算

??21??0(1)????53??11??1=?0???3?2?

5???0(2)??0(3)

2??11??0??????3??00??00? ?0??3??0??4????1????2???125=?0?

?12.计算??1???122?33???1?2???12????22434??2?3????6?1????341?25? 0??7??1511?22? 0???14???123???124??245??7197??245??5解 ??122??143???610???7120???610? =????????????1???1?32????23?1????3?27????0?4?7????3?27????33.设矩阵

?2A???1??031?1?1??1?1?,B???1?1???02113?2??1??,求AB。

解 因为AB?AB

电大秋经济数学基础复习资料-解答

1设矩阵?102?,?21?(2I?AT)BA????124?,求B???13?????311?????03??解:因为2I?AT=?100?T??102?2?010????124??001????????311??=?200??1?13?=?11?3???020
推荐度:
点击下载文档文档为doc格式
7exoa81rfz721et5ixox47ty70kcsx0050u
领取福利

微信扫码领取福利

微信扫码分享