線性代數(shù)同濟大學版 課后習題答案詳解

1、 111 cab解 課后習題答案詳解線性代數(shù)同濟大學版 cba222第一章行列式 222222 ?caba?abcb?bcac )?)(c?a(a?b)(b?c? ? 1?利用對角線法則計算下列三階行列式 120 yx?xy 1?14x?yyx (1) ? ?(4) 3?81yxxy?120 yx?xy 1?14解 xx?yy38?1 解 y?xyx8 ?0?(1)?(?11?1)?2?(4)?33331) (?0?13?2(?1)8?1(?4) ?x(x?y?)(x?y)yx?y)?y?x(x?y)?yx(x?y33233 ?816?44?24 ?y?y?3xxy?x(?3xyx?y)33

2、)?y?2(?xcab abc 求下列各排列的逆序數(shù)?2?按自然數(shù)從小到大為標準次序? ?(2) bca ?(1)1 2 3 40 解逆序數(shù)為 cab ?(2)4 1 3 2abc解 ? 32? 41? 43? 42 解 逆序數(shù)為4? bac (3)3 4 2 1?ccc ?bbb?cbabac?aaaacb ? 4 2? 4 1, 2 1? 解 逆序數(shù)為5? 3 2 3 1 333 ?c?b?abc3?a ?(4)2 4 1 3 4 3? 2 1 4 1? 解 逆序數(shù)為3 111 ? (2n)? (2(5)1 3 ?n?1) 2 4 cab ? (3)n?1(ncba222 解 ? 逆序數(shù)為

3、 21 64 / 解) 3 2 (1個 ) ? 5 4(2個5 224?1014142cc?3222102201) 個?7 2? 7 4 7 6(3?0521014103?2c?c7 ?7110000134) 1個?2)(n?1)(2 (2?n?1)6? (2n?n?(2n1)2? (2n?1)4101?4? )?1122?(?34? ?1) (2n) (2n?2) 2n?(6)1 3 (2 14?103 1) ?解 逆序數(shù)為n(n) 3 2(1個c?c10?14 109932) 5 4 (25 2?個0?0?20212? ?14103cc?1417171 ? 312) n?1)2(2n? (

4、2n1)4? (2?1)6?2)(n1個n (2n?1)(2?1421 ) 4 2(1個2?131) 個?6 2 6 4(2 ?(2)21322650 ?) (2n個n?)(2? (2nn2)(?1)6)4)2? (2n? (2n 214004211124 rr?c?c 的項a?3寫出四階行列式中含有因子a?2311242421?321213?123?2? 解 aa的項的一般形式為含因子 解 0231221301232311042125062560t aa1)aa?(s11233r4 4224即和? 這種排列共有兩個?42rs其中是和構(gòu)成的排列 ?0241 r?r aa的項分別是所以含因子23

5、11142?3120?1t aaaaa?(1)a?aaaaa1)(?a? ?02134423443211234423113211322t aaaaaaa1)(?aa?a?a?(1)a0000423434231142231123421134 ?4?計算下列各行列式 aeabac?4412 de?bdcd2021 ?(3)ef?cfbf (1)?05102701164 / 2 c?c babae?abacaec?b22 a?abaab?222212e?adfbcd?cdebd2?b2baaa2?bb?2a2a 解 ecbf?efbcf?cc?010111131?11 ab?ab?a 222ab?a

6、)?a?(b?a)(b)(?11?3abcdef?411?adfbce?13 ?b)?(a21 ?a?ab2b?21?11zxyaz?bxax?byay?bz 0a10 xzb)yby?(a?axay?bzaz?bx?33011?b; (2) ?(4)11c?0yx?azbz?bybxaxayz?d10?0 證明a1a00?ab010 arr? bx?bzazax?byay?211b?0110?1b?byaybx?bz?azax? 解 10c1c1?1?0bzay?bx?ax?byazd?10d000?1 bx?bzazy?xaybzaz?bxaydcc?0aab1? ada?1ab23bya

7、z?axbxazax?bxby?bza?y1?c)?1(?)(1112?cd1c1? 0?10d0?1bz?aybyax?bzbyzayaxx ad1?abbx?zbzyzazxay? )(?)(?112?3 ?abcd?ab?ad1?cdcd?11?byax?zyaz?a?bx?bxx22 bzaxz?yy?byxay: 5 ?證明 baba22xzyxzy bab2?a23yz?xbzx?ay33; b?a(1)?) 111zyxxzy 證明64 / 3 1111 zyyzxxdcabx?ayyzzx?b33 (4) dbac2222yxzyzxdbac4444 zyx); dc?d)(c

8、?d)(a?b)(?(a?b)(a?c)(a?db?c)(b?xz)yb?(a?33 證明 ?yzx 1111dcab dbca2222 )?2)(a3(a(a?1)a?2222dcba4444)3(b?b1b(b?)(?222220?; (3) 1111)?3c1(cc?)(?2)(c2222a?ac?ad0b)?)(d2)3d?(dd?12222?)b?a)c(dca(?d0b(a) )?a)d(daa0b(b?)c(c?222222222 證明 )(2)a?3?)?(aa1(a2222 111)3?(?)(bb?1(b2)b2222dbc)(b?a)(c?ad?a)(? ) c?c?(c

9、cc?c得)?c)(ca?a)d(b(bd?a222123324)()?)?c1(c2c3?(c2222)d?)?()?(d1d2(d32222 111ba0?d)c?b)(?(b?ac?a)(d? 5a?a31a2a?222)db?)(?c?b?a)d(dba)(c0c(?b5bb1b?2232?b?2?) c?c?cc(得52c1?2ccc3?2?2 23435222d?d1d3?d?211 )b?a)(da)(c?b)(d?(ba)(c)aa)d(d?b?(cc?b? 122a2a?2 21b?b2220? ?)(?)(?)(?)(?)(?)(?=( abacadbcbdcdabcd?)

10、 ?2c12?c222?d21d22 64 / 4 a? ?a 0? ?0?x?10 a ?a? ? n1110?0? x1?0? nn1na ? a?)(?1?D? ? ? ? ? ? ?1n?nn1n? ? ? ? ? ? ? ? ? ? ? 1?nn ? x?(5)a?ax?ax?1? ? ? ? ? ? nn?11a? ?a?1x0? ? 00n111a? ?a ?aa ?aaax? ?n2121n?22nn?1 a? ?a? n111a? ? ?a ? 用數(shù)學歸納法證明證明n221? ? ? (?(?1)?1)2n?1?naa? ? 1x? ?aDx?x?annn12?2時當n ?命

11、題成立? ? ? ? ? ?a?ax21212a? ? a?n313 n 假設對于(?1)階行列式命題成立?即)?1n(n21n?n? ?a?x?axx?D?a1?1n?2nn?1D1)?(?1)D?(?)1n?1?2? ? ?(n?2)?( 2 D按第一列展開有? 則 ?n 1?0 ? 0 0? ? 同理可證 x0 0 ?1? ? aa? ? ?)D?xD?a(?11n? )n(n?1n(n?1)?1n(n ? ? ? ? ? ? ? ? ? ? ? ? ? 1n11nn?1nD)(?1(?1)D? ? ? )D?(?1? ? ? ?T 2221 ?x1 1? ? ?2a? ? a?nn1n

12、1nn? ?x?a?a?xDax?axnnn11?1?n)?n1)n(n?1)n(n(n?1 n階行列式命題成立?對于 因此?D?(?D?D?(?1)?D?(1)1)?1)D)1n?n( 222 ? 23?、或依副對角線翻轉(zhuǎn)把det(a), D上下翻轉(zhuǎn)、或逆時針旋轉(zhuǎn)90?6?設n階行列式D? ij 依次得 ?計算下列各行列式(D為k階行列式)7?k 1aa? ? ? aa? ? ?aa?a? nnnnnn1nnn11? ?D?D? ? ? ? ? ? ?D? ?D? ? ? ? ? ? ? ? 0?, 其中對角線上元素都是a(1)?未寫出的元素都是 ?n321a?a ? a ?a ?a ?a

13、a1111n1n11111n 解 )1(n?nDDD)?1( 2證明 ?D?D321 )a?因為證明 Ddet(?所以ij64 / 5 a? ? ?xaa1a00? ? ?0 0?a? xx ?a?0?00?00 a? ?D0 ?aa ?x0x?0?000a ? ?n?D? ? ? ? ? ? ? ? n按第) (行展開? ? ? ? ? ? ? ? nax?000ax0 0?0?0?a 得再將各列都加到第一列上?a ?0010 ?a?a? ? 1x?(n?)aa ? 0000?1 0 ?0? ?0x?a?00? a00 ?D000x?a? ?1?n (x?a)?x?(n?1)a)?(?11n

14、?0 0 ?0a?0n? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?ax00?000? 0 0?a0?)?)(1n?1(n? )? ?(a?na(a?1)nnna ?)? ?(a?n?a(a1)? 1n?1?1n?n 1(?) a?n2? ?D? ? ? ? ? ? ? a; (3)1n?)n?1n(?)(?1na? aa ?1a 1? ? 11?a? 1(?) 1?)?(n1?nn?2?nn2n?2 ?a?aa?1)(a? 有 解 根據(jù)第6題結(jié)果?a1? 1?1? )n2?)(?2n(na ?aa?1? )n?1n(? ? ? ? ? ? ? ? )?1(?D ?

15、ax? a 2 1?na x? a? )?n(1aa(?)a1?1n?1n?n?D; (2)? ? ? ? ? ? ?n)? ?(a?n)(aa?1nnnx? ? aa 此行列式為范德蒙德行列式? 1)?(將第一行乘 解 得?分別加到其余各行)n?1(n?i)a(1j)?(D1a?1)(? 2 1?n1?i?n1j?64 / 6 0ba )1n(n?1nn?1?)?j?(?1)?(i 2? ? ? ?1j?i1n?ba11)1n(n?1n?(n?1)? ? ?dc?a? )(?i?(?1)?j)?1( 2n 112 ? ? ?1?n?1ij?dc0?1?1nn?)?j?(id ?00n1j?n

16、1i?0 ba 1?1n?n? ? ?ba ?nn? ? ? ba?11b?(?1)1n?2badc ?11?Dn11?; (4) ? ? dcn2?11? ? ? dc?1nn?1?0dccnnn 再按最后一行展開得遞推公式 解 Dbc)?即cD? D?(ad?DadDbba 22n?2nnn?22nnnn?2n2nnn2nnn?n ? ? ?DbcD?)(ad? 于是 ?2iniii2ba?D112i?) 按第(行展開1dcn211 ba? ? ? cb?a?d?D11?而 ?dc11211dc11nnn?)?bcD?(ad 所以 ?i2niii1i?|; i?j?det(5) D?a)其

17、中a|ijij |?|? 解 aij?ij64 / 7 1n? ? 0123?1 ? ?11?a 12?1102 n? ?1? ?11?a? ?D?210?1n? 3 2 ?)D?det(an? ? ? ? ? ? ? ? 4?23?10n ijna? 1?1?1 ? ? ? ? ? ? ? ? ? ? n0n?4?n1n?3n? ?21?0000? ? a 1a1? ? ?00a0? ? 11?111? cc?22?11111 ? 21001? ? ?a0arr?33211? 1?111 ? ? ? ? ? ? ? ? ? ? ? ? ?cc 11?11? ?1 32r?r1a0?a ? 0

18、0 ? ? ? ? ? ? ? ? ? ? 321?1n?n ? ? a1? ?0?a000? ?0? ? ?41nn?3n2?nnn a0?0?100? ?1 ? 0?0 001?1000 2? ?1?a010? ? ?0?11?cc?212? 0?2?21? ?0?a00 ? ?0?11?1? a?aa? ? ?0?21?2 ? 2?3 cc?n21? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?13 ? ? ?a 10?0? ?0?11?12?3?42?5n?n ?n?n12n 1?na?11? ?0? 000?1?2?1?nn 1)2n

19、?(1)(?n 1? 1?a1? a0?0?100? 1?111? ? a11?a? ? ?000011?D22 a?其中, (6)a0a?n12? ? ? ? ? ? ?na0? ? ?01001?3a? 11? ?1aa? ? ?a? ? ? ? ? ? ? ? ?n n21a00?0 ?01? 1? 1?n 解 n?a?10 000?01?i1?i64 / 8 nx?6?15x1?)?a?a)(1?(a21 ?x?6x?5x?0? n12a?3121i?i0?6x5x?x? (2)?423 0?6xx?5x? ?用克萊姆法則解下列方程組8534?1?5xx?545?x?x?xx?4123

20、 ?2?x?2x?x4x?1324? ?(1) 因為解 25?x?2x3x?x4312?06005 0x11?x3?x?2x?431205016665?D?05106 因為 解 ?650011111 5000141?21142?D? ?51?32? 00510100601112300100065601145?D?1507?D?0056001560 11155111 ?21651506000141?1?212?42284?D?142?D?5100110150 ?5?5?32?2?121?2111020231110005665011 0655001100 51111115395?D?703D?01

21、5006100022?1?12?214 ?43142?D?426?D?6061000005 ?23?152?2?3?2?43501100050101113032110650 DDDD11?x?2?x?x?x?3?341201605 所以 ? 4132DDDD212?D05016 ?50105010001 所以64 / 9 212703?39511451507?x?x?x?x?x? ? 矩陣及其運算第二章 44321665665665665665 ?0x?x?x?321 ?已知線性變換?1?0x?xx? 有非零解？?問?9齊次線性方程組取何值時?312y2y?x?2y?0x?2x?x?32113

22、21yyx?5?3y? ?3221 解 系數(shù)行列式為?y2?y3x?3y?3231 ?11 的線性變換?y?y?yx求從變量?x?x到變量332121?D?11 ? 解由已知?112yx?122?11? 得?令 D?0 ?yx53?1 ?22? 0或?1?323xy?32? ?1時該齊次線性方程組有非零解當?0或?于是? 1?yyx ?122?94?7?111?04x?1(?)x2xx1?53y6?y37? 故 ?321?22?2?0?2x(3)xx?3324?32xyy? 有非零解？?齊次線性方程組10?取何值時問?312323?0x)1?xx?(?321x?4x?9y?7x?3121 解

23、系數(shù)行列式為 x3x?7y?6x? ?3122?44?1?21?3? x4x?2?yx3?3132?11?32?D?21 ?01?1111 ?已知兩個線性變換2yy?x?2z?y?3z?3?) 3?2(1?4(13)?(1?)?(?)(?311211z?2z?3?2x?y2y?yy? ?23? ?3?2(1)?(1?)?3311222?z?z?35?x4yy?yy? D令 ?0?得 3122333? 2?0?或?3? ?的線性變換?x?到z?z求從zxx321123? 或?0?當 ?于是2?該齊次線性方程組有非零解?時3? 解由已知 64 / 10 zy7143x?0201?31201?111

24、?221?3z02?x22y?2?12?33 (1)? ?222?157030415415?1xzy?3323573141?2?1?4?73?z?31?6?1?12?)?72?3(?16?122?3解z?12?49 ? ?2?4917501?7?7?2?05?16?1?10?z?33?zx?z?36z?3112223)(1z9zz?12?4?x (2)?所以有 ?32121?zz10?z?16?x?31323?321111?(223)1? 解 ?(10)2?3?1)(1?3?2?A4?11?1?1B?2?T 及AB求? 3?2AA?3?設B?1?11?1150?2?111121311?)2?1

25、1(? ?(3)1?11142?1?23A3AB?2?11?解 3?11?011?1511?2)2?1?2(?42?2?22?211851013?)(122(1?1?)?1?2?1?1? 解 ?2017?22?651?3?01?1? ?63?32(3?1)?3?2141?129?290?113?853210111?2401?021?6B1A2?405?11?1?T? ? (4)?133?1141?015021?119?2?04? ?4?計算下列乘積11 64 / 31122?BA? ? 因為8?762?12140052? 解 ?65?20?3111?134?1482222?024?B)(A2?

26、 ?29214552?aaax? 但 1311121?xax(xx)aa ?(5)?22123232211610816380?xaaa?A?2AB?B22? ?2781215341143331323? 解 )?B所以(A222 ?2AB?AB?aaax? 嗎22B?)?B)(A?B?A(3)(A1311121?xa)xaxax( 22? ?B?)(A?B)?A( 解 A?B2121222323xaaa?20223332313?A?BA?B? ? 因為? 1520x?1?x x?axax?xxa?xa?(axaa?xaxa)?6020223331113 12213133 112122232223

27、?2?)?BA?)(A?B(x? ?9120503?x?x?ax?ax?axxa?xxa?222xa222 ?883102?323311112222233313112?AB?22? 而 ?7441113 ?0211?AB?A)?B)(A?B(故A22 ?B 設?5 問? ?2113? ? 6?舉反列說明下列命題是錯誤的 ? 嗎BAAB(1)?2 A?0?(1)若A?0 則 解?BA?AB10?A?2 ?0?但則? A?0? 取解 A2134?00?BA?AB? ?所以因為 ?ABBA?8643?2 EA0A?A(2)若?A則?或? 嗎222B?2?A?)B?A(2)(AB222 A( 解 2?

28、A?)B?B?AB?12 64 / ?113323?A?2 且A?E? 則A?A?但A?取 解 0?0030AA?A?2332? ?003? Y?則X?若(3)AX?AY?且A?0?64234? 取解 ?4AA?0A3443110111? ?YX?A? ?0041?10010?105354 Y?A?且?0?但X?則AXAY?50?A?A?A445501? ?A?00k235 ?設7?A?AA?求?1? ?011010?A?2?)1k?k(解 ?1112?k2k?k1k? 2?Ak? ?001110?k01k?k?A?A?A23? ?13112?00k? ? 用數(shù)學歸納法證明? 01? ?時?顯

29、然成立當 k?2 ?Ak? ?1k? 時，?1k 假設k時成立，則 )?1(kk?01?k2?k1kk?10 ?1?A0?2k ?設?8A求?1k0A?AA?01?k?k1kk?00 ?0000k? 解首先觀察?0011122?2?010?A0122 ?0000002?13 64 / k)?1(kAA2?5?)(k?11kk?1?k?1?A*2111? ?1?2 AA2?2212?)(k?10?1?1kk ?125?*?001k?A?A1?故? ?12? |A?sin?cos? 由數(shù)學歸納法原理知? (2)? ?cossin?)k?1k(?k2k?1kk? ?2sincos?A? 因為解存在?

30、1?A?1?0?故? |A|?k?0A1?kkk?cossin ?00k?AA?sincos?A*2111? ?cos?sinAA?2212T 也是對稱矩陣為對稱矩陣，證明ABAB?為設9?A?Bn階矩陣，且?1sincos?A證明因為T 所以?A*AA?1?所以 ?cossin? |A|?TTTTTTTT B?AB?BAB?)(BABB?B(A)B從而T ?AB是對稱矩陣112? 是對稱矩陣的充分必要條件是階對稱矩陣，證明都是BnABAB?BA?設10?A?24?3 (3)? ?A證明充分性因為?TT A所以 BAAB B?B?且?145?TTTT )BA?AB?AB?(AB()?1?12

31、即是對稱矩陣?AB?2A?4?3A必要性? 因為TTT 解存在?因為 ?)( ?且ABAB 所以B?AB1?A? |A|故?0?2?145?TTT )(?ABAB?BBAA? ?11求下列矩陣的逆矩陣? AAA?02?4?311121?21?1?A?A?136?A*A? ? (1) ?321222?52?214?32?AAA?33231321?A? 解存在因為?1?A1A |?|故?52?14 64 / 11?2?0?213?11?1113?02X1?* (2)?AA?1?3?243?所以? ?1?11 |A|22?1?167?1?11?2?3?11?a?X210?0a 解 1234?211?

32、1? a(a?a?(4) 0) n210?101?a?311?1?n?2?23? 243 ?3?a?033?01?a?A2?12?2解? 由對角矩陣的性質(zhì)知?28?0? ?5?a? 33n?1?411203?X?1 0a(3) ?1?2?11?10?1 a?A1?21?1? ?412031?X? 解 11?120?110? a?0?413121n? 211?101 ?解下列矩陣方程12? 12?645?2?11?X?06611 ?(1)?3121?1? ?0?2031 12? ?41?2?65423?5?3?462?X? 解 ?1123822?101?64 / 15 2x?x?x?34001?

33、0101?312?10?2X0011?001x?x?32x? ?(4) ? (2)?312010102010?0x?x?53x?2?31211? 方程組可表示為 解 030101?041?110?010020X?x?2?11?1? 解 ?1?02001101?012x?31? ?2?0?235x?0?1201031100?4?3?413?10200101?0?1? ?x?5?121?1?120010?0100?112?0?12?1?x3? 故 ?2?3?5032x ?13?利用逆矩陣解下列線性方程組?31x?3x2x?5x?132?25x?x?2x21? (1)? 0?x312? 故有 ?3?

34、x5?3xx?2?3x?321?3 解 方程組可表示為 1?k12k A?E?A?A設A?O(k為正整數(shù))證明(E?A)? 14?x?1132?A證明 因為kk ? 所以 E?A又因為?E?O?1?2522?x ?12k?k ?AA)?E?A(?E?A)(E?A?2?3135x?AA?E?A)(E?所以 (1k?2 E?)?A3 且A)可逆? (由定理2推論知E?1?x?11213?1?1?12k A?E?A?A?(E?A)?022?x25? 故 ? ?2?05331x?)(E?A證明一方面? 有E?1? ?A)?(E3A? 由另一方面k ? 有?O1?x?1k1k?1k?22) ?A?(?A

35、?AA(?E?(EA)?A?A)0?x? 從而有 ?212k? A)?A)(E?A?(E?A?0x?)A (E? 故 11k2?3 )A?()?(EA?EAA?)(EA?16 64 / )?A兩端同時右乘(E E?)?4 E?2)(A?3E? (A1? 就有?)2E(A?所以 11? E)?)?4(A?2 E(A?2E)(A?311?2k? (E?A)?A?A?E (E?A)?A 1)?A(3?2E)?E(A1? ?)2E及(A?121? 15?A?并求AA證明A及?設方陣A滿足A2E都可逆2E?O? 42 E?O得 證明 由AA?21?|A|2 2E?A(A?E)A ?A?2E?即1? ?5

36、|(2A為3階矩陣?A)A*|?求 16?設 21E)?A?EA?( 或 1 ? *2AA?1?解因為 所以? |A|1)(EA?A?1?且? 推論知由定理2A可逆 ?511 2|?|?|A|?AA?5|A|A|(2A)?5A*1?1?1?11? 222A由2 得?A?2EO11?13? 16?8?|?8|A2?|?A|2|?(?2)?|A?2 E?2即4A?A6E?E? (A?E)(A3E)41?1 ?且(A*)A)*?( 17?設矩陣A可逆?證明其伴隨陣A*也可逆?1EA?)?E(E?3?A)(21 或 * A?A41?證明由1? 有可逆時A? ? 所以當A?得A*?|A| |A|1)3(

37、AE)E?2A(?1?且2由定理推論知(A? 可逆E?2) ?11n?n? |A|A*|?|A?|A0?|?| 4 *也可逆?從而A A|A|因為A*?1? ?所以A證明由22 ?兩端同時取行列式得E?AOE?A2?得?A211? A|A (A*)?| 2 ?A| A?|?21*)*|( ?|EAA 即 |?2?A?AA?A1?1又 所以? |A1? A 故 |0?AE?A而可逆所以A?2?1?1?11?1 ?(AA?|A|)*|A|(A(A*)|?|A?)*222 E?A故0?A?A|E2A|?|?2?也可逆 ?*? 證明A18?設n階矩陣A的伴隨矩陣為A由2E A)?A(E2?O?2?A?

38、E ?0?|?0?則A*|若(1)|A1)A(?AE1?11? (AA?A2?)EA?E1n? ?|A|A(2)|*|? 2 證明A又由2E E2?A3(?AE2?A(?O?E2?A?)4?)1? 由此得*)*(則有?A假設用反證法證明(1)?|*|0 AAE?17 64 / 11? ?|E(A*)A?AA*(A*)O?|A102? ?0?|A|?0時? 有|A*|所以A*?O? 這與|A*|?0矛盾，故當003B?A?E? ?1210?*A?A1? A|E? 取行列式得到AA? 則*(2) 由于?| |A ?求B?8E? 2?1)?A*BA?2BA 21? 設A?diag(1?n |?|A

39、|A|A*| 得8E?2BA? 解 由A*BA |則A*|?|A|若A|?0? 1n? ? ?8EE)BA?(A*?2 A(1)知|*|?0? 此時命題也成立?若|A|0? 由11? )A?B8(A*?2E1? ?2E)?8A(A*|因此|A*|?|A1?n ?1 ?)*?2A?8(AA 1 ?)?2A?8(|A|E303?1 ?)?2A?8(?2E?01A1 ?求2?B19 ?設? BAB?A 1 ?)E?A?4(?3?12?1 ?2)1?4diag(2?11 ?2E?A故)B?A2AAB解由?E可得()1,diag( , ?4 221?302?333330? ?1)2?2diag(1?11

40、1?A0?01)2AB?(?E1?32?1 ?0001?31121?2?011?0100?A*? 的伴隨陣?已知矩陣 22? A110?0011?020A?8030?2? ?B B?求? 20 ?設?且? ABEA?110?ABA且1?1 ? BA求B?3E?2 B?E?由 解 ABA?得|?|A 由|A*|解3 2?得 |A|?8?2 (?ABEA?)?EABA由11? 得?3E?BA )E?A)(?A(B)EA( 即 ?E ?B?3AAB?100 11?1 A)?AB?3(?E)3A?A(EA00?|A?110?|E?1 ?A(所以因為 ?從而 ?)E可逆*)A2(E?6?*)(?3E?A

41、11? 010 218 64 / 11?00160000?*)PP(? ? |P00060100?6? ?0?100106622?0?2?11110?10603?0?03?320?00?20310? ?1?1?11120000?1?4?1?P?111? ?AP?其中?求? 23設PA2110?111?11?41P解由111111?1. ?A=?P得AP?A?P?PP? 所以A ?111?41411?P*P1? ?P |?3?11?1?1 ? 并求其逆陣也可逆31?1?B?A、B及A?B都可逆 證明A? 25? 設矩陣 因為證明 11 1?00?1?111?1?11?1?1? )B?BAB?AA

42、?(A?B?而 ?200211? ?A可逆A可逆? 即 而A是三個可逆矩陣的乘積?所以1?1?1?1?1?1?B?B)B(A(A?B)B?1?1?1?1?11?1 ?BB?)A)?B?A(B(A?B)(AA41?11320101?273227310?1?1?4 ?33?A11?故10201101? ?684?683?21011111? 計算? 26? 300020?21 33?30003?000?1?111?設解 ?120?1P? AP 24 ?設?其中 P?3?2312121?1?115?BA?A?B? ?302?11003?2121?A?A(求)28 6?E(5A?A)?B?AABAEEB

43、?( 解 ?)28) E(56?211111? 則 ?BOABOAO?825) 6?6?5)?5diag(?30)?diag(1?1?)diag(5?1?diag(1?522228 0)?12diag(1?0?)diag(125?diag(1?10)0?25?231213?ABB?1? )A(?(P?P)? 而 ?432?1?2001211?19 64 / 3?421?2343?AB?O? ?93?0?3003?422?A?48 A?求| 28?設A|及02 所 以 O?22?2512?B?AAEEBAB04234?102?A?A?211111?令? ? 解 ?224?3300?4BOAOAO

44、B21?22229?000?OA?A1? 則 ?251121100312?AO?24010102?1?2011? 即 ?834021200?30?00AOOA8?A?811? 故 ?903?000003?000AOAO8?22|A|B BA0110?|A|A|A|?|A|A|?8168888? ? ?ADC?B?2112驗證取? ?27?DC10|D|C?054?O?AO450400201001? ?A?41? ? 00220010110AB00A2O44 ?4?O?2 ? 解 1?CD10011001?0022246? 10?010110 求B都可逆?階矩陣A及s階矩陣29 ?設n |BA|

45、11 1?AO0? 而 ?11|D|C| ?(1) OB? |B|A|BA 1?CCAO? ? 故 ?21?DC?|D|C|解設 則?OBCC?43OEACCCACAO?n4123? ?OBCCEOBCBC?43s2164 / 20 0250AC?EAC?1?3n3?0201OC?OAC?44? 由此得 (1)? ?8300O?COBC?11?5020EBC?BC?1?s223852?1?AOBO1?BA? ? 設則? 解?2521所以 ?OBOA1?11?8?3252132?1?OA?A?B1?1? ?12525825? (2)? ?BC? 是 于 1?DDOA?1?21?設 解025?20

46、001? 則 ?BCDD?1?43AA1?02012500? ?80303002?OEADDDADBB1?OA?n1212? ?580?50200BCBD?DDBDCD?CDEO?443231s0100?A?D1?EAD?1n10201?O?DO?AD?22 (2)? 由此得 0213 ?O?CDBD?CA?DB1?1?313?4112?EBD?CD?B?D1?s244123010?1?CB?AOAOA1? 解 設?則?211241? 所以 ?BCBCA?B1?1?1? ? 求下列矩陣的逆陣 30? 1?0010?1?O21A00OA1? BC0213B?BCA1?1?1?4211?64 /

47、21 00011?102?1131?000?0) ? ?3(下一步?r 3? 223000?111? ?0?1?102? 326?3?0011511) r?(下一步 ?r?332?1000 ?412824? 1?102? 0100) ?r2)(下一步?r?(?r? r3121? 1000? 第三章 矩陣的初等變換與線性方程組0001? ?0001 ? 1 把下列矩陣化為行最簡形矩陣 ? ?1000?110?2?1230102?3? ?(1)?334?033?04? ?(2)?1?7?04?1012?1320132?0 解?) ?3)(?r?下一步(?r(2)r?r1321?3?430?3304? 解 ) ?2)rrr(下一步?2