工程數(shù)學(xué)線性代數(shù)第五版

第二章矩陣及其運算已知線性變換x12y12y2y3x23y1y25y3x33y12y23y3求從變量x1x2x3到變量y1y2y3的線性變換解由已知x221y1y2x2315x3323y2y12211x1故y2315x2y2323x3y17x14x29x3y26x13x27x3y33x12x24x3已知兩個線性變換x12y1y3x22y13y22y3x34y1y25y3求從z1z2z3到x1x2解由已知x1201y1x2232y2x3415y2

749y1637y2324y3y13z1z2y22z1z3y3z23z3x3的線性變換201310z1232201z2415013z3613z149z210116z3x16z1z23z3所以有x12z4z9z2123x310z1z216z33111123求3AB2A及設(shè)A111B124111051ATB3AB2A111123111解311112421111110511110581112132230562111217202901114292111123058ATB111124056111051290計算以下乘積(1)431712325701431747321135解123217(2)23165701577201493(123)213解(123)2(132231)(10)12(3)1(12)322(1)2224解1(12)1(1)121233(1)32362140131(4)01211341314022140131678解01211341312056402a11a12a13x1(x1x2x3)a12a22a23x2a13a23a33x3解a11a12a13x1(x1x2x3)a12a22a23x2a13a23a33x3(a11x1a12x2a13x3a12x1a22x2a23x3x1a13x1a23x2a33x3)x2x3a11x12a22x22a33x322a12x1x22a13x1x32a23x2x35設(shè)A12B10問1312ABBA嗎解ABBA因為AB34BA12所以ABBA4638(2)(AB)2A22ABB2嗎解(AB)2A22ABB2因為AB2225(AB)2222281425251429但A22ABB23868101016411812341527所以(AB)2A22ABB2(3)(AB)(AB)A2B2嗎解(AB)(AB)A2B2因為AB22AB022501(AB)(AB)220206250109而A2B23810284113417故(AB)(AB)A2B2舉反列說明以下命題是錯誤的(1)若A20則A0解取A0120但A000則A(2)若A2A則A0或AE解取A112A但A0且AE00則A(3)若AXAY且A0則XY解取A10X11Y11001101則AXAY且A0但XY7設(shè)A1023k1求AAA解A21010101121A3A2A10101021131Ak10k18設(shè)A10求Ak010解第一觀察1010221A2010102200000023323A3A2A033200344362A4A3A0443004A5A4A5541030554005kkk1k(k1)k2k02Akkk100k用數(shù)學(xué)歸納法證明當(dāng)k2時明顯成立假設(shè)k時成立，則k1時，kkk1k(k1)k210Ak1AkA02kkk10100k00k1(k1)k1(k1)kk10k1(k2k11)00k1由數(shù)學(xué)歸納法原理知kkk1k(k1)k2Ak02kkk100k9設(shè)AB為n階矩陣，且A為對稱矩陣，證明BTAB也是對稱矩陣證明因為ATA所以(BTAB)TBT(BTA)TBTATBBTAB從而BTAB是對稱矩陣10設(shè)

A

B都是

n階對稱矩陣，證明

AB是對稱矩陣的充分必需條件是

AB

BA證明

充分性

因為

AT

A

BT

B

且

AB

BA

所以(

AB)T

(BA)T

ATBT

AB即AB是對稱矩陣必需性因為

AT

A

BT

B

且(AB)T

AB

所以AB

(AB)

T

BTAT

BA求以下矩陣的逆矩陣1225解A12|A|1故A1存在因為25A*A11A2152A12A2221故A11A*52|A|21(2)cossinsincos解Acossin1存在因為sincos|A|10故AA*A11A21cossinA12A22sincos所以A11A*cossin|A|sincos(3)121342541121|A|20故A1存在解A342因為541A11A21A31420A*A12A22A321361A13A23A33321421A*210所以A11331|A|221671a1a0(4)2(a1a2an0)0ana10解Aa2由對角矩陣的性質(zhì)知0an110a1A1a201an解以下矩陣方程(1)25X461321解X2514635462231321122108(2)211113X2104321111132111解X2104321111113101232343233022185233(3)14203112X11011411201解X31201111243110121101121661011101230124(4)010100143100X00120100101012001011431001解X100201001001120010010143100210100201001134001120010102利用逆矩陣解以下線性方程組x12x23x312x12x25x323x15x2x33解方程組可表示為123x11225x22351x33x1123111故x222520x335130x11從而有x20x30x1x2x322x1x23x313x12x25x30解方程組可表示為111x12213x21325x03x1111152故x221310x332503x15故有x20x3314設(shè)AkO(k為正整數(shù))證明(EA)1EAA2Ak1證明因為AkO所以EAkE又因為EAk(EA)(EAA2Ak1)所以(EA)(EAA2Ak1)E由定理2推論知(EA)可逆且(EA)1EAA2Ak1證明一方面有E(EA)1(EA)另一方面由AkO有E(EA)(AA2)A2Ak1(Ak1Ak)(EAA2Ak1)(EA)故(EA)1(EA)(EAA2Ak1)(EA)兩端同時右乘(EA)1就有(EA)1(EA)EAA2Ak115設(shè)方陣A滿足A2A2EO證明A及A2E都可逆并求A1及(A2E)1證明由A2A2EO得A2A2E即A(AE)2E或A1(AE)E2由定理2推論知A可逆且A11(AE)2由A2A2EO得A2A6E4E即(A2E)(A3E)4E或(A2E)1(3EA)E4由定理2推論知(A2E)可逆且(A2E)11(3EA)4證明由A2A2EO得A2A2E兩端同時取行列式得|A2A|2即|A||AE|2故|A|0所以A可逆而A2EA2|A2E||A2||A|20故2E也可逆由A2A2EOA(AE)2EA1A(AE)2A1EA11(AE)2又由A2A2EO(A2E)A3(A2E)4E(A2E)(A3E)4E所以(A2E)1(A2E)(A3E)4(A2E)1(A2E)11(3)4EA16設(shè)A為3階矩陣|A|1求|(2A)15A*|2解因為A11A*所以|A||(2A)15A*||1A15|A|A1||1A15A1|222|2A1|(2)3|A1|8|A|1821617設(shè)矩陣A可逆證明其陪伴陣A*也可逆且(A*)1(A1)*證明由A11A*得A*|A|A1所以當(dāng)A可逆時|A|有|A*||A|n|A1||A|n10從而A*也可逆因為A*|A|A1所以(A*)1|A|1A又A11(A1)*|A|(A1)*所以|A|(A*)1|A|1A|A|1|A|(A1)*(A1)*18設(shè)n階矩陣A的陪伴矩陣為A*證明(1)若|A|0則|A*|0(2)|A*||A|n1證明(1)用反證法證明假設(shè)|A*|0則有A*(A*)1E由此得AAA*(A*)1|A|E(A*)1O所以A*O這與|A*|0矛盾，故當(dāng)|A|0時有|A*|0(2)因為A11A*則AA*|A|E取行列式獲得|A||A||A*||A|n若|A|0則|A*||A|n1若|A|0由(1)知|A*|0此時命題也成立所以|A*||A|n119033ABA2B求B設(shè)A110123解由ABA2E可得(A2E)BA故2E)1A2331033033B(A11011012312112311020101且ABEA2B求B設(shè)A020101解由ABEA2B得(AE)BA2E即(AE)B(AE)(AE)00110所以(AE)可逆因為|AE|010從而100B201AE03010221設(shè)Adiag(121)解由A*BA2BA8E得(A*2E)BA8EB8(A*2E)1A18[A(A*2E)]18(AA*2A)18(|A|E2A)18(2E2A)14(EA)14[diag(212)]4diag(1,1,1)222diag(121)已知矩陣A的陪伴陣A*

A*BA2BA8E求B11000010010100308且ABA1BA13E求B解由|A*||A|38得|A|2由ABA1BA13E得ABB3AB3(AE)1A3[A(EA1)]1A3(E1A*)16(2EA*)12100016000601000600101060600306030123設(shè)P1AP此中P1410求1102A11解由P1AP得APP1所以A11A=P11P1.|P|3P*14P111411311而1110111002021114101427312732故A1133110211116836843324設(shè)APP1111此中P10211115求(A)A8(5E6AA2)解()8(5E62)diag(1158)[diag(555)diag(6630)diag(1125)]diag(1158)diag(1200)12diag(100)(A)P()P11P()P*|P|1111002222102000303111000121111411111125設(shè)矩陣A、B及AB都可逆證明A1B1也可逆并求其逆陣證明因為A1(AB)B1B1A1A1B1而A1(AB)B1是三個可逆矩陣的乘積所以A1(AB)B1可逆即A1B1可逆(A1B1)1[A1(AB)B1]1B(AB)1A1210103126計算010101210021002300030003解設(shè)A12A21B310103211213B203則AEEBAABB111112OA2OB2OA2B2而A1B1B21231235201210324A2B2212343030309A1EEB1A1A1B1B21252所以0124OA2OB2OA2B200430009121010311252即01010121012400210023004300030003000927取ABCD10考據(jù)AB01CD10102000解01010101而|A||B|110|C||D|11故AB|A||B|CD|C||D|34O28設(shè)A4384O20求|A|及A22解令A(yù)34A20432212

|A||B||C||D|4則AA1OOA2故8A1O8A18OAOA2OA82|A8||A8||A8||A|8|A|810161212540O4A4O054AA424OO02262429設(shè)n階矩陣A及s階矩陣B都可逆求1OABO1C1C2解設(shè)OA則BOC3C4OAC1C2AC3AC4EnOBOC3C4BC1BC2OEsACECA13n3由此得AC4OC4OBC1OC1OBC2ECB1s2OA1B1所以O(shè)BOA1OAO1CB解設(shè)AO1D1D2則CBD3D4AOD1D2AD1AD2EnOCBD3D4CD1BD3CD2BD4OEsAD