线性代数第二章习题答案

1010120001解：31101400005001060100111111010021121004，求215124251098100420020220；2112122128012910040410124052；212112125216441010040421016215．12314321A0321，B53040321250123143213A-2B30321253040321250369382110550963106210156112096241001041961212A2121,B2121，求1234010112124解：1）3AB32121212340363646363236912012110111315218282；0137913210112124（2）2A3B2212132112340124241293144242636322468030321212413XAB2121212141234010116511140；35101025533Y2(AB)202020113324317473211355701577201492(1)22243(1)323613214001211341314026710313112512101240230043。00300091000解：A200020201010230B02120，304510102300226230022612014，AC12014；3045（3）BA(AB)T26T241461。311110311311311100f(A)A2AE312312312010110110110001133531110092414253120101103。0011100011121111A2AA0AE11110011001143；123343101cossinsincos；（3）210；32512341230123．789001200011170，故A1存在，A113A211A1241A*1317A414A11A*A（3）A21020，A1存在，A115,A1210,A137,A212,A222，325A2352,A311,A322,A3311A*5（4）A211A232,A314,A325,A332A11（5）A4A221,A230,A240,A311A322,A331,A340,A410,A421,A432,A441121A11A*0121A000012313A221,B21533433133523152112131131315201123152025210402104。25B46C2421，35161224（1）AXBXA354622312210XABXBA1463518321258AXBCXA1CB13524122111682452x1x2x343x2x4x11x12x23x313x5xx3211A342，Xx2，B11，则原方程组为AXB324332126611811118111123351323134A22515，A1138112033A110,ABA2B，求B．123x0233A2E110，A2E2，1211331113111133033033B(A2E)1A1113110123111123110101101001201100102若A0，则由AAAEAAAEAn，∴AAn1AA1(A1)*A1E1E，∴(A1)*1A，即(A1)*(A*)11AAAAT(A*)T(A*A)T(AE)TAE10011003011410000060。011414而P3，P1110111002021114100211113121142116836844213273127321112．设APP1，1111122233解：∵P1026，P*301P*10112111111361200000(5)000111120010200011100034441444214440131212420010630010213B124，12kEkA100kE0B10BBE，630，202Bk2k4k0k0；00k221A1212063002010121011100120E0BAE，B1BB1110B10B41AB12，211，320则ABBAB，10102433120111312433。1131，B1Ba1Ab10B其中A1=0a，21b，a0b01a1b，则AB11，ABA111A1B1A1aa2a1，A2B2A2b2b2b13b2a2a3a2a21b32b2b21344300002022AO8A18A2O34434320，令A3422A1AAA8A18A8A18A281016540O054A24OOAFACD113X4XX2BXBX0EAX3AX4E0AX3EX3A1AX0X0，即D11BXEXB1F11X3X4BXAX2CX4E0X1A1AXCX0BX0X0BXEXB1BB，则A1112，A2123255820011001A0830521202510A21023058A1AA1A1，A31A2A1116824A3126511124120001200130A1AA1214123441130243（1）2225；（2）01113；（3）33511120312230731333423137120241341732834723741470243rr1112r2r1112rr解：1）2225225001111202430243434120215101112r11r21r310100243（行阶梯形矩阵）20120（行最简形矩阵）0001(r23r3)0001123011130731234011100240010r12r2r301r2r3000011333522333411000000441213r3r15r20131r47r2001300r3008013126（行最简形矩阵）44132643r23r141r32r120r43r121113400480036005101011000000343122000（行阶梯形矩阵）0000231222312(4)322312010000137r12r212024r33r2010r42r20831rr07024r47r28912r3r47811024r12r210202r2r3011034（行阶梯形矩阵）r2(1)00014（行最简形矩阵）000006r4r314714713417r13r201314r37r201314473r2r3072145000143（行阶rr00147310001051050r14r201314r59r01301)0001r214r300010000000r33r2141414501122r2101122r45r2012rrr5r51051051005100510145103012000012000（行最简形矩阵）1400002020120r3r120r12r2106A1113320031213013013008800110001000110013．设001A001011001解：001A001010014710258，求A．013690014714710258可以写成E(2,3)AE(3,1(1))25801369369147147从而AE(2,3)1258E(3,1(1))1E(2,3)258E(3,1(1))369369147647369E(3,1(1))669258658101（1）210；3251233433201132（3）301232．1101100101100r2r101(A,E)21001012210325001r33r102230110100100511r2r3012210010511002721r32001711511A151121200123100rrr12312（2）(A,E)221010r22r10252103430010011111032110013352010r25r30203652001111r(1)00111111132100r3r1(A,E)30101021100131013210001111201043101r44r2000097310431011236100113r2r3010237001349113237349320111232001r3r40040r13r30212320103021034r22r11102r1r300341610011240113121610112411321610123253434312325rrr12(A,B)221313120251934343r22r100113101100302062201023r25r300113r(1)0011∴XA1B234236．求解矩阵方程XAA2X，其中A110。123211411rr21141132212212212213013031111110112332r3r111111103403111111011330016691003220108912001669386∴X2962129228912669三阶子式1000，1000。563123131111；2）A3153；3）A1121；428212213443213