線性代數(shù)復(fù)習(xí)筆記整理-精

線性代數(shù)復(fù)習(xí)整理AB1.設(shè)為×n矩陣，為n×m矩陣，≠n，則下列矩陣中為n階矩陣的是（B）A.B.C.D.TBTTABTABAABT為n×m，BT為×n解：由為×n矩陣，為n×m矩陣得：BATABT則T為×m矩陣；T為n×n矩陣；ABABA為×m矩陣，為×n矩陣；BABAB為n×n矩陣，為n×m矩陣A5A(A=2.設(shè)為n階方陣，n2則nkA得答案由公式n12A*A3.，則=-234(4(242AA1121*11AA212解：(3(13112221222AxbAx0，那么必有無(wú)窮多解4.設(shè)有無(wú)窮多解，則br()r(B)n0，故有無(wú)窮解解：由有無(wú)窮多解得r()r(B)n1，則br()r(B)nb則只有唯一解若5.無(wú)解，若有m個(gè)n維向量，若，則該m個(gè)向量(D)成立AB(A1B)1B1A=6.設(shè)均可逆方矩，則A7.8.為×n矩陣，則有（D）AB=BA，，那么ABC=BCA解：ABC=BAC=BCA12113274APAP9.，，則T=3401121011211021314130413274APT解：3411122A2t30有非0解，則t=210.設(shè)，若齊次方程組345122122122r2rA2t30t410t2002121r3rrr解：31233450210r()3∵齊次方程組有非0解，必有∴t-2=0得t=2101A110A1？，求11.012解：101100101100A|E110010011110rr21012001012001101100100211(rrr201111013010221E|Arrrr13223001111001111211A2211∴=111注：變換時(shí)用“~”代，中間步驟不少于3步，老師特別說(shuō)明是踩分點(diǎn)來(lái)的aa2)aaT，求向量組的秩和一個(gè)極大線性無(wú)關(guān)組？12.向量組T，T，T，12341031103103300110103100000110rr1301217221r2r3123r2rr3r解：430004000442140得向量組的秩r()3，其中a,a,a為一個(gè)極大線性無(wú)關(guān)組134xxxx112343x2xxx013.求線性議程組的通解（用基礎(chǔ)解系和一個(gè)特解表示）1234x2x2x3234解：對(duì)方程組的增廣矩陣施以初等行變換1111111111(A|b)321100122301223r3r21012231111110112(r20122301223rrrr32120000000000r()r(A|b)24kx,kx，得原方程組的通解為得，故方程有無(wú)窮多解。設(shè)1324112x1223xkk(k,k為任意實(shí)數(shù))，22x1100123010x4注：變換時(shí)用“~”代，中間步驟不少于3步，且上邊步驟一個(gè)不能少，老師特別說(shuō)明是踩分點(diǎn)來(lái)的xx01214.若有非0解，則k=(-1)x0121111A~0rA()2，故k+1=0，則k=-1解：，因有非0解，則k1k1015.沒抄16.沒抄17.沒抄18.沒抄19.沒抄AB,20.A.為同階可逆方陣，則下面錯(cuò)誤的是（C）AB1B11B.C.(AB)AB111ABBATTTD.AA2(A)*1，則21.為三階矩陣，且（A）A.1/4B.1C.2D.411*An12314(A)4*11解：，44a,a,,a22.向量的秩為r，且則r<5，則（D）125a,a,,aA.B.C.D.23.A.C.線性無(wú)關(guān)125a,a,,ar個(gè)向量線性無(wú)關(guān)任意r+1個(gè)向量線性相關(guān)任意r-1個(gè)向量線性無(wú)關(guān)125a,a,,a125a,a,,a125a,ab0的解，是對(duì)應(yīng)齊次線性方程的解，則（B）設(shè)是12a0(aa)0是的解是的解B.112aabaab是的解是的解D.1212A3，則2A=6A24.為三階方陣且2A2A3此題抄得有問(wèn)題，感覺像是要考AAx0的基礎(chǔ)解系所含向量個(gè)數(shù)是325.為4×5的矩陣，且r(A)=2，則齊次方程A∵為4×5的矩陣，可見未知變量個(gè)數(shù)為5個(gè)∴自由變量為5-2=3a2)aaaaa12326.設(shè)向量，，，則的秩是2123132132000000r2r312760aa(4,a4,7)的秩是227.，，123141141141r2r2130950950181009500021rrrr154r3r31r2r32414236700011AA28.A為3階方陣，則，=A33A.-9B.-3C.-1D.91111A()3AA9A解：由得3327311，1010BA29.已知矩陣，則=01121111110111100110(1110111(01(100(121111021111011111021a2130.已知行列式2300，則a=3111思路：將行列式化為下三角后，再求解a21a130a300230230230(a310得a=3rrrr由1312111111042111333201BAB357AT31.，，則=35711392120(324(522(7042TB010013041502173571310(314(512(73333579111902111a8aa332.已知，，的秩為2，則t=3（本題無(wú)法求解）5t141220AA2A*33.若為3階矩陣，，則=4A*An12314解：34.為4階方陣，r()3；則=0AAr()3AA0得內(nèi)有一行全為0值，故由k1111k11Ar()3k35.矩陣，，則=-311k1111kk1111k1111k1111kk3k3k3k3111k111k111kArrrr123411111k1111k1111k11110rr0k1021(krr(k0000k10k131rr041(kk03k3時(shí)，r()3成立當(dāng)當(dāng)k1時(shí)，r()3不成立，故舍去11570k150，則應(yīng)滿足kk036.設(shè)00k9000111570k1500k900011kk1k2021112111237.行列式=4AB,A2B3，ABT38.若都為n階方陣，且，求=-6ABABAB2(6TTA39.40.若為3階對(duì)稱矩陣，則AT=01A1AA1AA）1為n階可逆矩陣，若行列式，則=-n（公式nBABABT41.為任一n階方陣，為n階實(shí)對(duì)稱矩陣，證明為實(shí)對(duì)稱矩陣(BAB)BABBABBTAB證明：∵TTTTT∴為實(shí)對(duì)稱矩陣ABABBA42.（必考題）若是對(duì)稱矩陣，是反對(duì)稱矩陣，是否為對(duì)稱矩陣，請(qǐng)證明。AABB證明：由定義對(duì)稱矩陣有T，反對(duì)稱矩陣TAB)(AB)()BAAB(B)A(B)ABBA∵∴TTTTTTTABBA是對(duì)稱矩陣120261A231120()ABT？B43.（必考題）設(shè),，求013210210122(AB)BA621231TTT10301047521120022130122110061221062231162211010191411023012033112013015244.判斷下列向量的線性相關(guān)？aaaa；；；1234解：1021120110211021rrr022001102312r1r2rr20112410112A2130rr2512514113105560556011201121021102101100110rr32r5rr3r420002430002rrrr52530006000200000000r()34a,a,a,a線性相關(guān)得(向量的個(gè)數(shù))∴1234r()r(A)用以下解法也可以，因?yàn)門，但為了不混亂，建議全倒置成豎排來(lái)解題1122111022121021512031351r2rA131rr402151002221104111221021510000000222rr32r()34a,a,a,a線性相關(guān)得∴1234123A221，用初等變換的方法求A1？45.設(shè)343123100123100Ar2r|E22101002521021r3r31343001026301102110100132rrr2r1200111112026301rrrr232302630100111113210013210035r02036520103E|Ar6r12232120011110011113235A312211146.用基礎(chǔ)解系的形式求下列方程組xxxxx7123453xx2xx3x2123452xx2x6x23458x3x4x3xx12345解：對(duì)方程組的增廣矩陣施以初等行變換111117111117312132021262383431120212623r3r(A|b)21r8r4102126230545914111117102081800000002126230111925000000rrrr2314r3rr2r0030122743340111925102081810000000000000104901119250000000010490101516r2rr313rr3431000000101516001049000000rr42r()r(A|b)35kx,kx得，故方程有無(wú)窮多解。設(shè)，得原方程組的通解為1425000516x11x2kxk049(,為任意實(shí)數(shù))kk，12311200x4010x5注：變換時(shí)用“~”代，中間步驟不少于3步，且上邊步驟一個(gè)不能少，老師特別說(shuō)明是踩分點(diǎn)來(lái)的11511123a1a2a3a447.，，，求其秩及最大無(wú)關(guān)組？31811397解：11511151115rr11230274027421r3rrrrr31r3r32A31814102744200000000139704148r()2aa,得，最大無(wú)關(guān)組是（首元為非0的對(duì)應(yīng)的列的向量組成最大無(wú)關(guān)組）1248.無(wú)題49.無(wú)題2x4x17x6x01234xx2x3x0123450.用基礎(chǔ)解系的方式求方程組3xx17x5x012343xx8xx01234解：對(duì)方程組的系數(shù)矩陣施于初等變換241761123311751123112306024048r2r241763117521r3rr3r31Arr124131813181rr3112302740223411231811030204001000007rr21630274rr3rr2320016016rA()34，故方程有無(wú)窮多r42r3得20000041481103100101020102r212rr00102001000000000kx解。設(shè)，得原方程組的通解為141x12xkk(為任意實(shí)數(shù))2，x10130x4注：變換時(shí)用“~”代，中間步驟不少于3步，且上邊步驟一個(gè)不能少，老師特別說(shuō)明是踩分點(diǎn)來(lái)的線性代數(shù)重點(diǎn)整理1.1)矩陣的運(yùn)算ABBA(AB)CA(BC)2)AOOAAA()O3)4)1AA5)6)()A()()AAA7)8)(ABAB)kan121121222n矩陣數(shù)乘：1m22.矩陣的乘法ABAB為×n矩陣，為n×m矩陣得為×m矩陣ABBA(AB)AB，一般kkkA0ABACBCABOBO不能推出若，從不能推出，特別從a11a12aanbb11bm12aabbbAB21222n21222ma1abab1111aabbb1n2nmm2mn即A的第1行乘于B的每abab1112ababn1abab1nn2ab1221n1122211111221ababababab2112abababab211122212n122222nn2211122212n1ababababababababab111m221mn1112m222mnn2111m221mn1一列得AB的第1行對(duì)應(yīng)列的值A(chǔ)的第2行乘于B的每一列得AB的第2行對(duì)應(yīng)列的值??A的第n行乘于B的每一列得AB的第n行對(duì)應(yīng)列的值(ABC()1)(BC)AC(ABC2)3)()()()ABAB3.矩陣的轉(zhuǎn)置AAT或'為×n矩陣，將行列依次互換，所得到的n×m矩陣稱為矩陣的轉(zhuǎn)置矩陣，記作設(shè)(A)A1)TT(AB)AB2)3)4)5)TTT()kATT（k是常數(shù)）(AB)BA(ABC)CBATTTTTTTr()r(A)T4.行列式2n行列式一定是nn列切記不存在×n的行列式（≠n）AA1)2)Tn階行列式等于它的任意一行（列）的各元素與其對(duì)應(yīng)的代數(shù)余子式的乘積之和AaAaAaA行列式行展開定理行列式列展開定理11i2i2ininAaAaAaA1j1j2j2j3)4)5)行列式的某一行（列）的公因子可提到行列式外a11a12na11aa12nkakakakaaa例：1i2in1i2inaaaaaa1n2nn1n2nn把行列式的某一行（列）的所有元素乘以數(shù)k加到另一行（列）的對(duì)應(yīng)元素上去，行列式的值不變a11aaakaakaaka111212n1i21nina1aaa1ai2ain例：i2inaaaaaa1n2nn1n2nn交換行列式的兩行（列）的位置，行列式變號(hào)aaaaaaa221112n212naaaaa21222n1112na31aa