线性代数第一章习题答案

1、利用对角线法则计算下列行列式解:(1)(2)(3)(4)bb2(2)习题一A组bb2(4)152.计算下列个行列式。(1)12bc22x3ca3xyx2x3ab22y3cb22acba2cb26x2(2)精选资料，欢迎下载o精选资料，欢迎下载(3)(5)b4(1)解:(2)解:(3)解:(4)解:a?b30b2a30b1a,3111131111311113riC|c2C3cc4abcd(4)(6)4810cd1abcdabadcd11223344103102341-2-10341041b1c1b2c2c1a1c2a2a2bb1b216321041101221214123341322111 23

2、41 2(5)解:02003r3r20201r4r2034132204160a100b40a2b300b1b20a300a4a2ab30b2a3000a40b0b4a22b3a300a2b3b2a3a2b3b2a3aa4a2%b3bb)4a2a3aa2a3a4a1b2b3a4b1a2a3b4bib2b3b4bc(6)解：b1c1b2C2caC1a1C2a2aa1a2bbib22aqC2c32a12a2bb1b2CC1C2cac1a1C2a2aba1b1a?b2abc2a1ble1a2b2c2cac1a1c2a2aa1a2bb1b2bqc2b1c3qb2cac1a1c2a2aa1a2bc2c3b

3、1b2caGa1c2a2bc1a2ca1b2ab1c2ba1c2cb1a2ab2cl3.设行列式31050402227003-22求M412M42M43M44,其中M4j(j=1,2,3,4)为第4行元素的余子式解：M412M42M43M44A412A42A43A4430401222070012117A32c2c17c3c1精选资料，欢迎下载4.已知7219842x求x3的系数.解：fxAnA12A132A142x2x的展开2x2x式中只有xA11中含有项，又xA)1=xx214x2x2所以2xxA1中含有的项为2x2x3的系数为-1.5.证明:(1)bb3C2证明：（1）bb3(2)b2d2

4、acb2abacabb2证毕精选资料，欢迎下载o精选资料，欢迎下载(2)2ab22a12b12c12cd12a22b22c22d2a32a2a14a42cqb2b3Qq2b14b4c23c4q2c2c14c4d23d22d14d426a96b96c96d9a22a126c2c2b22b126c4沁c22c1262d22d1266.用克莱默法则解下列线性方程组.证毕2x1X25x3x48,X13X26X49,2x2x32x45,X14、7x36x40;解：(1)方程组的系数行列式(2)211302146r12r2107513r12r2130642021207712c12c2Q2c28159305

5、21047x1x12x13x1751330621277127513212771262127,11c13c20c42c211247x2x32x2-x33x2-x3x2x45,4x42,5x42,311x40.1131121c2q2c32q211311262100522102262138152210513则X128510D21906r12r210512r4r2010760959D3D4D1D121059210113C15c21520C32c2122651751789501013r3r152912271532621013965291210137c3c252215807309ri?13r2020750

6、171095910810353103,x2(2)方程组的系数行列式11231059c12c2C35c24名11213112D3D515107261,X45c4q2113015215272600233718精选资料，欢迎下载23C22q10037C33C15138182514152142精选资料，欢迎下载10182852723D227 32232214211235 r4 2r1 30211151130530410 0923333310r1 2r3r2 3 r3233310133198423 1135111511102214c32c2225182315C411c223528012110100Di1

7、10032022115111511224r2r3r12r101732325r43r10512731011021583D2681438141747291oo11白37871215r2r1D442h10C2C310C12333231033142则X1D1,X22,X3D3D3,x47.齐次线性方程组ax2X30，有非零解时,解：D2X1x2-bx30,axoXq0.23b必须满足什么条件?8.已知齐次线性方程组有非零解,齐次线性方程组有非零解，则0,即取何值?3a(12X1X1)X1(3X22x2)x2(13a0,从而ab解：D齐次线性方程组有非零解时,4x30，X3)X30,0.0,即=2或=3

8、.1.设，是三次方程X3px0的根，试计算解：因，是方程x3pxq0的根，则clc2c3所以02.计算下列行列式.1X11111x11(1)111y11111ya1(3)DnO；1ax123L1x21L(5) 11x3LMMM1 11Ln11；Mxn解：1x111x1111111y11xr2r1r31xr4r1x(4)(6)1111abcd2222abcd4444abcdDn133L323L333LMMM333L333L111L22223LMMM23nnnL111x000y000y2nnn333333.MMn133no精选资料，欢迎下载C1C2(2)将行列式D1D1ba3x3的系数为a2a4a

9、bb2b4c2c4cdd2d4加边，得新的行列式a2a3a4abb2c2c3c4cdd2x2x3x4x则由范德蒙行列式知将行列式D1按第五列展开,D1A15xA25xAx3A45x4A55，所以D1中x3的余子式M45即为所求的行列式D,将Dn按第二列展开,DnaDn1a2Dn2D2(4)列，得行列式除对角线上以外，其他元素都相等，将第三列的-1倍分别加到其他各Dnrir1原式i2,L,nrir1i2,3,L,n(6)原式n!12212n1n!由范德蒙行列式知,所以所求行列式3.证明:a01a10(2)(3)(4)a2M0ManDnDn证明:an1an2222n332222332M2n1M3n

10、11!2!La1a2La2n!aniaon11aiM3n12!L1!其中a1a2Lano；n1a1xLan1xan；1a。11L11a10L010a2L0MMMM100Lani2,3,L,nn1c1q1aiaiaLan%ai证毕.(2)将行列式按最后一行展开,anAn1an1An21andn31an2a1ananan1a?a2An2ana10MaAnna2a2Man精选资料，欢迎下载an证毕1anan1xn1a1xan1xan2Xa2xan1xan(3)将行列式Dn按第一列展开，得Dnx(4)将行列式又因为D2Dn一行展开,Dn2Dm2Dm3Dn24Dn33,D12n1a12an2xDnna?

11、xnax2Dn33Dn2,Dn3n12n222Dn32Dn3n1.证毕Dn3Dn41d2Dn2Dn3n2D1精选资料，欢迎下载3.求三次多项式23fxa0alxa2xa3x,使得f10,f14,f23,f316.解：因fxa0axa2x2a3x3,f10,f14,f23,f316则a。&a2a30a。&a2a34a02a14a28a33a03a19a227a316111该方程组系数行列式为D1 112341214802 483 927D104316111111248392700104212364121612936336422361216123621124124439D2101114111348

12、1169271000140213391168282400151310 2 7402339168282012411342724102D3101141238316271111100012421339141628242441628242400010所以5.记行列式D416164816242424967,ai20,a2Dfx75x22x3.D3D5,%x2x1x2x32x22x12x22x33x33x24x53x54x4x35x74x3为fx,求方程fx0的根的个数x2x1x2x3x2101解：2x22x12x22x32x21013x33x24x53x53x31x224x4x35x74x34x3x73x21002x2