1利用对角线法则计算下列三阶行列式(精)

1、（i）i第1.利用对角线法则计算下列三阶行列式:201(1)1-41；(2)-1 83111(3)abc；(4)2.22abc201解(1)1 -4_1=2工( (_4)_ 1830 x1X 3 _：=- 24十8 +1 -4bc3 0 (-1) (-1)118(-1) 8-1 (- 4) (-1)i - 4abcbcacabxyx + yyx+ yxx + yxy卩行列式(2)=acb bac cba - bbb- aaa - ccc(3)333二3abc - a - b - c12 2 2 2 2 2=bc + ca + ab - ac - ba -cb=(a _ b)(b _ c)(c

2、_ a)x十y=x(x y)y yx(x32=3xy(x+y)-y - 3x y-3y =2(x3十y3)3.计算下列各行列式：3y)+(x+ y)yx-y3x(x y)3- x33-x4124214111202； （2）312111015202320117一5062一-f-（i）23=0-abacae1bd-cdde；(4)bfcf-efJ-1(3)1000240121010104 3(一1)10-1410C21010121心4 - 93_ 1221232123050625062231417171421402140r423-12 24-13-122123012302

3、1400000-abacae_ bcebd-cdde=adfbcebfcf-efbc-_ 111=adfbceJ1-11=4abcdef11-1=01044.证明:a10001十aba0-1b10A + ar?1b100-1c10-1c100-1d00-1d1a=(D( 1)21=(1)(1)3 22 2a ab bab-101 ab-1c-1ad1 + cd01d1 ab-10ad1 + cd0=abed ab cd ad 12aa + b2b=(a-b)3J111ax + byay+ bz az + bxay+ bzaz + bx ax +byaz + bxax+ by ay + bz3=

4、(ax+ b3)yza2(a1)2(a2)2(a3)2b2(b1)2(b2)2(b3)2c2(c1)2(c2)2(c3)2d2(d1)2(d2)2(d3)2=0;1a2a111bcdb2c2d2444b c dx0-1x0-1-000anan =an-24a证明0000=xn+ a1xn+ an=x+ anx-1a2X + aq二( (a - b)(a - c)(a - d)(b - c)(b - d) (c - d)(a b c d)；56(1)左边二幺M-1)312aab - a2ab- a2b2a2b2ab2- a22b 2a按第一列x ay + bzaz + bxyay十bzaz十bx

5、左边八 F分开ay az + bxax + by +bzaz + bxax + byz ax + byay + bzxax+ byay+ bz分别再分2xay + bz zyz az + bxa yaz + bx x+ 0 + 0+b zxax+ byzax + by yxyay + bz3二右边=(a _ b)= (b_a)(b_a)12分别再分xyzyzx3ayzx + b3zxyzxyxyz(1)2xyzxyz3ayzx + b3yzxzxyzxy(2a1)2a2a右边左边b22cd2b22cd2C3一Ci2ab22cd2(2b 1)(2c 1)+ 1) (d十2)24a+ 46a +

6、94b+46b+ 94c+ 46c + 94d + 46d + 9(2d2a 12b 12c 1(d(a 2)2(b 2)2(c 2)2(a 3)2(b 3)2(c 3)23)22aa4a + 46a+ 92a14a +46a +9b2b4b +46b+ 9+b214b +46b+92cc4c + 46c + 92c14c +46c +9d2d4d +46d + 9d214d +46d +92d 1按第二列分成二项278=(b _ a)(c _ a)(d _ a) b + ab2(b + a)=(b _ a)(c _ a)(d _ a)0c- bc2(c a)-b2(b a)a)(d _ a)

7、(c _ b)(d -0 d - b d2(d +a)-b2(b+a)b)(c=(a - b)(a - c)(a-d)(b - c)(b- d) (c - d)(a b c d)(5)用数学归纳法证明假设对于（n - 1）阶行列式命题成立，即Dn_ xn最-2则Dn按第1列展开：Dn= XD-1為(-1)11所以，对于n阶行列式命题成立第一项 5 一4c2c 6c2第二项c 4c2c4- 9c21a2a 4 9 b2b 49 c2c 4 9 d2d4 9 ) 0ab - ac a左边 =2.2222ab -ac -a4.4444ab -ac -ab - ac-a2222=b - ac -ab2

8、(b2a2)c2(c2a2)a21b21c21d214a 6a4b 6b a =04c 6c4d 6dd - a.2 2d -a.44d - ad - a.2 2d - ad2(d2-a2)0c ac2(c + a)d + ad2(d + a)1b + ab2(b + a)=(b- a)(c -1 1bc+ b2) + a(c+ b) (d2+ bd+ b2) + a(d + b)当 n = 2 时,D2=a2-1x a1=x2a1xa2,命题成立.an _2Xan_1,xDn -1an5.设n阶行列式D二det（aij）,把D上下翻转、或逆时针旋转90、或依59an1annam annann

9、a1nD1 =aa，D2 =aa，D3 =aana1nanan1an1a11副对角线翻转, 依次得n(n J)证明证明DiP二D2十 1)D,D3二D = det(ajj)anainaniannDiH-i)n-ianiannaiia21a2n同理可证D2（_1）2（_1严a2iania31(-1 严(-1)2 (_1)ai1ania3nainannn(n/)I)1 22) )(f 十 j2Dn(n/)ai1anin(n /)n(n /)十 1)2=(1)2D (1)2Dalnannn(n/)n(n 4)n(n /)D3十 1)2D2(1)2(1)2D = (i)n(ni)D = D6.计算下列

10、各行列式（Dk为k阶行列式）：10Dn(a1)n(a - n)nn-1an -1(a-1)n-1(a - n)n 1提示：利用范德蒙德行列式的结果.bnD2na1bid1dnDn二 det(aj 其中 aj1 a1Dna2，其中a1a an = 0.an按最后一行展开11从第n 1行开始，第n 1行经过n次相邻对换，换到第 1 行，第nn(n + 1)行经( (n - 1)次对换换到第 2 行，经n (n - 1) 1次行2交换，得000a00卜 1 严0a0000(再按第一行展开)0(2)X2)十 1)a+ (-1)2nan n n _2a a - an _22a (a -1)Xaaaa x

11、 x _ a00Dn = a x0 x a0a x000 x a再将各列都加到第一列上， 得x + (n -1)aaaa0 x a00Dn =00 x - a00000 x aa( (n_2)(n_2)(2)将第一行乘(-1)分别加到其余各行，得n 1二X (n -1)a(x -a)12andnan -1dn-1dnan1(1)2n1bn111aa -1a nna (a-1)n1(a-n)na(a-1)n(a-n)nn(n 1)Dn 1二( (1)2此行列式为范德蒙德行列式Dn 1二( (1)n(n 1)2(a i 1) (a - j 1)n 1j _1n(n 1)n(n：；1)(1)2(ij

12、) 十 1) (1)j)n (i -n1j _1j)按第一行a1bi展开and1a1b1bnD2na1C1d1bn_1CnCn Abn-113dn_10Cn14都按最后一行展开 a“dnD2nd -由此得递推公式:D2n=(and.- QCn)D22D2nn-Ji.i (aidi- biCi)D2i =2D2a1di二a1d1-4GD2nn=ni =1(aidibe)012310122101Dn =det(aij)=3210n - 1n-2 n-3n -1111B-1-1111 -r-1-1-11a BE - r3 -1-1-1-1 Bn -1n-2n-3n - 4-1000-0-12000-

13、12-200-12-2-20n - 12n32n 42n - 5n -4101111(5)= i -=(1)n1(n - 1)2n2151 + ai111 + a2Dn =.1 1a100-a2a200- a3a300- a4 SB B B B0000001 + an000000001111C4,按最后一列 展开（由下往上）(1 anXaa a-a100-a2a200- a3a3an_1an_110-an1 + ana100 00- a2a20 000-a3a30000a400000 - andan-2000 00000000+00000-an000000-a2a200- a3a300- a4

14、-an_1an_10-an000000-an-1an_10-anaa=(1an)(a1a2an_1) )an -3an -2an十a2a3an16n1二(aia2an)(1)i g7.用克莱姆法则解下列方程组：XiX2X3X4= 5,Xq+ 2x2_ x3+ 4x4= -2, (1)2xq_ 3x2_ x3_ 5X= -2,3xqx22x311x4二0;5xi6x2=1,xq5x26X3二0,(2)X25X36X4二0,X35X46X5二0,iX4+ 5x5= i 1 11 21-1解D=2-313 12111101-2300-13800_ 51451111711111401 -23_ 50-

15、5 -3-7110-2 -18111101 _ 23=-14200 _ 1-54000142511115_ 1915_ 1-9050901211013_ 323050901211013_ 3 -23012 110121118-1-11-5-1-91-5-1-9012110121100-10-46-00-1380023120000142二142151115111-2-140 -7-232-2-1-50 -12-3-7302110 -15-185111511-1320-1320231100-1-1903931000-28411511224-4262-32-53101110001115-284D3D

16、2142D4=19（D为行列式D中an的余子式,D为D中叭的余子式,D ,D 类推）D11,X2X4-1按最后一行展开5D -5D6D5(5D6D )- 6D9D-30D65D114DFFF F=65 佃 114 5=66520二D 64二19D - 30615075 6 0315 6 - 5汇63= -65 -1080= -11450 155610011500560 015000按第三列01601500D3 = 01060+展开00560160000560015005600115Di1000651000 06 05 61 50 1按第一列D展开0 0 06 0 05 6 01561 6 05

17、 6 00 5 6+ 615 00 150 1 6=19 6 114二703D25 10 010 6 00 0 5 60 0 150 10 1按第二列展开16 0 00 5 6 00 15 60 0 155 0 0 016 0 00 5 6 00 15 6D45 6 0 115 6 00 15 00 0 100 0 0 1按第四列展开51006 05 61 50 1000621560-5156=39501522以 + x2+ x3= 08问人，卩取何值时，齐次线性方程组 乙+X2+X3= 0有非零解?X + 2 咲 + x3= 0人11解D3= 1卩1 =卩一収，1 2卩1齐次线性方程组有非零

18、解，则D3= 0即 - -0得二0或二1不难验证，当- 0或二1时，该齐次线性方程组确有非零解(1 _ 几冈 _ 2x2+ 4x3= 09问取何值时,齐次线性方程组2X1 (3 - ) )X2 x3= 0X1 + X2+ (1 -巧 X3= 0有非零解?解1 - &-241 h-3 +人4D =23-扎121 -二二1111 k101 -扎二(1 -) )3( -3) 4(1 -) )- 2(1 -) )( 3 -)=(1 -) )32(1 -) )2- 3齐次线性方程组有非零解，则D = 0560011156015600按最后列0156D5= 01560展开0015001500001

19、00011212150765x2=-665X3665X4-395665X4212665D = 1211二114570323不难验证，当=0, = 2或=3时，该齐次线性方程组确有非零解矩阵习题111f123】1.设A二11-1B =-1-24 1,-11051求3AB 2A及ATB解111f123、卩58ATB=11|-11-1-240-5611-1110512902.计算下列乘积:卩11、f123卩11、3AB - 2A = 311-1-1-24-211P-1-11丿051-11二34151 73 2;0川-30a12(Xi,X2,X3) )ai2ai3卩2101010K11 II00200

20、03A1000100a23031,2,3 2；1112a13X1X2；a23a33八X3丿1-13一31 - 1,2;3丿卩58、卩11、1f-213220-56-211-1-2-1720290丿1 -111yi49丿( (1 2 3) 2丨=(1 x 3 +2x 2 +3汇1)= (10)J丿公2疋(_1) 22(-2 4、11 1(-1 2)=仆(1) V 2-1 2 1。丿亠(-1)3汽2,3 6丿AB二BA吗？( (A B)2二A22AB B2吗?( (A B)(A- B)二A2- B2吗？解1 21 0、A =B =1 31 2311广2 14 0A0-12 11广6-78、1 -1

21、3 41-3120-5- 640211a12a13X：X1X2X3:a12a22a23x 11 1 1la13a23a33八X3丿ai2X1a22X2a23X3ai3X1a23X2a33X3XX2uaXq1X3+ a33xj +2ai2XqX2 +2a1sX1x 2023X2X331卩2521012-233100403丿0 00f1 2、3设A =H 3丿1 0、B =，问:11 2丿a11x1ai2X2azX1220 10 10 10 0 2 1 0 010 0 03丿i0 02I-432638、5 02 8、A2- B2=l -=14 11丿&4J 7故( (A B)(A- B)

22、= A2- B24 举反列说明下列命题是错误的：2(1)若A = 0,则A = 0；(2)若A2=A,则A=0或A =E；(3)若AX二AY,且A = 0,则X二Y.2A = 0,但02A2二A,但A二0且A二( (11、1X =Y =1 1丿0AX二AY且A = 0但X Y(15设 A =(A+ B)(A- B)=2 20 2S 6、=2 50 10 9故( (A B)2二A22AB B2则AB(A B)2二(2(1 2) BA =3 8丿2计2 2、 8 5八2 5厂但A22AB B20481161丿10利用数学归纳法证明：Ak=31当k = 1时,显然成立,假设k时成立,则k +1时1

23、0、10、1 0Ak= AkA =W 1a1Jk+1110A3=A2A =282(k 1)k1k 1由数学归纳法原理知6设 A =0扎11,求Ak00解首先观察510、p. 10fa 2口1A2=010扎1=0扎22A00切（0 000210九I 23 A3入1A A2A =0n 3A3扎21100入31丿01kk-2k*-1ik(k -1)由此推测2k_1k.k(k-2)用数学归纳法证明：当k=2时，显然成立.假设k时成立，则k 1时,kkAk41=AkA=0k(k -1)心-K2k1knk0ZII00&丿29(k 1)(k 1)k -1kk_1由数学归纳法原理知：Akk(k1) *

24、2AI2k z7设A,B为n阶矩阵，且A为对称矩阵，证明TAB也是对称矩阵.30证明 已知：AT= A则（BTAB）T= BT（BTA）T= BTATB =BTAB从而BTAB也是对称矩阵.&设A,B都是n阶对称矩阵，证明AB是对称矩阵的充分必要条件是AB二BA.证明由已知：AT= A BT= B充分性：AB = BA =AB = BTAT= AB = （AB）T即AB是对称矩阵必要性：（AB）T= AB二BTAT= AB二BA = AB.9.求下列矩阵的逆矩阵：A =1A11 =5, A21= 2(1),码 2=2 汇（-1）,血护A11A215-2A_11AA=A = |AVA12

25、A22丿21丿|A|_152故A=-21丿A =1式0故A1存在每=cos A21= sid2= -siMA22= cost coSsin9从而AT =sin日coS丿A= 2,故A1存在*岂f52000rijrf210 0 10；(5)008341 0 01 2 02131 2 11 2coS - sin日;2 5.sin日cos丿卩2一1】134-2 115-41丿ai0解a2(aan= 0)an31A11二24A22二12A33二8A44二610 0人2=（-1）3230=-1211412 0人4十1）521 3=3121100人24=(-1)6213=-5121_11屮A = A|A|

26、Af10011022故A1=_111263151 82412A = 1鼻0故A1存在而人1=1A21 = -2A12 =_2A22= 5A-3二0A230人4= 0A24= 0120A13= (-1)42 10 = -12124100A23=(-1)52 10 = -4124100人34=(-1)712021210、0014丿A31=0A41=0A32二0A42二0A33= 2忑=-3A34 = 5A44二8A11 = -4A21=2A31=0A12= -13A22 =6A32=-1A13= -32A23 =14A33=-2r-210、A-XA*1331|A|一22i 71A =2 11 2A

27、l = 24I 101 2000 03 01 4二03211- 2从而A =0 0-200I50002-3io设A*= O(k为正整数)，证明(E - A)二E A A2AkJ.证明一方面，E = (EA)(E A)另一方面，由Ak= O有E =(E - A) (A - A2) A2- - Ak( (Ak1- Ak) =(E A A2Ak)(E - A)故( (E - A)/(E - A) = (E A A2Ak1)(E - A)两端同时右乘(E - A)，就有( (E - A)1二E A A2Akii 设方阵A满足A2- A - 2E二O,证明A及A 2E都可逆,并求A，及(A 2E).证明

28、由A2 A 2E = 0得A2- A=2E两端同时取行列式：A2- A = 2即|A|A E = 2,故|A鼻02所以A可逆,而A 2E二A2 2A + 2E=A = A工0故A十2E也可逆.由A2-A 2E=O二A(A E)=2E=A=A(A _ E) = 2A=E二AIAE)aiAa2A二0由对角矩阵的性质知10I. IIan丄a1宀I01a2丄an丿33又由A2- A-2E =0= (A 2E)A-3(A2E)二-4E=(A 2E)(A - 3E) = -4E(A 2E)(A 2E)(A-3E) = 4(A 2E)1(A 2E)1二(3E - A)412 .设n阶矩阵A的伴随矩阵为A，证明：若|A| = 0,则A