利用對(duì)角線法則計(jì)算下列三階行列式

1、第一章 行列式1利用對(duì)角線法則計(jì)算下列三階行列式2 0 1(1) 1-4 -1 ；-1832 0 1 解 1-4 -1-183=2 (-4) 3 0 (T) (-1) 1 1 8-0 1 3-2 (T) 8T (_4) (_ 1)=-24 8 16-4 = -4abcbcacababc解bcacab二acb bac cba-bbb-aaa-ccc =3abc_a3_b3_c31 1 1 a b c ； a b c111解abc2 . 2 2abc=bc2 ca2 ab2-ac2-ba2-cb2 =(a_b)(b_c)(c_a)x=x(x y)y yx(x y) (x y)yx-y3-(x y)

2、3-x3= 3xy(x y)-y3-3x2 y-x3-y3-x3一 2(x3 y3)2按自然數(shù)從小到大為標(biāo)準(zhǔn)次序求下列各排列的逆序(1) 1 2 3 4解逆序數(shù)為0(2) 4 1 3 241-43-42-323 2- 3 1- 4 2- 4 1, 2 12 1 - 4 1 - 4 3解逆序數(shù)為4(3) 3 4 2 1解逆序數(shù)為5(4) 2 4 1 3解逆序數(shù)為3(5) 1 3- (2n-1) 2 4- (2n)解逆序數(shù)為凹© :23 2 (1 個(gè))5 2 - 5 4(2 個(gè))7 2 - 7 4 - 7 6(3 個(gè))(2n-1)2. (2 n-1)4. (2n-1)6.(2 n-1)(

3、2 n-2) (n-1 個(gè))(6) 1 3(2n-1) (2n) (2n-2)2解逆序數(shù)為n(n-1):3 2(1 個(gè))5 2- 5 4 (2 個(gè))(2n-1)2- (2 n-1)4. (2n-1)6，(2 n-1)(2 n-2) (n-1 個(gè))4 2(1 個(gè))6 2 - 6 4(2 個(gè))(2n )2.(2 n) 4(2 n) 6，(2n)(2 n-2) (n-1 個(gè))3 寫出四階行列式中含有因子ana23的項(xiàng)解含因子ana23的項(xiàng)的一般形式為(T)a11a23a3a4s其中rs是2和4構(gòu)成的排列.這種排列共有兩個(gè)即24和42 所以含因子ana23的項(xiàng)分別是t1(T) a11a23a32a44

4、 = (T) ana23a32a44二-ana23a32a44(T)tana23a34a42 = (T)2a11a23a34a42=ana23a34a424 計(jì)算下列各行列式：4 12 41 2 0 2；10 5 2 00 1174124C2 - Q4-12-101202120210520C4 - 7 C31032-1 10122 X 1)44310 3 144110-12310-214901710 -214=03-121123250622141C4 - c221404- D21403-1213-1223-1221232123012305062506221402114

5、23101-12042300200一 ab ac ae(3) bd- cddebf cf- ef-abacae-b c ebd-cdde=a d fb -c ebfcf efb c -e-111=adfbce1-11= 4abcdef11-1a-1001b-100 0 1 0 c 1 -1 da100A + ar2解-1b100-1c100-1d2-H1 + ab <= (1)(-1)-1 (0 -= (-1)(-1)3 421 +ab-1 101 + ab a 0-1b100-1c 100 -1 d0c3 + dc21 + ab a ad1-1c 1 + c1 d0-1 0ad + c

6、d=abcd+ab+cd+ad+15.證明:2a(1)2a1abab22b1= (a-b)3;證明=(-1)3 12-ab -aabb2 _a2b -2a=(b _ a)(b _ a)aba1 2= (a-b)3 ax +by(2) ay +bz az +bxay bz az bx ax byaz bxax byay bzx= (a3+b3)yz2 aabb2c2 - c12 a.2.22ab - a b -a2aa +b2b2ab-a 2b-2a111C3 _ c110 0證明axbyaybzazbxaybzazbxaxbyazbxaxbyaybzx ay +bz =a y az + bx

7、z ax +byaz +bx yax + by + b zay +bz xay bz az bx ax byaz bx ax byay bz2x ay + bz zy z az + bxy az + bx x+ b2z x ax + byz ax +by yx y ay +bz二 axyzyzx3=ayzx+ b3zxyzxyxyzxyzxyz3,3=ayzx+ byzxzxyzxy-(a3 b3)2 a b22Cd2證明2(a 1) (b 1)2 (c 1)2 (d 1)22(a 2) (b 2)2 (c 2)2 (d 2)23)23)23)2 (d 3)2(a(b(C=02ab22Cd2(

8、a 1)2(b 1)2(c 1)2(a 3)2 (b 3)2 (c 3)22(d 1)2 a b22 Cd2(a 2)2(b 2)2(c 2)22(d 2) (d 3)(Q-C3 C3_C2 C2 Ci 得)2 a b22 C d22a+ 12a+ 32a+ 52b+ 12b432b+ 52c+ 12c+ 32c+ 52d+ 12d+ 32d+ 52a+ 1222b+ 1220 2c+ 1222d+ 1222(C4_C3 C3-Q 得)11a2 a4 a1bb2b4c2 c4 c1dd2d4=(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)(a b c d);證明1a2a4a1bb

9、2b41c2c4c1dd2d410001b -ab(b -a)2 22、b (b -a )1c -ac(c _ a)2 22、(c -a )1 d -a d(d - a) d2(d2-a2)=(b-a)(c -a)(d -a)=(b-a)(c - a)(d - a)1bb2 (b a) c2(c a)1dd 2(d a)1 10c -b0 c(c 一 b)(c + b + a) d (db)(d + b+a)=(b -a)(c-a)(d -a)(c -b)(d -b)1 1c(c b a) d(d b a)=(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)(a bed)X-1000

10、0X-100* * * * * * * * *000X-1anan 4an 2“2X + 6二X門乜靈一1 . . . . anX an 證明用數(shù)學(xué)歸納法證明當(dāng)n=2時(shí).d2X -1a2 xa1x2 a1 a2 -命題成立假設(shè)對(duì)于(n-1)階行列式命題成立即n_1n-2Dn /二Xai Xan/X an_1則Dn按第列展開有-1 00 0Dn=XDn+an(-1)“x -10 01 1 x 1f 丄n丄 nV=xD nj+a n=x +a1x +'+an_1x+an，因此對(duì)于n階行列式命題成立6設(shè)n階行列式D二det(aj),把D上下翻轉(zhuǎn)、或逆時(shí)針旋轉(zhuǎn)90、或依副對(duì)角線翻轉(zhuǎn)依次得a n

11、1'''anna1 n'"anna nn'''a1nD1 =a11a1n- D2 =a11'''an1-D 3 =an1a11n(n)證明 D廣 D2 =(-1)D DD -證明因?yàn)镈=det(aij).所以a n1T T fO6n=(-1)2a11an1T T TOa1nT T TOa nna11T T fOa1na21a- -r -rOa2nDia11a1 na21'''a2na n1''anna31'''a3nn -2=(-1)2(-1

12、)n(n -1)1 2；.(n -2) (n L)= (T)D =(T)同理可證n(na11'1 ' an1n (n J)2=(-1) 2a1 nannn(n _J)n(n /)n(n _J)2D2=(一1)2 (-1) 2D2=(-1)D3 =(-1)D=(-1) 2 D .= (_1)n(n)D =D dtn( n J)7 計(jì)算下列各行列式（Dk為k階行列式）:(1)Dn,其中對(duì)角線上元素都是未寫出的元素都是0 -n 1= (T)(T)n an =a n_an_2二an-2/ 2(a1)a00010a000Dn =00a00（按第n行展開）000a01000a00001a0

13、000a/八n F= (T)0a000+ (-1)2n a*a000a0(n 4)n )(n -J) (n-1)a(n，)(n /)xaaaxaaaxDn =解 將第一行乘（-1）分別加到其余各行得再將各列都加到第一列上x (n -1)aDn 二=x (nT)a(x-af= (T) 2丨J 一 j)nz八na (a 1) (a_ n)nn/八a(a 1)n -1 (a-n)n(3)Dn 十*Jaa -1*a _ n1 1*1解根據(jù)第6題結(jié)果-有11 1n (n 比)aa -1a nDn寺=(-1) 2n -1 a(a-1)2(a-n)n'n a(a-1)n(a-n)n此行列式為范德蒙德

14、行列式-n (n 出)D卄= (T) 2n(a i+1) (aj+1)0000x an(n 1)n 1.j _1n 1rj _1n （n 二）亠 “1= (T) 2(T)an2nI 丨(i - j) n 1.j _1bnaibiCidi2naibiCn -J0(i)Ciai 0Ci di2n ibn0 an J（按第1行展開）bn_i 0dn/00 dnbn ai biCi diCn -4Cn再按最后一行展開得遞推公式D2n_andnD2n-2 _ bn CnD2 n-2 即 ID2n - (and n_bn Cn) D 2in-2于是而所以D2n" (aidi-bidD2 =a1C

15、1bid1二 ad _ bCD2n Jll佝dj -b&)'(5) D-det(aj)其中 aj = |i-j|;解 aj=|ij1-1-1111-1-1-111-1-1-1-11n -1n -2n 3n 40-11111C2 C1C3C1-200 0-2-20 0-2-2-202n 3 2n 42n 5n 一0000n -11-1-1-1-10123n -11012n _ 22101n_33210n _4* * * * * * * * *n -1n - 2n 3n -4 0Dn =det( aj)二= (T)n"(門-1)2心a1Dn1a2,其中 a1aan=0an

16、1 +a1111a2c1a100 001a；2a20 0010-a3a3 001000-ana n -二 1000 0- an 1 +10000_1a1-11000-Aa2'an0-1100-Aa3000 一11-A an Ja101000-Ja2'an00100-Ja300001an 二n00000V Z aimnA an)(1+zi m-)ai 1 anC2 一 C3=品2二 aa2二 a?an8 用克萊姆法則解下列方程組乂 + x2 + x3 + X4 =5X1+2X2-X3+4x4 = -2；2x - 3x2 - x3 - 5x4 = _23

17、x1 x2 2x311 x4 =0解因?yàn)?mH寸 8CXIH9000tqi 999"OH OHOH90009000:xg十X X9 +X9+X X9 + exg+XX9 +xg +><“ Zx9;g9CXI寸 HCXI寸lhCXI寸 l hH ox0CXI01二<N)-5610015000D3 =01060= 703000560011560015560101560001500= -395 .0010600015所以x115071145665665X3 =703665X4 =-395665X4212665片 x2 x3 = 0 9 問(wèn)丄取何值時(shí)-齊次線性方程組 % % X3 =0有非% + 2 卩 x2 + x3 = 0零解？解系數(shù)行列式為人11D =1 卩1 =