第一次作业 1计算下列行列式 183 =2×(-4)×3+1×8×1-1×(-4)×(-1)-(-1)×8×2=-24+8-4+16=-4 54 2)15123|=215123|=103123|=0 20321220321243212 bd - cd de=adf b -c e=abcdef 1 -1 1=abcdef 0 -2 0=4abcdej C 0 2证明下列各式 b2 1)2aa+b2b=(a-b) 111 a2 ab b2a2-b2 b(a-b) 2ab+b200 证明:2aa+b2b=2a-2ba-b0=2a-2ba-b0=(a-b X 2)yx+yx|=-2(x3+y2) ty x 证明: y x+y 2(x+y) y x+ y xty x (x+y)1 x+y x+yx x+I x =2(x+y)0xy=2(x+-x2x-y)=-2(x2+y) x 3求下列各排列的逆序数 )13478269(1347826=0+0+0+0+4+2=6 2)987654321t(987654321)=1+2+3+4+5+6+7+8=36
ԛྡྷұᆴྜ 1.䅵ㅫϟ߫㸠߫ᓣ 1) 201 1 41 18 3 - - - = = ´ - ´ + ´ ´ - ´ - ´ - - - ´ ´ =- + - + =- 2 ( 4) 3 1 8 1 1 ( 4) ( 1) ( 1) 8 2 24 8 4 16 4 2) 10 8 2 15 12 3 20 32 12 = 5 4 1 14 1 2 15 12 3 10 3 12 3 0 20 32 12 4 32 12 = = 3) 11 1 11 1 1 11 0 20 4 1 1 1 00 2 ab ac ae b c e bd cd de adf b c e abcdef abcdef abcdef bf cf ef b c e - -- - = - = -= - = -- - - 2.䆕ᯢϟ߫ᓣ 1) 2 2 3 2 2( ) 111 a ab b aab b ab + =- 䆕ᯢ˖ 2 2 22 2 2 3 ( )0 2 0 0 2 2 2 2 0 2 2 0( ) 1 1 1 1 1 1 1 11 a ab b a b b a b a ab b aab b a b ab a b ab ab - - -+ + = - - = - - =- 2) 3 3 2( ) x y xy y xy x x y xy x y + + =- + + 䆕ᯢ˖ 2 2 33 2( ) 1 2( ) 2( ) 1 2( ) 1 1 2( ) 0 2( )( ) 2( ) 0 x y xy xy y xy y xy y xy x xy xy x xy xy x xy x y xy x y x y y xy x y x y x y x xy y x y xy x ++ + + + = + + =+ + + + + = + - = + - + - =- + - - 3.∖ϟ߫ᥦ߫ⱘ䗚ᑣ᭄ 1) 13478269 t(1347826)=0+0+0+0+4+2=6 2) 987654321 t(987654321)=1+2+3+4+5+6+7+8=36
3)135..(2n-1)(2n)(2n-2)42 τ(135..(2n-1)(2n)(2n-2)42)2+4+6+.+2(m-1)=m(n-1) 4在五阶行列式中,下列各项之前应取什么符号? 1)a3a24a244a5解:由于τ(34215)=2+3=5,所以-a1224a24145 2)a211y4a542解:由于(21354)(134522+3=5,所以-a21a13414a5a42 5写出四阶行列式中含有因子a1a23的各项 F:-a1a23a32a4, a1a23a3a42 6计算下列各行列式 3-12 056-2 0 506 056 056 41 105 2021 4 12021202 202 10520105201-152-200-152-20 011 01170117 1202 1202 001785 0015 0acca 9ddbb
3) 1 3 5 … (2n-1) (2n) (2n-2)… 4 2 t(1 3 5 … (2n-1) (2n) (2n-2)… 4 2)=2+4+6+… +2(n-1)=n(n-1) 4.Ѩ䰊㸠߫ᓣЁ,ϟ߫乍ПࠡᑨপҔМヺো? 1) 13 24 32 41 55 aaaaa 㾷˖⬅Ѣt(34215)=2+3=5ˈ᠔ҹ 13 24 32 41 55 -aaaaa 2) 21 13 34 55 42 aaaaa 㾷˖⬅Ѣt(21354)+ t(13452)=2+3=5ˈ᠔ҹ 21 13 34 55 42 -aaaaa 5.ߎݭಯ䰊㸠߫ᓣЁ᳝ᄤ 11 23 a a ⱘ乍. 㾷˖ 11 23 32 44 11 23 34 42 -aaaa aaaa , 6.䅵ㅫϟ߫㸠߫ᓣ 1) 214 1 3 12 1 123 2 506 2 - - - = - - 㾷˖ 2 1 4 1 1 24 1 1 2 4 1 3 12 1 132 1 0 5 6 2 0 1 2 3 2 2 13 2 0 3 5 0 5 0 6 2 0 56 2 0 5 6 2 -- - - -- - - =- =- = - - -- -- - 2) 4 124 1 202 10 5 2 0 0 117 = 㾷˖ 4 124 1 202 1 2 0 2 1 2 0 2 1 202 4 124 0 7 2 4 0 7 2 4 10 5 2 0 10 5 2 0 0 15 2 20 0 15 2 20 0 117 0 117 0 1 1 7 0 1 1 7 12 0 2 1202 01 1 7 0117 17 9 0 0 0 17 85 0 0 1 5 0 0 9 45 0 0 1 5 -- -- =- =- =- -- -- = =´ = 3) cadb acdb acbd cabd =
00 d b0 0 d-b b-do 0 d-b b-d 100 1b10 解 00|01+aba 1b10|-1b100-1 11|0-1 0-1 0 1+ab a 00 ab a+c 1dl00 1 b 0 1 b 0 00 a+c+abc 1+ab0 0 00 00 a+c+abc 1+ab 1 b 1+abad t dc +abcd 00-1 000 1+ab+d(a+c+abc 7计算下列各行列式 x+(n-1)ax+(n-1) x+(n-1)a [x+(n-l)a] [x+(n-l)a] =(x-ax+(n-1)a] 00 x-a
㾷˖ 00 00 00 00 0 00 0000 cad b caac caac acd b dbbd dbbd ac bd acca cabd c a b d c a b d -- -- -- -- === - - 4) 1 00 1 10 01 1 00 1 a b c d - = - - 㾷˖ 1 0 0 01 0 1 1 0 1 1 0 1 10 1 10 0 1 1 0 1 1 0 1 1 0 1 1 01 0 0 1 00 1 0 0 1 0 0 1 00 1 1 1 01 1 0 01 1 01 1 00 1 00 1 00 1 00 1 11 0 01 1 1 00 1 0 0 01 ( ) a ab a b b bb cc c c ab a ab a c d d dd b b c c a c abc ab d d a c abc ab b c ab ad dc abcd d ab d a c abc +- - -- - - === -- + + - - -- - - - - = =- ++ + - - ++ + - - =- = + + + + - + + ++ 7.䅵ㅫϟ߫㸠߫ᓣ 1) ... ... ... ... ... ... ... xa a ax a aa x = 1 2 1 ... ( 1) ( 1) ... ( 1) 1 1 ... 1 ... ... ... [ ( 1) ] ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... 1 1 ... 1 0 ... 0 [ ( 1) ] ( ) [ ( 1) ] ... ... ... ... 0 0 ... n i i r r n xa a x n ax n a x n a ax a a x a ax a xn a aa x a a x aa x x a x n a xa x n a x a = + - +- +- +- å = =+- - =+- =- +- -
b 2 f(x)=a bbb c2 d =(x-a)(x-b)(x-c)(x-d)(d-a)d-b)(d-c)(c-a)(c-b)(b-a) (x(a+b+c+d)x'++abcdd-ad-bd-cc-ac-b(b-a) b b d f(x)=a2 b 1C8 b a b3 c3 d3x b d 6. c b c d+ x b b c d b'c d b2 c2 d 1111 (a+b+c+dd-a((c-a(c-b(b-a) 000 000 0 解
2) 222 2 444 4 1111 abcd abcd abcd = 㾷˖䆒 222 22 333 33 444 44 4 3 11111 ( ) ( )( )( )( )( )( )( )( )( )( ) ( ( ) ... )( )( )( )( )( )( ) abcd x f x abcd x abcd x abcd x x ax bx cx d d ad bd cc ac bb a x a b c d x abcd d a d b d c c a c b b a = =- - - - - - - - - - = - +++ + + - - - - - - 㗠 222 2 222 2 222 22 333 3 333 3 333 33 444 4 444 4 444 44 234 333 3 222 2 222 2 444 4 444 4 333 3 11111 1111 ( ) 1111 1111 1111 abcd abcd x abcd abcd fx x abcd x abcd abcd abcd x abcd abcd abcd x abcd abcd abcd xxx abcd abcd abcd abcd abcd abcd = =- +-+ 222 2 444 4 1111 ( )( )( )( )( )( )( ) abcd a b c d d ad bd cc ac bb a abcd abcd = +++ - - - - - - 3) 0 ... 0 0 0 ... 0 0 ... ... ... ... ... ... 0 0 0 ... 0 0 ... 0 x y x y x y y x 㾷˖
00 00 00 0 x 0010x 000 000 000 y00….0 000 x n-1) =x2+(-1)+y 123 11 n(n+1) 0 1-n1|= 11.1-n1 11 n(n+1) 1111 0…0-11.…1 n(n+1) n n(n+1) 00 000 000 (-1) i n(n+D) 000 00.0 0 2xn(n+1)n2 8证明下列等式 a1 a2 11|=aa0+∑1 11+
1 ( 1) ( 1) 1 0 ... 0 0 0 ... 0 0 0 0 ... 0 0 0 ... 0 0 0 ... 0 0 0 ... 0 0 ... ... ... ... ... ... ... ... ... ... ... ... ... . ( 1) .. ... ... ... ... 0 0 0 ... 0 0 0 ... 0 0 0 ... 0 0 0 ... 0 0 0 0 ... 0 0 0 0 ... ( 1) n n n n nn xy xy y xy xy xy x y xy xy y y x x xy x y + - - + = +- = +- 4) 1 2 3 ... 1 1 1 1 ... 1 1 1 1 1 ... 1 1 ... ... ... ... ... ... 1 1 1 ... 1 1 n n n n n - - - - 1 ( 1) 1 2 3 ... 1 2 3 ... 1 2 1 1 1 ... 1 1 0 1 1 ... 1 1 1 1 1 ... 1 1 0 1 1 ... 1 1 ... ... ... ... ... ... ... ... ... ... ... ... 1 1 1 ... 1 1 0 1 1 ... 1 1 1 1 1 ... 1 1 1 1 1 ... 1 1 ( 1) 1 1 1 ... 1 1 2 ... ... ... ... ... ... 1 1 1 ... 1 1 1 ( 1) 2 n n n n n n n n n n n n n n n n n n n n - + - - - - - = - - - - - + = - - + = 1 1 1 2 1 1 ... 1 1 1 1 1 ... 1 1 1 1 1 ... 1 1 0 0 0 ... 0 ( 1) 1 1 1 ... 1 1 0 0 0 ... 0 0 2 ... ... ... ... ... ... ... ... ... ... ... ... 1 1 1 ... 1 1 0 0 ... 0 0 0 0 0 ... 0 0 0 0 ... 0 ( 1) ( 1) ( 1 0 0 0 ... 0 0 2 ... ... ... ... ... ... 0 0 ... 0 0 n n n n n n n n n n n n n n n - - + - -- - - - - + = - - - - - + = - =- - - ( 2)( 1) 2 ( 1) 2 ) 2 n n n n n n - - + - 8.䆕ᯢϟ߫ㄝᓣ 1) 1 2 3 12 1 1 1 1 ... 1 1 1 1 1 ... 1 1 1 1 1 1 ... 1 1 ... (1 ) ... ... ... ... ... ... 1 1 1 ... 1 1 n n i i n a a a aa a a a = + + + =+ + å
1+a,1 000 11+a3…11 a3…00 11 a1 400 000 (1+ 4012…012 000 n 000 000 21.00021.00 000.12000 000 21 +012 000 000 000….01000 n1=1+ln2…,12=1+l1,1=2 Ⅰ=n+1 9用 Cramer法则解下列方程组 x+x2+x3+x4=5 x,+5x,+6x 0 x1+2x2 x2-x3-5x4 +5x,+6x,=0 3x,+x,+2x2+1lx1=0 x4+5x4=0
1 1 2 12 3 13 1 11 1 1 2 3 2 3 1 1 1 ... 1 1 1 1 1 ... 1 1 1 1 1 ... 1 1 0 ... 0 0 1 1 1 ... 1 1 0 ... 0 0 ... ... ... ... ... ... ... ... ... ... ... ... 1 1 1 ... 1 1 0 0 ... 0 1 ... 1 1 ... 1 1 0 0 ... 0 0 0 0 ... 0 0 ... ... ... ... ... ... 0 0 0 ... 0 n n n a a a aa a aa aa a aa a a aa a a a a + + + - + - = + - + + + ++ = 11 1 1 2 12 2 3 1 1 (1 ... ) ... ... (1 ) n n n n n i i aa a a a a aa a aa a a = =+ + + ++ = +å 2) 2 1 0 ... 0 0 1 2 1 ... 0 0 0 1 2 ... 0 0 1 ... ... ... ... ... ... 0 0 0 ... 1 2 = + n 1 1 1 0 ... 0 0 1 1 0 ... 0 0 1 1 0 ... 0 0 1 2 1 ... 0 0 1 2 1 ... 0 0 0 2 1 ... 0 0 0 1 2 ... 0 0 0 1 2 ... 0 0 0 1 2 ... 0 0 ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... 0 0 0 ... 1 2 0 0 0 ... 1 2 0 0 0 ... 1 2 1 1 0 ... 0 0 0 1 1 ... 0 0 0 0 1 ... 0 0 ... ... ... ... ... ... n I + = =+ = 1 1 1 2 2 11 2 1 0 ... 0 0 1 2 1 ... 0 0 0 1 2 ... 0 0 1 ... ... ... ... ... ... 0 0 0 ... 0 1 0 0 0 ... 1 2 1 ,..., 1 , 2 1 n n n n n I I I I II I n - - - - + =+ =+ =+ = = + 9.⫼ Cramer ⊩߭㾷ϟ߫ᮍ㒘 1) 1234 1 23 4 1 23 4 12 3 4 5 2 42 23 5 2 3 2 11 0 xxxx x xx x x xx x xx x x ì +++= ï ï + - + =- í - - - =- ï ï î ++ + = 2) 1 2 123 234 344 4 4 5 6 =1 5 6 0 5 6 0 5 6 0 5 0 x x xx x xxx xxx x x ì + ï ++ = ï ï í ++ = ï ++= ï ïî + = 㾷 1)
100 142 000 2110 1005 237 11 2-141-14-2|0-23-701105 D2 30211321100 0018-10 D3 0001 426 5-12-700-478 31011 15800-291 =142 0-5-3-1200-13-47 D D, D Da x4= D D 解2)
1 1 1 1 1 1 1 1 11 1 1 1 2 1 4 0 1 2 3 01 2 3 142 2 3 1 5 0 5 3 7 0 0 13 8 3 1 2 11 0 2 1 8 0 0 5 14 D - -- = = = =- --- --- - -- - 1 5 1 1 1 1 1 1 11 1 1 1 11 1 1 1 11 2 2 1 4 2 2 1 4 0 4 1 26 0 0 7 18 2 3 1 5 2 3 1 5 0 1 1 17 0 0 3 28 0 1 2 11 0 1 2 11 0 1 2 11 0 1 2 11 1 1 1 11 0 1 2 11 142 0 0 3 28 0 0 7 18 D - - - - -- = = == ---- ---- - = =- 2 15 1 1 11 1 5 15 1 1 15 1 1 1 2 1 4 1 1 4 2 0 2 3 7 0 1 10 5 2 2 1 5 2 1 5 2 0 3 7 12 0 3 7 12 3 0 2 11 3 2 11 0 0 1 8 15 0 1 8 15 15 1 1 0 1 10 5 284 0 0 23 3 0 0 18 10 D -- - - - - === = --- --- --- --- -- -- = = - 3 1 1 5 1 1 1 5 1 11 5 1 1 2 2 4 0 1 7 3 01 7 3 426 2 3 2 5 0 5 12 7 0 0 47 8 3 1 0 11 0 2 15 8 0 0 29 14 D - -- = = = =- --- -- - - -- - 4 1 1 1 5 1 1 1 1 11 1 1 1 2 1 2 0 1 2 7 01 2 7 142 2 3 1 2 0 5 3 12 0 0 13 47 3 1 2 0 0 2 1 15 0 0 5 29 D -- - - - - == = = --- --- - - --- - - 12 4 3 12 3 4 1 2 3 1 DD D D xx x x DD D D \ = = = = = = = =- 㾷 2)
065 0065 65100 065 5 65 000 0005 0000 510 0651 09 03 06510 0 0 14 6 065 006 065 000 1 065 0 10065 0000 006565 665 6 0000 065 006 651 006 00 6510 000 5100 5 065 65114 D2 0005 1000065 0065 1 000 6510 6 65 0 5 D3 010000 006511 0006 5100 0651 9 0005 0065 0651 0000 0006 0 000 0 6510
56000 15600 1 5 6 00 15 6 0 0 15600 01560 0 1 5 60 01 5 6 0 01560 00156 0 0 1 56 00 1 5 6 00156 00015 0 0 0 15 00 0 1 5 0 0 0 1 5 5 6 0 0 0 0 19 30 0 0 0 0 65 114 0 156 0 0 1560 0 015 6 0 0156 0 001 5 6 0015 6 665 000 1 5 0001 5 0 0 0 211 390 0 0 0 0 665 D === = - - = == - - 1 16000 5600 1560 1 5 6 0 05600 1560 0156 0 1 5 6 01560 0156 0015 0 0 1 5 00156 0 0 1 5 5 6 0 0 0 19 30 0 00015 15 6 0 01 5 6 211 00 1 5 0 0 65 114 D = = =- =- - - =- =- 2 51000 1600 10600 560 156 1 5 6 0560 00560 1 5 6 0 1 5 0 1 5 65 0156 00156 0 1 5 5 6 0 0 19 30 0015 00015 D = =- =- = = = - - 3 56100 1500 15000 0160 01060 19 0056 00056 0015 00015 D = == 4 56010 1560 15600 0150 01500 5 0016 00106 0005 00005 D = ==
6001 1560 15600 D=01560 00 00150 00010 D1_-2 D2 D,19 D45 D665 D665 D665 D665 Mx+x+x 0间、p取何值时,齐次方程组{x+Hx2+x=0有非零解 +2x,+x2=0 01-41-元 01 解:D=141=14 (1-) 12 0 当(1-)=0.,即=0or=1时,该方程组有非零解
5 56001 1560 15600 0156 01560 1 0015 00150 0001 00010 D = == 12 4 3 5 1 23 45 211 65 19 5 1 665 665 665 665 665 DD D D D x xx xx D DD DD - \= = = = = = = = = = 10.䯂¬ǃপԩؐᯊˈ唤ᮍ㒘 123 1 23 1 23 0 0 2 0 xxx x xx x xx l m m ì ++= ï í + += ï î + += ᳝䴲䳊㾷. 㾷˖ 1 1 01 1 0 1 1 1 1 1 (1 ) 1 1 12 1 0 0 D l lm l l m m m ml m m - - - = = =- = - ᔧ ml m l (1 ) 0, 0 1 -= = = े or ᯊˈ䆹ᮍ㒘᳝䴲䳊㾷DŽ