第一章行列式课后的习题值得我们仔细研读.本章建议重点看以下习题：5.（2)，（5)：7：8.（2).（这几个题号建立有超级链接）若您发现有好的解法，请不各告知。1.利用对角线法则计算下列三阶行列式：201bca(1)1-4(2)-1bc-183ba1121+yy(3)b(4)acya+yrc2a262ayT+y解： (1)2011-4 -1-193=2×(4)×3+0×(-1)×(-1)+1×1×8-0×1×3-2×(-1)×8-1×(-4)×(-1)=-24+8+16-44(2)0aacb+bac+cba-bbb-aaa-ccc=3abc-a3-b3-c36n6Ca(3)111b= bc2 + ca2 + ab2 - ac2 - ba? - cb2= (a - b)(6 - c)(c- a)aC262c2(4)rT+yya(r+y)y+yr(a+y)+(a+y)ya-y-(r+y)3-r3yr+yr+yTy=3ry(+)-3-3r32-2393-23= -2(23 + g).2.按自然数从小到大为标准次序，求下列各排列的逆序数(1) 12 3 4;(2) 413 2;(3)3421;(4) 2413;(5) 13...(2n -1) 2 4... (2n);(6) 1 3.. (2n -1) (2n) (2n - 2)...2.解(1)逆序数为0.(2)逆序数为4:41,43,42,32.1
第一章 行列式 课后的习题值得我们仔细研读. 本章建议重点看以下习题: 5.(2), (5); 7; 8.(2). (这几个题号建立有超级链接.) 若 您发现有好的解法, 请不吝告知. 1 . 利用对角线法则计算下列三阶行列式: (1) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 0 1 1 −4 −1 −1 8 3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ; (2) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ a b c b c a c a b ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ; (3) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 1 1 a b c a 2 b 2 c 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ; (4) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x y x + y y x + y x x + y x y ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ . 解: (1) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 0 1 1 −4 −1 −1 8 3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 2 × (−4) × 3 + 0 × (−1) × (−1) + 1 × 1 × 8 − 0 × 1 × 3 − 2 × (−1) × 8 − 1 × (−4) × (−1) = −24 + 8 + 16 − 4 = −4. (2) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ a b c b c a c a b ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = acb + bac + cba − bbb − aaa − ccc = 3abc − a 3 − b 3 − c 3 . (3) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 1 1 a b c a 2 b 2 c 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = bc2 + ca2 + ab2 − ac2 − ba2 − cb2 = (a − b)(b − c)(c − a). (4) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x y x + y y x + y x x + y x y ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = x(x + y)y + yx(x + y) + (x + y)yx − y 3 − (x + y) 3 − x 3 = 3xy(x + y) − y 3 − 3x 2 y − 3y 2x − x 3 − y 3 − x 3 = −2(x 3 + y 3 ). 2 . 按自然数从小到大为标准次序, 求下列各排列的逆序数: (1) 1 2 3 4; (2) 4 1 3 2; (3) 3 4 2 1; (4) 2 4 1 3; (5) 1 3 · · ·(2n − 1) 2 4 · · ·(2n); (6) 1 3 · · ·(2n − 1) (2n) (2n − 2)· · · 2. 解 (1) 逆序数为 0. (2) 逆序数为 4: 4 1, 4 3, 4 2, 3 2. 1
2第一章行列式(3)逆序数为5:32,31,42,41,21.(4)逆序数为3:21,41,43.(5)逆序数为n(n=1);.1个32....2个52,54.3个72,74,76...(n-1)个(2n-1) 2, (2n-1) 4, (2n-1) 6, ..*, (2n-1) (2n-2).(6)逆序数为n(n-1):.1个32.2个52,54.3个72.74.76.(n-1)个(2n-1) 2, (2n-1) 4, (2n-1))6, ...,(2n-1) (2n-2).1个422个62.64(n-1)个(2n) 2. (2n) 4, (2n) 6, ***, (2n) (2n - 2)3.写出四阶行列式中含有因子a11a23的项.解：由定义知，四阶行列式的一般项为(1)a1pia2paa3paa4p4,其中t为p1p2p3P4的逆序数。由于p1=1，P2=3已固定，P1P2P3P4只能形如13，即1324或1342.对应的逆序数t分别为0+0+1+0=1,或0+0+0+2=2.所以，-a1a23432a44和a11a2334442为所求4.计算下列各行列式2N1241423-121-20(1)(2)23210512002011756010a-ahacae0b1(3)de(4)bd-cd01cbfcf-ef00d解： (1)202V021242+40120a1224-7r2-4ri1010520-1522052r3-10r100111011177
2 第一章 行列式 (3) 逆序数为 5: 3 2, 3 1, 4 2, 4 1, 2 1. (4) 逆序数为 3: 2 1, 4 1, 4 3. (5) 逆序数为 n(n−1) 2 : 3 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 个 5 2, 5 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 个 7 2, 7 4, 7 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 个 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2n − 1) 2, (2n − 1) 4, (2n − 1) 6, . . . , (2n − 1) (2n − 2). . . . . . . . . . . . . .(n − 1) 个 (6) 逆序数为 n(n − 1): 3 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 个 5 2, 5 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 个 7 2, 7 4, 7 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 个 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2n − 1) 2, (2n − 1) 4, (2n − 1) 6, . . . , (2n − 1) (2n − 2). . . . . . . . . . . . . .(n − 1) 个 4 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 个 6 2, 6 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 个 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2n) 2, (2n) 4, (2n) 6, . . . , (2n) (2n − 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (n − 1) 个 3 . 写出四阶行列式中含有因子 a11a23 的项. 解: 由定义知, 四阶行列式的一般项为 (−1)t a1p1 a2p2 a3p3 a4p4 , 其中 t 为 p1p2p3p4 的逆序数． 由于 p1 = 1, p2 = 3 已固定, p1p2p3p4 只能形如 13¤¤, 即 1324 或 1342. 对应的逆序数 t 分别为 0 + 0 + 1 + 0 = 1, 或 0 + 0 + 0 + 2 = 2. 所以, −a11a23a32a44 和 a11a23a34a42 为所求. 4 . 计算下列各行列式: (1) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 4 1 2 4 1 2 0 2 10 5 2 0 0 1 1 7 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ; (2) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 1 4 1 3 −1 2 1 1 2 3 2 5 0 6 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ; (3) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ −ab ac ae bd −cd de bf cf −ef ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ; (4) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ a 1 0 0 −1 b 1 0 0 −1 c 1 0 0 −1 d ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ . 解: (1) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 4 1 2 4 1 2 0 2 10 5 2 0 0 1 1 7 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ r1↔r2 ====== − ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 2 0 2 4 1 2 4 10 5 2 0 0 1 1 7 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ r2−4r1 ======= r3−10r1 − ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 2 0 2 0 −7 2 −4 0 −15 2 −20 0 1 1 7 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
线性代数（同济四版）习题参考答案3202122102021117700110117T4+7T2r2→r4= 17 × 9=0.0-152-20r3+15r200178501500-7200450０15-49(2)221411402140321322322-1-1-1C4-C2T4-T21232123023015222065064012140223-1T4-T1 0.23100000(3)-abacae-bCLbdb-cdde= adfadfbce11-C1ebbfcf-efc1-1-11120r2+T102-adfbce4abcdef.adfbce00r3+r12200(4)1000001+ababb10-110-1ri+ar20-1011-1c00d00-1d-11 + ab01 + abadaa第 1 列(-1)(-1)2+1Ca+dea-1-1c1 + cdc展开0-1 d0-10ad1 + ab按第3行(1)(1)3+2abcd + ab + cd + ad +1.-11 + cd展开5.证明：62a2ab(a - b)3;(1)2aa+b2b=111证明Q262a?ab-α262 - a2abC2262a2b - 2a2aa+bb-aC3-C1011110Qb-α262-a2b+aa=(-1)3+1(α - b)3.(b - a)(b -a)2b-a2b-2a1ay+bzaz+brar+byayz(2)= (α3 + 63)ay + bzaz+brar + byy之5ay+bzaz+brar +byZy
线性代数 (同济四版) 习题参考答案 3 r2↔r4 ====== ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 2 0 2 0 1 1 7 0 −15 2 −20 0 −7 2 −4 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ r4+7r2 ======= r3+15r2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 2 0 2 0 1 1 7 0 0 17 85 0 0 9 45 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 17 × 9 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 2 0 2 0 1 1 7 0 0 1 5 0 0 1 5 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 0. (2) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 1 4 1 3 −1 2 1 1 2 3 2 5 0 6 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ c4−c2 ===== ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 1 4 0 3 −1 2 2 1 2 3 0 5 0 6 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ r4−r2 ===== ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 1 4 0 3 −1 2 2 1 2 3 0 2 1 4 0 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ r4−r1 ===== ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 2 1 4 0 3 −1 2 2 1 2 3 0 0 0 0 0 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 0. (3) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ −ab ac ae bd −cd de bf cf −ef ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = adf ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ −b c e b −c e b c −e ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = adfbce ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ −1 1 1 1 −1 1 1 1 −1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ r2+r1 ===== r3+r1 adfbce ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ −1 1 1 0 0 2 0 2 0 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = −adfbce ¯ ¯ ¯ ¯ ¯ 0 2 2 0 ¯ ¯ ¯ ¯ ¯ = 4abcdef. (4) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ a 1 0 0 −1 b 1 0 0 −1 c 1 0 0 −1 d ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ r1+ar2 ====== ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 0 1 + ab a 0 −1 b 1 0 0 −1 c 1 0 0 −1 d ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 按第 1 列 ======== 展开 (−1)(−1)2+1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 + ab a 0 −1 c 1 0 −1 d ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ c3+dc2 ====== ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 + ab a ad −1 c 1 + cd 0 −1 0 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 按第 3 行 ======== 展开 (−1)(−1)3+2 ¯ ¯ ¯ ¯ ¯ 1 + ab ad −1 1 + cd ¯ ¯ ¯ ¯ ¯ = abcd + ab + cd + ad + 1. 5 . 证明: (1) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ a 2 ab b2 2a a + b 2b 1 1 1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = (a − b) 3 ; 证明 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ a 2 ab b2 2a a + b 2b 1 1 1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ c2−c1 ===== c3−c1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ a 2 ab − a 2 b 2 − a 2 2a b − a 2b − 2a 1 0 0 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ =(−1)3+1 ¯ ¯ ¯ ¯ ¯ ab − a 2 b 2 − a 2 b − a 2b − 2a ¯ ¯ ¯ ¯ ¯ = (b − a)(b − a) ¯ ¯ ¯ ¯ ¯ a b + a 1 2 ¯ ¯ ¯ ¯ ¯ = (a − b) 3 . (2) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ax + by ay + bz az + bx ay + bz az + bx ax + by az + bx ax + by ay + bz ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = (a 3 + b 3 ) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x y z y z x z x y ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ;
第一章行列式4证明：az +braz + brar + byay+bzay+bzyay+bzaz+brT按第1列ay+bzaz + baar + byyaz+brar +by+6az+brzar + by分裂开az+brar +byay+bzay+bzaar+byay+bzzar+byzaz+bray+bz242再次+0+0+62az+barar+by22y裂开ar+byyayay+bzZ42yyzr再次a3+ 63Zzryy裂开ryz2yIy2yzZ92=α3+63(-1)2= (a3 + 63)y2ryZT4ZT2工yTzyry此题有一个“经典”的解法：bybzbaraz+brar + byay+bzara2aybzbrbyay+bzaz+brar+byayazarXbrby bzaz+brar+byay+bzazarayT22ayzryyzay+63Ov+ 63(-1)2a3ZTZryZ2yzryyZTayz2工yZEyTy=(a3 + 63)yzrzay这个解法“看上去很美”，实则是一个错解！我们强调，行列式不能作这种形式上的加法：b11bina1aina1 +b1rain+bin.t.::：t：XAbn1bnnanlann...ani+bnlann+bnn....Q2(α + 2)2(a + 1)2(a + 3)2b2(b + 2)2(6 + 1)2(6 + 3)2(3)= 0;c2(c + 1)2(c + 2)2(c + 3)2d2(d + 1)2(d + 2)2(d + 3)2证明：a2(a + 1)2(a + 2)?(a+3)2a?2a+114a+46a+96262(6 + 1)2(b + 2)2(6 + 3)226 +146+466+9cj-c1c2c2(c + 3)2(c+1)2(c + 2)2j=2,3,42c+14c+46c+9d2d2(d + 3)?(d+1)2(d + 2)22d +14d+46d+9a22a+11266226+1.26两列成比例0.C32c2C22c+126c4-3c2d22d+1.26
4 第一章 行列式 证明: ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ax + by ay + bz az + bx ay + bz az + bx ax + by az + bx ax + by ay + bz ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 按第 1 列 ======== 分裂开 a ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x ay + bz az + bx y az + bx ax + by z ax + by ay + bz ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ + b ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ y ay + bz az + bx z az + bx ax + by x ax + by ay + bz ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 再次 ==== 裂开 a 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x ay + bz z y az + bx x z ax + by y ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ + 0 + 0 + b 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ y z az + bx z x ax + by x y ay + bz ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 再次 ==== 裂开 a 3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x y z y z x z x y ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ + b 3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ y z x z x y x y z ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ =a 3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x y z y z x z x y ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ + b 3 (−1)2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x y z y z x z x y ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = (a 3 + b 3 ) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x y z y z x z x y ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ . 此题有一个 “经典” 的解法: ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ax + by ay + bz az + bx ay + bz az + bx ax + by az + bx ax + by ay + bz ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ax ay az ay az ax az ax ay ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ + ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ by bz bx bz bx by bx by bz ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ =a 3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x y z y z x z x y ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ + b 3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ y z x z x y x y z ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = a 3 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x y z y z x z x y ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ + b 3 (−1)2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x y z y z x z x y ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ =(a 3 + b 3 ) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x y z y z x z x y ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ . 这个解法 “看上去很美”, 实则是一个错解! 我们强调, 行列式不能作这种形. 式. 上. 的加法: ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ a11 . . . a1n . . . . . . . . . an1 · · · ann ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ + ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ b11 . . . b1n . . . . . . . . . bn1 · · · bnn ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ a11 + b11 . . . a1n + b1n . . . . . . . . . an1 + bn1 · · · ann + bnn ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ . (3) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ a 2 (a + 1)2 (a + 2)2 (a + 3)2 b 2 (b + 1)2 (b + 2)2 (b + 3)2 c 2 (c + 1)2 (c + 2)2 (c + 3)2 d 2 (d + 1)2 (d + 2)2 (d + 3)2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 0; 证明: ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ a 2 (a + 1)2 (a + 2)2 (a + 3)2 b 2 (b + 1)2 (b + 2)2 (b + 3)2 c 2 (c + 1)2 (c + 2)2 (c + 3)2 d 2 (d + 1)2 (d + 2)2 (d + 3)2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ cj−c1 ====== j=2,3,4 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ a 2 2a + 1 4a + 4 6a + 9 b 2 2b + 1 4b + 4 6b + 9 c 2 2c + 1 4c + 4 6c + 9 d 2 2d + 1 4d + 4 6d + 9 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ c3−2c2 ====== c4−3c2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ a 2 2a + 1 2 6 b 2 2b + 1 2 6 c 2 2c + 1 2 6 d 2 2d + 1 2 6 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 两列成比例 ========= 0
5线性代数（同济四版）习题参考答案1111bdca(4)(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)(a+b+c+d)62c2d2a2a4b4 c4d4证明：00011111bdb-ad-aacac-aCj-c1Q262c2dR62 - a?c2 - a2d? - a?j=2,3,4a2d41a464c4a4b4 - a4c4- a4d4-a4b-ad-ac-a展开r162 - Q22 - α2d2-a?(d2-2)62(62 - d2) c2(c2 - a2)111= (b-a)(c-a)(d - a)b+ad+ac+ab2(b+a) c2(c+a) d2(d+a)100- (b - a)(c - a)(d - a)c-bd-bb+aC3-C16b2(b+a) c2(c+a) -b2(b+a) d2(d+a) -b2(b+a)11展开 (b-a)(c-a)(d -a)(c-b)(d -b))(d2+bd +62)+a(d+b)(c2+bc+62)+a(c+b)=(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)(a+b+c+d)0-1005....00-10a... .......:目(5)=rn+airn-lI+...+an-ir+an.000-1T...+ a1anan-2...02an-1证明：方法.设法把主对角线上的变为0，再按第一列展开，0000-1T..0-1000T.............:...Dn =00002-1...0000-12...anr+aian-1an-2a3a20000T-10000T-1........8：1:-1+TC000-10T...00000-1...+r+2r+a1ana3an-1an-2
线性代数 (同济四版) 习题参考答案 5 (4) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 1 1 1 a b c d a 2 b 2 c 2 d 2 a 4 b 4 c 4 d 4 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = (a − b)(a − c)(a − d)(b − c)(b − d)(c − d)(a + b + c + d); 证明: ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 1 1 1 a b c d a 2 b 2 c 2 d 2 a 4 b 4 c 4 d 4 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ cj−c1 ====== j=2,3,4 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 0 0 0 a b − a c − a d − a a 2 b 2 − a 2 c 2 − a 2 d 2 − a 2 a 4 b 4 − a 4 c 4 − a 4 d 4 − a 4 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 展开 r1 ====== ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ b − a c − a d − a b 2 − a 2 c 2 − a 2 d 2 − a 2 b 2 (b 2 − a 2 ) c 2 (c 2 − a 2 ) d 2 (d 2 − a 2 ) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = (b − a)(c − a)(d − a) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 1 1 b + a c + a d + a b 2 (b + a) c 2 (c + a) d 2 (d + a) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ c2−c1 ===== c3−c1 (b − a)(c − a)(d − a) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 0 0 b + a c − b d − b b 2 (b + a) c 2 (c + a) − b 2 (b + a) d 2 (d + a) − b 2 (b + a) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 展开 r1 ====== (b − a)(c − a)(d − a)(c − b)(d − b) ¯ ¯ ¯ ¯ ¯ 1 1 (c 2 + bc + b 2 ) + a(c + b) (d 2 + bd + b 2 ) + a(d + b) ¯ ¯ ¯ ¯ ¯ = (a − b)(a − c)(a − d)(b − c)(b − d)(c − d)(a + b + c + d). (5) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x −1 0 · · · 0 0 0 x −1 · · · 0 0 . . . . . . . . . . . . . . . 0 0 0 · · · x −1 an an−1 an−2 · · · a2 x + a1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = x n + a1x n−1 + · · · + an−1x + an. 证明: 方法一. 设法把主对角线上的 x 变为 0, 再按第一列展开. Dn = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x −1 0 · · · 0 0 0 0 x −1 · · · 0 0 0 . . . . . . . . . . . . . . . . . . 0 0 0 · · · x −1 0 0 0 0 · · · 0 x −1 an an−1 an−2 · · · a3 a2 x + a1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ cn−1+xcn ======== ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ x −1 0 · · · 0 0 0 0 x −1 · · · 0 0 0 . . . . . . . . . . . . . . . . . . 0 0 0 · · · x −1 0 0 0 0 · · · 0 0 −1 an an−1 an−2 · · · a3 x 2 + a1x + a2 x + a1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯