1+a0a,a000-aa00-a,a.(属于交行列式)=a,a.(1+a,i=2a21+1+2-)=aa,...a,(a=2a,=,(1+2→)要求(a,+0)台a,p101.19@2-23232 -125-133o-6-4-6 2-303-33-600=3D=-40-4=-18-52=-70210-3-1-430100-3-33003-1-11-3Di=1 1=1,2,3, 4= D4 = 70 ..且=D,=D3 x, =DP101,19②12-2|23-2]315810012-3-5-81-1-2013654D=1083654842283-1-4-1018 360574-18-3652-3-42107=182|2 3/=182 =3243 2且D,=324，D=648，D,=-324，D4=-648，D/D./% =2,x =-1,x4 =-2% =1,x2 =.. X =12-222403O-1626-222082-1 3 -415-85-88-18P10119 (3),即D=3 1-12-18 38-243414227 50-42023
1 1 2 1 3 1 1 1 2 2 1 1 2 1 2 1 1 2 2 1 11 1 0 0 0 00 0 0 ( )(1 1 ( 1 (1 ) n n n n i n n i i n i n i i a a a a a a a a aa a a aa a a a a aa a a = = = + − = − − − = +− = ++ ≠= + ∑ ) ∑ ) ∑ " " " " " " 属于交行列式 要求( 0) p101.19① 2 23 2 2 13 2 3 6 4 0 6 13 3 33 2 30 6 4 0 4 0 0 4 3 18 52 70 3 3 12 1 0 40 1 0 310 3 3 13 1 1 0 0 3 D − − −−− −− − − −− = = =− = − = − − − − − − − −− − = − 且 70 1 1, 2, 3, 4 1234 Di DD DD x i i D = = = =− ∴ = = = P101，19② 12 3 212 3 2 581 0 0 1 2 1 2 30 5 8 1 36 54 4 10 8 36 54 8 3 2 1 2 0 4 10 8 18 36 7 4 5 18 36 5 2 32 1 0 7 4 5 D − − −−− − − = = == − − = − −− − − − − − 且D1=324，D2=648，D3=-324，D4=-648， 2 3 2 2 18 18 324 3 2 = == 1 2 1, 2, 1, 2 1 2 34 D D x x xx D D ∴ = = = = =− = − 03 1 2 2 0 16 26 22 1 3 4 2 01 5 8 8 1 5 88 31 1 2 1 0 4 2 4 8 0 18 38 24 4 3 4 2 2 0 7 8 6 14 0 27 50 4 1 1 12 3 1 1 1 2 3 −− − − − − − − − − −= − = − − − − − − −− − −− − 12 2 4 1 2 P101 19 (3),即D= 2
2-220-20[1014|=-(-2) (x-1)1 +2|10 4/1028=24-18 28-2413 413 1-922-1813224同理算出d,=96,d2=-336,d,=-96,d4=169,ds=312即得 = -4,x2 =-14, g =-4,x4 =7,x, =13(消元法解)(1 2-2 4-1-1)2-3-3(1 -1 -15行移最上2-13-4209522再相减3行移到第2行再相减A=l31-1 2-10-8814S43422-2042-48121 -1-1 2 -3-3-6 14 101781(1045-611(3)-(4)后再乘2015814-8(5)×(4)的2倍-4000-162622-24-4400-182800-275042 885104-6118814015(4)+(5)的2倍后乘以再用3行的相反倍加到各行12001440 0-18 282460009-6000-1-9 13100-20100-14(5))后再乘相应倍加到各行34010-300042001:130000-000-3-11000-60000401(1) +(4)-14000(5)+(4)后再交换(4)，(5)即得0-4010000130000P101,19
2 22 0 20 10 4 10 1 1 2 18 28 24 10 28 4 ( 2) ( 1) 8 24 13 4 13 1 9 22 18 13 22 4 x − − + =−− − =− =− − ⋅ − =− = − − 96, 336, 96, 169, 312 2 3 45 4, 14, 4, 7, 13 1 2 3 45 d ddd x x x xx = =− =− = = =− =− =− = = 同理算出d1 即得 ( ) 12 2411 1 1 12 3 3 5 2 13 428 0 1 1 222 3 2 3 112 13 0 1 5 8 8 14 4 34 2 2 2 0 4 2 4 8 12 1 1 1 2 3 3 1 7 8 6 14 10 A ⎛ ⎞⎛ ⎞ − −− − − −− ⎜ ⎟⎜ ⎟ − − − = − − ⎯⎯⎯⎯⎯⎯→ − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ − − ⎝ ⎠⎝ ⎠ −− − − − 消元法解 行移最上 再相减 行移到第 行再相减 1 0 4 6 5 11 (3) (4) 0 1 5 8 8 14 (5) (4) 0 0 16 26 22 40 0 0 18 28 24 44 0 0 27 50 42 88 1 0 4 6 5 11 0 1 5 8 8 14 (4) (5) 2 3 001 1 1 2 0 0 18 28 24 44 009 6 6 0 ⎛ ⎞ − − ⎜ ⎟ − × − − − ⎯⎯⎯⎯⎯⎯⎯⎯→ − −− ⎝ ⎠ − − ⎛ ⎞ − ⎜ ⎟ − + − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ → − −− ⎝ ⎠ − 1 后再乘 2 的2倍 1 的 倍后乘以 再用 行的相反倍加到各行 4 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ ( ) 1000 1 9 100 2 1 3 1 0 1 0 0 0 14 5 010 3 3 4 3 0010 0 4 001 1 1 2 0 0 0 0 1 13 0 0 0 4 3 11 0000 1 6 ⎛ ⎞ − − ⎛ ⎞ − ⎜ ⎟ ⎜ ⎟ − × − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ − − ⎜ ⎟ ⎝ ⎠ − − ⎝ ⎠ − − 后再乘相应倍加到各行 10000 4 (1) (4) 0 1 0 0 0 14 (5) (4) 00100 4 00010 7 0 0 0 0 1 13 ⎛ ⎞ + ⎜ ⎟ − + ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ − ⎝ ⎠ 后再交换(4),(5) 即得. P101，19④
0F60006263626=665D=2-3e015见例2,α+βαβα+βDα-β取α=2,B=3且,D, =1507 D,=-1145D,=703D-395=212D34537792121507-229.xX35133665133665P101.20解：-+..·Cn-1=bin-2+Cn-1=b2tc1acoun-2n-1+..Cn-1=bn+c,aco"nun代入各a于f(x)系数行列式：-3-al.aiaoardsnnan由于aja2a,两两不同，故V，0,d0由Cramer法则，存在n-1唯一解Co.C.C2cn-1.即有(x)=Cxn-1-i(唯一地)使(a)=bi=01%=13.60°% +10q, +100, +1000a, =13.57+20a,+400a,+8000a,=13.55[ +30a + 900a, + 7000a, =13.52例P101.21解：
5600 0 1560 0 6 6 2 3 6 6 0156 0 3 2 665 2 3 0015 6 0 0 0 1 15 D − = == − = − , 1 1 1 2, 3 n n D n α β αβ α β α β α β α β + − − − =+= − = = % % % 见例2 取 1507 1145 703 395 212 1 2 34 5 1507 229 37 79 212 , , , 1 2 34 5 665 133 35 133 665 D D DD D x x xx x = =− = =− = − ∴ = = = =− = 且, P101.20 解： 代入各a于f(x) 1 2 1 1 01 11 1 2 1 2 02 12 1 2 0 1 1 n n c a ca c b n n n c a ca c b n n n c a ca c b n n n n ⎧ − − + += − ⎪ ⎪ − − ⎨ + + − ⎪ ⎪ − − + + − ⎩ " " " = = 系数行列式： 2 1 1 2 1 11 1 1 11 1 2 1 1 2 1 2 2 2 1 ( 22 2 1 2 1 2 1 1 n n n aa a aa a n n n aa an da a a C V n n n n aa a nn n n aa a n n n − − − − − − = == − − − " " " " """ " " ， −1) , 0, 0 1 2 1 1 ( ) () 0 aa a d n n n i C xi i f a b i ≠ ≠ − − − ∑ = i = " " ， 由于 两两不同,故V 由Cramer法则,存在 n 唯一解C ,C ,C c ,即有f(x)= 唯一地 使 0 1 2 n-1 例P101.21 解： 13.60 0 10 100 1000 13.57 01 2 3 20 400 8000 13.55 01 2 3 30 900 7000 13.52 01 2 3 a aa a a aa a a aa a a ⎧ = ⎪ ⎪ ++ + = ⎪ ⎨ ⎪ ++ + = ⎪ ++ + = ⎪⎩
00010310210=1.2×107d=20 202203=(la,20,30-(20-10)(30-10)(30-20)30230330-50000,d,=1800,d,=-40d。=1.632×108n25×10-3,11-do/x10-5=13.6, a,=1.5×10ao=a3/d63000(113.6)00013.6)1041051030-31010013.5711000用消元法：5x5052x1034x10420400800013.551-5030(19002700013.52(0 3x30327×1059×104-800000013.6)113.61105103104105103104-3-300→(略)6×1052x1046x1052×1040010016x10524×105(o00-206x10401253×10-32×10-4×,21g/x10.5×th=13.6c.h=x t+cm.6233故当t=15℃时，h=13.56(精确）-（g/T当t=40℃时，h=13.46（书上答案13.48是错的）P102 补1aianayan..a2ja2元D=设D=alJij2"jamian..am(-1)(j.j).. D = Z (-1)()D:>D=0取通hi-J.(n≥2奇偶排列各半）当 n=1 时
0 1 3 10 0 0 2 3 1 10 10 10 7 1.2 10 2 3 1 20 20 20 (1 ,20,30 (20 10)(30 10) (30 20)) 2 3 1 30 30 30 8 1.632 10 , 50000, 1800, 40 2 d a d d dd = × = = ⋅− −⋅− = × =− = =− 5 0 25 3 4 1 13.6, 10 , 1.5 10 , 10 1 23 6 3 do a a aa d − − − = = =− × = × =− × 1 0 0 0 13.6 34 5 1 10 100 1000 13.57 0 10 10 10 3 1 20 400 8000 13.55 34 5 0 2 10 4 10 5 50 5 1 30 900 27000 13.52 34 5 0 3 30 9 10 27 10 8 ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ − → × × ×− ⎝ ⎠ ⎝ ⎠ × × ×− 1 0 0 0 13.6 用消元法: 1 0 0 0 13.6 1 0 0 0 13.6 34 5 34 5 0 10 10 10 3 0 10 10 10 3 ( ) 4 5 4 5 0 0 2 10 6 10 1 0 0 2 10 6 10 1 4 5 5 0 0 6 10 24 10 1 0 0 0 6 10 2 25 3 3 42 3 1 13.6 10 10 10.5 ( , ) 62 3 3 g h ttt t c h cm ⎛ ⎞ ⎛ ⎜ ⎟ ⎜ − − ⎜ → → ⎜ × × ⎜ × × ⎜ ⎜ ⎝ ⎠ × × ⎝ × − − − = − × × + × × − × × =° = 略 ⎞ ⎟ ⎟ ⎟ → ⎟ ⎟ ⎟ ⎠ ° ° 故当t=15 c时,h=13.56(精确)-( ) g 3 cm 当t=40 c时,h=13.46(书上答案13.48是错的) P102 补 1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 11 1 22 2 1 ( ) 1 | | ( 1) 0 n n n n n n n jj j jj j ij jj j nj nj nj jj j jj j jj j aa a aa a D D aa a D D τ τ = = ⎛ ⎞ a ∴ =− = ⎜ ⎟ D = ⎝ ⎠ ∑ ∑ ∑ " " " " " " " " """ " （j j j ） 12 n 取遍 设 （-1） (n≥2 奇偶排列各半) 当 n=1 时
P102 补2 :D= Z (-1)(a,()a2,(0).-am ()iijndakk (t):D= Z (-1)(Zax(0)-(0)am()ak+1jstdtdtihik=lday (t)M(-1)(..)a,(t)a(t).ak-ami. (t))tak+1jstdt台a2(t)a.(t)ain(t)ddd"Wa(t)ak2(t)-am(t)edtdtan(t)an2(t)amm(t)a(t)azi(t)am(t)ddd-(a(t))a2 (t)-am()dtdhdtdd同理也有dtD=dtD'=da(t)ailaindtdZ(转置)=2(t)a2ra21dtk=lda()anlaadt000+x+x a+x..a.12Ina1+xa12+xain+xa21+xa22+x2n左=a21+x+x°2n+xa 22+an+x a+x2+°2u+xautxau2 +xP102补3①1-X-r. a.aj ayj-an011a12a12lair...ann(-x) ≥ (-1)(j+1)la,a..a222anl:.a..i=1n1a....amam++..amlaO0n2nn
1 2 1 2 1 1 1 1 2 1 2 11 1 1 102 2 ( ) ( ) ( )) ( ) ( () () ( n n k k n j j nj jj j n kjk j kj kj nj jj j k P D a ta t a t d da t D a t a t a dt dt τ τ − + − + = = )) n ∴ = a t ∑ ∑ ∑ " " " " ∵ " " " 12 n 12 n （j j j ） （j j j ） 补 （-1） （-1） 1 2 1 2 1 1 1 2 (, ) 12 1 1 1 , ( ) ( ( 1) ) ( ) ( ) ( ) ( n k k k n n jj j kj j j kj k j nj k jj j da t a ta t a t a a t dt τ − + − + = = − ∑ ∑ " " " " )) n = 11 12 1 1 2 1 1 2 () () () () () () () () () n n kk k k nn nn at at at ddd at at at dt dt dt at at at = ∑ n 同理也有 d dt D= d dt D T = 11 21 1 1 2 1 () () () ( ( )) ( ) ( ) n n k k n at at at d dd at at at dt dt dt = ∑ " "" nk 11 1 1 21 2 2 1 1 ( ) ( ) () ( ) k n n k n k n nk d a at a dt d a at a dt d a at a dt = 转置 = ∑ nn P102 补 3 ① 10 0 0 11 12 1 1 11 12 1 21 22 2 1 21 22 2 12 3 1 12 2 a xa x a x n a xa x a x n a xa x a x n a xa x a n a xa x a x nu u x x a xa x au u u ++ + + + + ++ + = = + + ++ + + + + + " " " " 左 ( ) 11 1 1 1 1 1 11 1 1 1 1 11 1 1 1 11 12 12 ( 1) 1 21 22 2 1 1 1 1 2 j j n n nn n nj nj nn xx x aa a a aa a a a n x aa a a a n i aa a a aa a n n nn − + + − + −− − = =+ − ∑ − = " " " " " " " # " " " （j+1） n