数学外文翻译外文文献英文翻译范德蒙行列式的相关应用

1、范德蒙行列式的相关应用（一）范德蒙行列式在行列式计算中的应用范德蒙行列式的标准规范形式是:11IIIX1X2IIIDn =2X12X2IIIIIIlhIIIn _1 X1ndX2IIIXn2XnIIInxn-(xi - Xj )nj_1根据范德蒙行列式的特点,将所给行列式包括一些非范德蒙行列式利用各种方法将其化为范德蒙行列式，然后利用范德蒙行列式的结果，把它计算出来。常见的化法有以下几种：所给行列式各列（或各行）都是某元素的不同次幕，但其幕次数排列与范德蒙 行列式不完全相同，需利用行列式的性质（如提取公因式，调换各行（或各列）的次序，拆项等）将行列式化为范德蒙行列式 例1计算2 223 32I

2、IIIIIIIIHI2n3nDn中各行元素都分别是一个数自左至右按递升顺序排列，但不是从0变Dn而是由1递升至n如提取各行的公因数，则方幕次数便从0变到n -1.111in1123IIIn12232III2nIllIIIIIIinin12门-13nAIIInn!(2 -1)(3 一1)丨1|( n-1)(3 - 2)|l|( n-2)|l|1)1=n!(n -1)!(n -2)!川2!1!例2计算n a(a-1)III(a _n)nnd a,八 n J.(a -1)III/(a - n)Dn 书=IIIIIIIIIaa -1IIIa -n11III1解本项中行列式的排列规律与范德蒙行列式的排列

3、规律正好相反，为使Dn 1中各列元素的方幕次数自上而下递升排列， 将第n 1列依次与上行交换直至第1行，第n行依次与上行交换直至第2行第2行依次与上行交换直至第n行, 于是共经过n （n 一 1） （n 一2） |（ 2 1 二凹卫2次行的交换得到n1阶范德蒙行列式：11卅1n(ndf)aa -1a nDn 十=(-1)2IIIIIIhiIIIn A. a/八n(a-1)(a-n)nn a(a -1)卅(a- n)nn(n 1)n=(-1) 2 (a-1-a)(a-2-a) (|(a -n-a)a-2-(a-1)川a -n-(a-(n-1)l：i 丨 k!若Dn的第i行（列）由两个分行（列）所

4、组成，其中任意相邻两行（列）均含相同分行（列）；且Dn中含有由n个分行（列）组成的范德蒙行列式，那么将Dn的第i行（列）乘以-1加到第i1行（列），消除一些分行（列）即可化成范德蒙行列式：例3计算1111D1 +sin 希1 +si n 21 +si n 31 +si n 4si n +si n21sin 2 +sin22si n 3 +si n2 3si n 4 +si n2 sin2 育 +si n 3.sin2 2 +s in3 2sin2 3 +sin3 3sin2 4 +s in3Q解将D的第一行乘以-1加到第二行得:1111sin為sin 2sin 3sin 4sin 药 +sin

5、2 為sin 2 +sin2 2sin 3 +si n2 3si n 4 +si n2 4sin2 為 +si nsin2 2 +si n3 2sin2 3 +s in3 3sin2 4 +si n3 4再将上述行列式的第2行乘以-1加到第3行，再在新行列式中的第3行乘以-1 加到第4行得：1sin迟sin2 尬1.3 ,sin 鬥1sin d2sin2 述 2.3 ,sin *21si n 3sin2 尬3.3 ,sin *31si nd 4sin2 症 4sin4 冲 4二(sin 处i 一 sin j)1包:凶例4计算1X-I 21X1III,n1X11X21 X；III1 X；IIII

6、IIIIIIII1Xn1 X2III1 X：D =(1)解先加边，那么100III01-1-1III_111+x11 + x2III1+x；1X12X1IIIx1n11+x21 + x2III1+x；=1X22X2IIIX；IIIIIIHIIIIINIIIHIHIIIIIII11+冷1 + x：IIId . n1+Xn1XnX：IIInXnD =再把第1行拆成两项之和,(xk -Xj ) - I i=1n(Xi -1)丨 I (兀-Xj)1勺逑弐200III0111III11X12X1IIInX11X12X1IIInX1IIIIIIIIIIIIIHItIIIIIIHIIH1Xn2XnIIInX

7、n1Xn2XnIIInXnD =(Xk Xj)2口 X -口 (Xi -1)勺： ni d加行加列法各行（或列）元素均为某一元素的不同方幕，但都缺少同一方幕的行列式, 可用此方法：例5计算11III12X12X2IIIxjD =3X13X2IIIx3IIIHIIHHInX1nX2IIInXn解作n 1阶行列式:由所作行列式可知通过比较系数得:z2z3zIIInz2X13IIInX1X22X23X2IIInX2IIIIIIIIIIIIHIIIIXn2Xn3XnHinXnn二口(X z) 口(Xj -Xk)i|迷:j兰z的系数为-D，而由上式可知z的系数为:n /(-1)22皿2必(7 丄)丨(X

8、j - XjiXi泮豐n 1DXIlXj) H(Xj-Xk)i 二 Xj n _j k 丄拉普拉斯展开法运用公式ID =M“A +M2A2 +|MnAn来计算行列式的值:例6计算10X10川n 1X1 0010y1川0n 1y1 10X20川n 1X2 0D =010y2川0n 1y2 -III川IH川MlIH川10Xn0川n 1Xn 0010yn川0n 12n -1 行，第1,3,川2n-1列展开得：1X1IIIn -1X11y1IIIn_1y11X2IIIn 4X21y2IIInJy2=n(XIII10HIHIHIIIIHIIII诣沦1XnIIIn 4Xn1ynIIInJyn解取第1, 3

9、j -人)厲- y)乘积变换法Xk JIl Xk 八x：(k =0,1|2n-2)，计算行列式i =1S0S1川Sn4S1S2IIISnIIIHI10IIISn 4SnIIIS22nnZ XiIIIi丄nnD =送Xiz X2IIIi丄i 二川HIIIInn寸nZ Xi丁nZ X2III7i 二7n一 n xii丄IIIn- 2n N Xii J1iIII112IIIn AIIIXXXiXiX2Xn12inn Axi22X2X2入i川X2HIII!IllXi 川|IH*nIII2HlIIIn _1nndIIIn _11XnXninXnxiX2Xn1 / 、2=n (% -x)例8计算行列式i岂

10、:j岂(a。b)n (a。by |l( (a bn)n(ai bo)n(abjn |l| (abn)niii hiin hi解在此行列式中，每- 根据行列式的乘法规则,从而变成乘积的和。(an bo)n (an bi)n 山(a. bn)n=CnCn *Cn(ai0_j :i in1a。川na。11川11a川a?n(n * )Ll(-i )2 Lb。bIIIbnIIIinIIIIIIIIIHIIHIII1anIIIa；bobnIIIb：D - Dib tc；C：n (b-bj)0:i_n一 aj)_(T)n(n -1)2CGa。IIIn nCn a。bobininbnCCnaiIIICnna；

11、D2 =b畀brinb：IIIIIImHIIIIinIHCC：anIIICnna；11in1Di =对D2进行例2中的行的变换，就得到范德蒙行列式，于是个元素都可以利用二项式定理展开,D 二 D1LD2，其中=C：C汕|C： 口 佝-aj)(b-bj)0弐成宝升阶法 例9计算行列式1X12X1IIIn 2X1_nX11X22X2IIIn_2X2nX2IIIIIIIIIIIIIHHI1Xn2XnIIIn_2XnjLnXn1xn2xnIIIn_2xnnxnD =解 将D升阶为下面的n 1阶行列式1X2X1IHn_2nX1nX11X22X2IHn_2X2nX2nX2IIIHIinIIIHiHI1Xn

12、A2XnIHn_2 沁nXnnXn_11Xn2XnIHn_2XnnXnnXn1X2 XIHn_2XnXnX此处X是x的n次即插入一行与一列，使L n 1是关于M,X2,lHXn,X的n 1阶范德蒙行列式， 变数，于是 1 =(X -xj(x -X2)(X - xn) I (X-Xj)故 L n 1 是一个关于1应d多项式，它可以写成L 1：丨丨(Xi -Xj)n (-1)(X1 X2Xn)Xn J Y1应也另一方面，将Ln 1按其第n 1行展开，即得L._.n2n:：1n1= . (Xi Xj)x (1)Dx 1：!_n比较L n1中关于XnJ的系数，即得D%川召)【(3 - X1X3IIII

13、IIIIIXn xnJn2 HYPERLINK l bookmark71 o Current Document x2- x 冷n 4n -2X3- X1X3x2 - x1x2沧-人 III x-XiX； X3 III XnfXnn2-魁XlX3n2IIIn_1X.n-2-XlXn1III1X3+III4d*n-23IIIn-2Xn1X2= (X2-x)(X3-m)|(人-Xi)n-2IThe latter determ inant is a n -1 Van Dear Mi nd determ inant, accord ing to the in duct ionassumption, it

14、 is equal to all possible differeneexi - Xj 2 _ j ： i _ n ； Containsxi differe nee all appear in front of the con seque nt con clusi on van dear Mend determ inantof the level n the establishme nt of mathematical in duct ion, the proof is completedThis result can be abbreviated as even the multiplica

15、ti on sig n11HI1X2X3HIX2X22X3HI2Xn= (Xj _Xj)11h+1刍匕勺InVndHI+n 二X1X2XnImmediately by the results obta ined n ecessary and sufficie nt con diti on for van der Mond determ inan t is zero,X2, X3，H I , XnAt least two equal nu mber nThe applicati on of the Van der monde determ inant() Van der monde determ

16、 inant in the determ inant calculatio nThe Van der monde determ inant sta ndards form11III1X1X2IIIXnDn =2X12X2III2Xn=n 区一）IIIIIIIIIIIIndIIIX1X2xnAccording to the characteristicsof theVaridermonde determ inant give n determ inantusing various methods, in cludi ng some non-Van der monde determ inant i

17、nto the Van dermonde determ inant,and the n use the results of the Van der monde determ inant,itcalculated. The com mon method of follow ing1. Given determ inant of the colu mns (or rows) are differe nt powers of an eleme nt, but the nu mber of power arra ngeme nt with the Van der monde determ inant

18、 is not exactly the same, the n eed to use the n ature of the determ inan t (such as extracti on com mon divisor, cha nge each line (or colu mn) order, the dissoluti on of items, etc.) as the determ inant of the Van der monde determ inant.Example 1.2232IIIIIIIIIIII3nSolutions ofDn elements of each r

19、ow are a number from left to right in ascendingorder, but not from 0 to n . But by a delivery ros . The com mon factor, such as extracti on of each line number of a power from zero change to n1.111III11222III2门4Dn = n!1332III3nA= n!(2 1)(31)川(n 1)(32)川(n 2)川n (n 1)IIIIIIIIIIIIIII1n2 nIIIn二n!(n -1)!(

20、n -2)!川2!1!Example 2nan 1aDn 1 =|l|(a-1)n 1(a -1)IIIa1IIIIII in in in(a _n)n/(a - n)Soluti on The law of the law of the determ inant arran ged with the arran geme nt of theVan der monde determ inant on the con trary, to make the colu mns in theDn d elements ofa power of frequency from top to bottom

21、in ascending order, the n 亠 1 columns uplinkswitch in turn until the first line,n rows sequentially exchanged with the uplink until thesecond row 2nd row sequentially with uplink switch until the firstn rows, so after a total ofn (n 一 1) (n 一2) |(2 1 二n(nSub-l ine excha nge Van der monde determ inan

22、t of ordern(n 1)11卅1aa -1a nIIIIIIhiIIIn A. a(a-1)(a-n)nn a(a -1)卅(a_ n)nn(n 1)n(a -1 -a)(a - 2 - a) (|(a -n -a)a - 2 - (a -1)川a - n - (a -(n -1)1 - k!kTIf Dn i th row (column) consistsof two bran ches (colu mn), any two adjace nt lines(columns) contain the same branch(column); and contains the Vand

23、er monde ranks nbran ches (colu mn) type, the n theDni-th row (column) multiplied by -1 added to the-row (column) to eliminate some of thebran ches (colu mn)into the Vander mondedeterm inant :Example311 sin ：sin sin2 ：、sin2 已 sin3 耳11sin2sin ：2 sin22sin2 冲2 sin3 邛211 sin 门3sin3 sin2 ：3sin2 % sin3 %1

24、1 sin4sin 4sin24sin2 冲4 sin3：； 4Solution will be the first line of the d multiplied by -1 to the second line we have:si n為 sin sin $ 住 sin彳勒 sin 命sin 处2si n疗 2sin2住 2sin2 2si n茂 2si n 处 3sin 3 sin2住 3sin23 sin3 茂3si n 处 4sin 住 4 sin2处 4sin2 4 si n364sin %sin 尬 22不2不sinsin :-.j2sin 3迟si n3 住2The n th

25、e sec ond row of the above determ inant is multiplied by -1 added to line 3, the new determ inant line 3 line 4 was multiplied by -1 added:si n 3 sin 4.2币.2不= (sini sinj) sin OP3 sin 41 勺心sin3 3 sin4 4plus line to add lawEach row (or colu mn) the differe nt square a power of the whose eleme nts are

26、all the of an eleme nt, but are the lack of the the determ inant of the same party a power of, available this method:Example 411III12X12X2III2Xn33I 1 JX1X2IIIX3nIIIHIIHHInX1nX2IIInXnDSoluti on for order determ inant111川1zX1X2川XnDn+ =2 z2X12X2III2Xn333III3z片X2Xn川IIIIIIHIHin zXnX；IIIx：n二口(x - z) 口 (Xj

27、 - Xk)i dl 乞k : j nz see n by made determ inant coefficie nt-D , by the coefficients of the above equationz：(-1)2nX1X2|“XnC 丄)H(Xj - Xji m Xi n 3 % 注By comparing the coefficients obtained:D人( 丄) (Xj - Xk)i =1 Xi n _ij k _l.Laplace expa nsion methodsApply the formulaM1AM2A2 JHMnAnto calculate the val

28、uethedeterm inant :Example 510X10HIn 1X1 010y1HI010X20HIn 1X2 =010y2川0ill川IHHIIH10Xn0川n 1Xn 一010yn川02n 1,1,3,川 2 n-1seriesDexpa nsionDereferenee lines1, 3,0n丄 yi on 1y2 一IIIon丄 yn 1X1IIIn A.X11y1IIIny11X2IIIn A.X21y2IIIny2IIIinHIHIHIIIIHIIII1XnIIIn A.Xn1ynIIInYnD =(XjXi)(yj - yj n：_j iproduct tra ns

29、formatio n methodExample 6nSet up sk 二 x1 x；八 x：(k =0,1j|2n -2) , Compute the determinanti=1s0qs1S2川IIID =IIIHIHisn 4SnIIIsn_1snIII%二2Solutionnn v 川i =1nnXi、x2 IHi di =1ziTn -1xinXiIII IH IHIHIIIn2n -2Xii =1 TOC o 1-5 h z HYPERLINK l bookmark109 o Current Document nn n -1l n HYPERLINK l bookmark252

30、o Current Document XiX2i di =111IIIIII11X2为IIIn _!X1XiX2IIIXn2n 1221 i j21X2屜IIIX2_XiX2IIIXnIIIHIIIIIH|IHIIIIII2HIIIIn AnndIIIn 二1XnXnIIIXnXiX2Xn=n (召-x)2i 岂:j n.Ascending OrderThe solutionD l order for the followingExample 7 compute the determi nant1X2X1川n_2nX11X22X2IHn_2%nX2IIIHIHIIIIHin十=12IHndnJXn4Xn铝Xn1XnXIHndXinJXn1Xx2IHndXnXnX1nX2HInXn_1nXnInsert a row with a l_n 1，X, X2,丨 11焉，x,n 1 order Van de