大學-----行列式經典例題例1aij=||的n.解1a11=0a121，L,a1nn1,LDn10LnMO2Ln101Ln10111r1Lrii1MOin,n1,Ln1n2LL1011n1nLLn1ccjn02LL1MOOL(n12n2(nj,n1M20L1000n謝謝閱讀2DnLn11L011110Ln11L1rrMOMO2ii1i,n1n1n2Ln1n2L

00ccj1j,n1L0012L0OM=(1)2n2(n1)n1n12n3Ln1例設,,c是互異的實數(shù),證明:謝謝閱讀謝謝閱讀的充要條件是a+b+c=0.證明:考察范德蒙行列式:=行列式即為2前的系數(shù).于是=所以的充要條件是a+b+c=0.x00

1K0xK0例3計算D1=MnMMMKxaaaan1

n1n2111D=xDn）n1n1ax1nx1OO=xD+an1n由于Dana，D11=x+a22D+an2n=xaxLaxann11nn1x1n1x+a=L=xn1D1+axn2++an1n1n2xa2x+xa1，于1是nD=xD++a）+n1an=x（xDn1n22第2x3x2倍，nxn11精品文檔放心下載Dncxc0001Kx2xK0100xK012MMMMaxaKx

aaan

1n1n1

n20000

1K0x00

1KK0cx2c10x1x33MMMMManxaxaKxn1n2n32aa2aaaa2aaan2n11101

xx1

1OO按rx1n(1)n

1f=LL==OfxOOx1n1xnaxn1Laxa

1n

n13Dn1ccc21x1c32xKx00000K0x00xK0MMMMcn1xLcaann1n1anxan2aaKknn1按cn展開xn1k=xn(n1nxx2aaa+++=anaxLaxxn1nn11n2n1xn1xxn214Dn按rn(1)n1an10Kx1KMM0000MM+00Kx1(n2a2an1x0Kx1K000001K000xK00++(1)MMMMMM2n1aMM200Kx100K01+(1)(a2n1x)x1K0xKMM0000MML000x）n1）an）xn2n1n2）an1+）ax）(axn12n12

n22n1=aaxLaxxnn1

nn1

1例4．nLaaab2Laa11D(babnbn10)22nMMLbM12nLaaa1nb2n解a,a,L,a12n精品文檔放心下載Dn0a11cc1jbj1j,n1L1aaaLa120aLaanbn112na20aMba122MMMLan1a1b1L00Mrrr21r31Lrrn11a1b1b100n111MLa1b10La2000Lbn0b2MMDaLxa12aaLxa2LananManx=xn1(xani1in12MMa1例5．nK000D=nK10001＋K00MMMMM000K1解1Dn按c1()D—n11MK00KMM0000MM000K1(n1)按r()D－Dn1n2D=(n故DnDn而D1()，D2＝α+β1－Dn1n23))DDD)＝(Dn2n1n1D＝D)＝(Dn22(Dn2n1n1＝L＝n2(DD)21αβ＝α+β得D22nD)n3Dn1nDnDn1n）DnDn1n＝(Dn2n1)α2D＋n1n＝Lnn2βn－α

n1，當αβ

時＝＋＋＋＝β－α謝謝閱讀nn1n1n，當α＝β時

(n1)αn12把Dn12個n10Dn=01K000＋0101MM000M00K000M0MMM00M000K＋K0K0M0K0MK11Kn1K101MM00000M00K000M00KciK0i1i,n0000000=β00MM1MK1K101KMMMK000KKnDnDn1n31Dn=()D又因為當時D1==－Dn122n2D213=()==2220D=3101=()3()=()()=2244Dnn1n1當nk當謝謝閱讀D=()D－Dk1k1kk1k1kkk2k2=()—=nN例6．nDn1a1L111aL2MM11L11M1anaa12Lan0．解1Dnrri11a1a1M1La2O1ac1c1jab1L0a2MO1i,nja1anj,n0anb1a11111．1nnnaaaLaDa

aaa

11n12

ni2ii1i1ii2LL1111111101aL110L0

a11rDL110aL01rn201MMMMMi11ai,n12MMM01L1a100La1nn1L

1a11n1j1cci1a1aj11aaLa2O11jaaj,n1Lan12ani1ian3DnDn1a1L111aL2MM1010M111Lan01a1L11a1L11011aL111aL按cn拆開2+MMM2MMML111La11n1a1a1L111a11aLr2MMMraLaina212n1i1,L,n1L111L1111a1L111aL2MM00M按n展開aDnn11a1LnDan=aDa12nn1Dn=aDnn1