行列式乘法規(guī)則的證明方法及其應(yīng)用2講解

ProofMethodofDeterminantMultiplicationRuleandItsApplicationName:StudentNumber:Advisor:asistosolvethee rr同 n 11 n Mr1nnMnLLL2rΜ=2rΜ=,MrrmmMmr1mmMmLLLmm2rMmrrr2 c 11 c r2rrrikijjk j=1 D==1.D==1.nnnr1100nnn010nnrrmrr1r1rrr1rrnnnr1nnnmmmrrrrmrr22rr22..mmrrrr2rrrrrr21m11111101m=m=rnrn2r2rn01r01rrr1102r2r1rrr1rrr1nnnn1001mm12,i+12,i+1 2,i1n000100n00nn00n0100100k1018kr–kr2r21222.r=「kr–kr2r21222.r=k2223k22233n2212i–,1i+2,1k2212i–,1i+2,1kk=i1|]nn]nnnnr1rr–r1rr–nnnr1nnnmrrrrnnmr1rrrrr1rrrrnnnrc00c=210rr00rr00,0mr2rr2r2rrrrrrikijjkj=1:r1r+1r+2rr0000==D0000==Dnnnr1100nnn010n1jj11jj21jjrnrnm2jj1rnm2jj2rnm2jjrnrrrnmrjj1rnmrjj2rnmrjjr00001000nnnr1100nnn0100ccnnr1rr,r1rr,001 n 1r n 1r n 2r =rcccr21222r=2r= ccc rrr1r2rrr1rr0r1rrqr)aaqr)aap)+T(qqr12pq1122rpq1122rrq,q,q,q,p,p,rr+1r+1rr+1r+1r+2r+2pqa,(pqnnp,p,p,r+1r+2n又p,p,,p,p,p,,p又p,p,,p,p,p,,p與q,q,12rr+1r+212rr+1r+2n121122rrr+1r+1n1122rrr+1r+1nn+12rr+1r+2n12rr+1r+2n12rT(ppp)T(ppp)=T(pp12rr+1r+2n12rr+1r+2n12r+12rr+1r+2n12rr+1r+2nT(pp12rr+1r+2n12rr+1r+2np,pp,p,r+1r+2nr+1r+2npprrr+nn12rrnrr+212nrr+1r+12n+12rr+1r+2n12rr+1nT(qqq)12rr+1r+2n12rr+1nr+r+1rr+1rrrr+1r+1nr+21r+212rr+2nTpppqqq12r12rr+1r+2nr+1r+2n=T(pppppp)[(p1)+(p2)++(pr)]12rr+1r+2n12rqnqrrnr+1r+2np)+T(qqqqqp)+T(qqqqq)(p+p++p12r)n12rr+1n12r12rr+1r+2+r)2+(q+q++r)2+(q+q++q)]12rp)+T(qq12rp)+T(qqqqq)+2(1+2++r)(p+p++p)n12rr+1n12r12rr+1r+2(q+q(q+q+12raaapqpqpqaaapqpqpqrrr+1rrrr+1r+1nn11221122rrr+1r+1n1122rrr+1r+1nnpqpqpq1pqpqpq1122 aa1122rrr+1r+1n1122rrr+1r+1nn 1122rrr+1r+1nn1122rrr+1r+1nn 12tttD=MA+MA++MA1122tt.iiMA(ij)jjiit=Cr=n!nr!(nr)!,MA與ii=1=1D則D1nnnn==pppppppk2rk2rpppppr1rrr1rr1nnnr1nnnnmmmrrrrrrnmmmrrrrrrmmmrr2rrrr2rr2m21pDDnpDDnnpppr1nnnpppr2rrrnpmmmkk1pmmm令p]「pn]p]「pn]p|11||||||11|||||11pD|n21DpD|n21D|m21pp2r2r2r2r||,3LpppLppp」rrr1r1r1rrr1r1r1rrD=D=1D3oDD1D3oD=DD.2DDy1yy1a1a2a321y1yy2b1b2b322x0x01x1210x0x01x2220x0x01x3230y1yy3c1c2c323供解例解例y1yy1a1a2a321y1yy2b1b2b322y1yy3c1c2c323x0x01x1210x0x01x2220x0x01x330=y1yy121a1a2a3y1yy222b1b2b3y1yy33c1c2c3=y1yy121y1yy222y1yy323x1xx121x1xx222x1xx323=(yy)(yy)(yy)(xx)(xx)(x3x0x001x121x0x001x22x).20001x3x2x3==b0.zxy=(ab2+a20.zxy=(ab2+a2b)zxy.byzxyzx10111=211=2cba.ac0sin.000sin.00sinysin00da dD2=DDbadccdabdcabadcbadc=