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2024年4月12日 星期五

113年暨大電機碩士班-工程數學詳解

 國立暨南國際大學113學年度碩士班入學考試

科目:工程數學(線性代數、微分方程)

解答: (a)det(A)=0+12+1230+24=45(b)[AI]=[223100216010120001]R12R3R1,R2+2R3R2[063102036012120001]R3R3,R2/3R2[06310201201323120001]R16R2R1,R22R2R3[00151220120132310402313]R1R3[10402313012013230015122]R3/(15)R3[1040231301201323001115215215]R1+4R3R1,R22R3R2[1004152151501021511525001115215215]A1=[4152151521511525115215215](c)det(AλI)=(λ+3)2(λ5)=0λ=3,5λ1=3(Aλ1I)v=0[123246123][x1x2x3]=0x1+2x2=3x3v=x2(210)+x3(301),choose v1=(210),v2=(301)λ2=5(Aλ2I)v=0[723246125][x1x2x3]=0{x1+x3=0x2+2x3=0v=x3(121), choose v3=(121)eigenvalues: 3,5 and eigenvectors: (210),(301),(121)(d)AB=[223216120][582]=[0621](e)rref(A)=[100010001]Ax=0[100010001][x1x2x3]=0x=[000]



解答{kx1+x2+x3=2kx1+kx2+x3=3x1+x2+kx3=1[k111k111k][x1x2x3]=[2k31]Ax=b(a)det(A)=k33k+2=(k1)2(k+2)0k1,k2(b)k=1{x1+x2+x3=2x1+x2+x3=3x1+x2+x3=1,no solution (c)k=2A=[211121112]rref(A)=[101011000],infinitely many solutions(d)x1=|2k113k111k|det(A)=2k36k+4k33k+2=2,x2=|k2k113111k|det(A)=k2+k2k33k+2=1k1x3=|k12k1k3111|det(A)=k2k+2k33k+2=1k1[x1x2x3]=[21/(k1)1/(k1)](e)k=0[x1x2x3]=[21/(k1)1/(k1)]=[211]


解答: (a)y1=xAy1=AxA1y1=A(A1)xA2(x2x)y1+Bxy1+y1=(A2A+AB+1)xA(A2A)xA1=0{A2A+AB+1=0A2A=0{A=1B=1(b)y2=uy1=uxy2=ux+uy2=2u+ux(x2x)y2xy2+y2=(x3x2)u+(x22x)u=0Let v=u, then v+x2x2xv=01vdv=x2x2xdxlnv=lnx1x2+c1v=c2x1x2u=c2x1x2dx=c2(lnx+1x)+c3y2=ux=c2(xlnx+1)+c3xy=y1+y2y=c2(xlnx+1)+c4x

解答: (a){P(x,y)=ex+y+2xyeyQ(x,y)=x2ey2ey{Py=ex+y+2xey+2xyeyQx=2xeyPyQxP=ex+y+2xyeyex+y+2xyey=1u=uintegrating factor u(y)=ey(b){uP=ex+2xyuQ=x22e2y{(uP)y=2x(uQ)x=2xexactΦ(x,y)=(ex+2xy)dx=(x22e2y)dyΦ(x,y)=ex+x2y+ϕ(y)=x2y+e2y+ρ(x)ex+x2y+e2y=C

解答: (a)y1=e2xy1=2e2xy1=4e2xy+y+Ey=(2+E)e2x=0E=2(b)y+y2y=0λ2+λ2=0(λ+2)(λ1)=0λ=1,2y=c1ex+c2e2x

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