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+12−3−0+24=45(b)[A∣I]=[−22−310021−6010−1−20001]R1−2R3→R1,R2+2R3→R2→[06−310−20−3−6012−1−20001]−R3→R3,−R2/3→R2→[06−310−20120−13−2312000−1]R1−6R2→R1,R2−2R2→R3→[00−151220120−13−2310−402313]R1↔R3→[10−4023130120−13−2300−15122]R3/(−15)→R3→[10−4023130120−13−23001−115−215−215]R1+4R3→R1,R2−2R3→R2→[100−415215−15010215−115−25001−115−215−215]⇒A−1=[−415215−15215−115−25−115−215−215](c)det(A−λI)=−(λ+3)2(λ−5)=0⇒λ=−3,5λ1=−3⇒(A−λ1I)v=0⇒[12−324−6−1−23][x1x2x3]=0⇒x1+2x2=3x3⇒v=x2(−210)+x3(301),choose v1=(−210),v2=(301)λ2=5⇒(A−λ2I)v=0⇒[−72−32−4−6−1−2−5][x1x2x3]=0⇒{x1+x3=0x2+2x3=0⇒v=x3(−1−21), choose v3=(−1−21)⇒eigenvalues: −3,5 and eigenvectors: (−210),(301),(−1−21)(d)AB=[−22−321−6−1−20][582]=[06−21](e)rref(A)=[100010001]⇒Ax=0⇒[100010001][x1x2x3]=0⇒x=[000]解答: {kx1+x2+x3=2kx1+kx2+x3=3x1+x2+kx3=1⇒[k111k111k][x1x2x3]=[2k31]≡Ax=b(a)det(A)=k3−3k+2=(k−1)2(k+2)≠0⇒k≠1,k≠−2(b)k=1⇒{x1+x2+x3=2x1+x2+x3=3x1+x2+x3=1,no solution (c)k=−2⇒A=[−2111−2111−2]⇒rref(A)=[10−101−1000],infinitely many solutions(d)x1=|2k113k111k|det(A)=2k3−6k+4k3−3k+2=2,x2=|k2k113111k|det(A)=k2+k−2k3−3k+2=1k−1x3=|k12k1k3111|det(A)=−k2−k+2k3−3k+2=−1k−1⇒[x1x2x3]=[21/(k−1)−1/(k−1)](e)k=0⇒[x1x2x3]=[21/(k−1)−1/(k−1)]=[2−11]
解答: (a)y1=xA⇒y′1=AxA−1⇒y″1=A(A−1)xA−2⇒(x2−x)y″1+Bxy′1+y1=(A2−A+AB+1)xA−(A2−A)xA−1=0⇒{A2−A+AB+1=0A2−A=0⇒{A=1B=−1(b)y2=uy1=ux⇒y′2=u′x+u⇒y″2=2u′+u″x⇒(x2−x)y″2−xy′2+y2=(x3−x2)u″+(x2−2x)u′=0Let v=u′, then v′+x−2x2−xv=0⇒∫1vdv=∫−x−2x2−xdx⇒lnv=lnx−1x2+c1⇒v=c2⋅x−1x2⇒u=∫c2⋅x−1x2dx=c2(lnx+1x)+c3⇒y2=ux=c2(xlnx+1)+c3x⇒y=y1+y2⇒y=c2(xlnx+1)+c4x
解答: (a){P(x,y)=ex+y+2xyeyQ(x,y)=x2ey−2e−y⇒{Py=ex+y+2xey+2xyeyQx=2xey⇒−Py−QxP=−ex+y+2xyeyex+y+2xyey=−1⇒u′=−u⇒integrating factor u(y)=e−y(b){uP=ex+2xyuQ=x2−2e−2y⇒{(uP)y=2x(uQ)x=2x⇒exact⇒Φ(x,y)=∫(ex+2xy)dx=∫(x2−2e−2y)dy⇒Φ(x,y)=ex+x2y+ϕ(y)=x2y+e−2y+ρ(x)⇒ex+x2y+e−2y=C

解答: (a)y1=e−2x⇒y′1=−2e−2x⇒y″1=4e−2x⇒y″+y′+Ey=(2+E)e−2x=0⇒E=−2(b)y″+y′−2y=0⇒λ2+λ−2=0⇒(λ+2)(λ−1)=0⇒λ=1,−2⇒y=c1ex+c2e−2x
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