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2023年12月23日 星期六

112年成大測量-線性代數詳解

 國立成功大學112學年度碩士班招生考試

系所:測量及空間資訊學系
科目:線性代數

解答:{3x1+x2+x3+2x4=1x1+3x2+2x3+3x4=42x1+x2+3x3+3x4=3x1+2x2+x3+2x4=2⇒[3112132321331212][x1x2x3x4]=[1432]≡Ax=b[31121000132301002133001012120001]R1/3→R1→[113132313000132301002133001012120001]−R1+R2→R2,−2R1+R3→R3,−R1+R4→R4→[1131323130000835373−131000137353−230100532343−13001]3R2/8→R2→[113132313000015878−1838000137353−230100532343−13001]−R2/3+R3→R3,−5R2/3+R4→R4→[113132313000015878−18380000178118−58−181000−38−18−18−5801]8R3/17→R3→[113132313000015878−1838000011117−517−117817000−38−18−18−5801]3R3/8+R4→R4→[113132313000015878−1838000011117−517−1178170000217−417−11173171]17R4/2→R4→[113132313000015878−1838000011117−517−11781700001−2−11232172]−2R4/3+R1→R4,−7R4/8+R2→R2,−11R4/17+R3→R3→[11313053113−1−173015801388316−2116−119160010172−12−1120001−2−11232172]−R3/3+R1→R1,−5R3/8+R2→R2→[113004352−56−236010013−1−40010172−12−1120001−2−11232172]−R2/3+R1→R1→[1000132−12−52010013−1−40010172−12−1120001−2−11232172]⇒A−1=[13/2−1/2−5/213−1−417/2−1/2−11/2−2−11/23/217/2]⇒x=A−1b=[13/2−1/2−5/213−1−417/2−1/2−11/2−2−11/23/217/2][1432]=[1/225/2−5/2]⇒{x1=1/2x2=2x3=5/2x4=−5/2
解答:|012−125−730362−2−54−2|R2↔−R1→|25−730−1−210362−2−54−2|R1+R4→R4→|25−730−1−21036200−31|3R2+R3→R3→|25−730−1−21000500−31|−R3↔R4→|25−730−1−2100−31000−5|=2⋅(−1)⋅(−3)⋅(−5)=−30


解答:(a)A=[12348765910111216151413]⇒rref(A)=[10−1−2012300000000]Ax=0⇒{x1−x3−2x4=0x2+2x3+3x4=0⇒[x1x2x3x4]=s[10−32]+t[01−21]⇒A basis is {[10−32],[01−21]},and dim(N(A))=2 (b)rref(A)=[10−1−2012300000000]⇒Col(A)={a[18916]+b[271015]∣a,b∈R}⇒rank(A)=2 (c)Ax=b,假設擴增矩陣為(A|b),基底代表線性獨立的個數,因此若rank(A)=rank(A|b)=未知數的個數⇒有唯一解若rank(A)=rank(A|b)<未知數的個數⇒有無窮多解若rank(A)<rank(A|b)⇒無解


解答:令{a1=x3a2=x2a3=x1,藉由Gram–Schmidt Process求正交基底e1=a1‖a1‖=1√2[1001]b2=a2−(a2⋅e1)e1=[1200]−12[1001]=[1/220−1/2]⇒e2=b2‖b2‖=[√2/62√2/30−√2/6]b3=a3−(a3⋅e1)e1−(a3⋅e2)e2=[1230]−12[1001]−[1/220−1/2]=[0030]⇒e3=b3‖b3‖=[0010]⇒orthogonal basis ={e1,e2,e3}={[1/√2001/√2],[√2/62√2/30−√2/6],[0010]}

解答:A=[3−24−262423]⇒det(A−λI)=0⇒−(λ+2)(λ−7)2=0Thus, the eigenvalues of A are λ=−2,7,7,the corresponding eigenvectors are v1=[−1−1/21],v2=[−1/210],v3=[101],so P=[v1,v2,v3], but v2⊥̸v3We choose{u2=−v3−v2u3=v3/2−v2⇒u2=[−1/2−1−1],u3=[1−11/2]Q=[v1‖v1‖,u2‖u2‖,u3‖u3‖]=[−2/3−1/32/3−1/3−2/3−2/32/3−2/31/3]⇒Q−1=QT=Q⇒QAQ−1=[−200070007]

解答:A=[011110]⇒B=ATA=[2112]⇒the eigenvalues of A are 1,3,and the corresponding eigenvectros are u1=[−11],u2=[11]⇒u1‖u1‖=[−1/√21/√2],u2‖u2‖=[1/√21/√2]⇒∑=[100√3],VT=[−1/√21/√21/√21/√2]⇒A=U∑VT⇒AV=[011110][−1/√21/√21/√21/√2]=[1/√21/√202/√2−1/√21/√2]=[1/√21/√602/√6−1/√21/√6]⇒U=AV∑=[1/√21/√602/√6−1/√21/√6][100√3]=[1/√21/√202/√2−1/√21/√2]=[1/√21/√602/√6−1/√21/√6]⇒A=U∑VT=[1/√21/√602/√6−1/√21/√6][100√3][−1/√21/√21/√21/√2]

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