國立臺灣大學112學年度碩士班招生考試
題號:295
科目:線性代數(C)
解答:(a)A=[uT1∣ut2∣uT3]=[123−23851−3−3−4−5]⇒rref(A)=[10−1012000000]⇒ basis of W:{(1,−2,5,−3),(2,3,1,−4)} and dimension of W=2(b)B=[uT1∣ut2∣uT3∣eT1∣eT2∣eT3∣eT4]=[1231000−238010051−30010−3−4−50001]⇒rref(B)=[10−10041711701200−317−517000102179170000111]⇒extended basis: {u1,u2,e1,e2}={(1,−2,5,−3),(2,3,1,−4),(1,0,0,0),(0,1,0,0)}

解答:A=[sin2αsin2βsin2γcos2αcos2βcos2γ111]R1+R2→R2→[sin2αsin2βsin2γsin2α+cos2αsin2β+cos2βsin2γ+cos2γ111]=[sin2αsin2βsin2γ111111]⇒det(A)=0,∀α,β,γ⇒A is not invertible,Q.E.D.

解答:‖1−25−121−21301−21−14−1‖=17
解答:(A−I)(A−2I)=0⇒mA(t)=(t−2)(t−1)(B−I)(B−2I)≠0⇒mB(t)=(t−1)(t−1)2
解答:[A∣I]=[1−32100−37−50102−58001]3R1+R2→R2,3C1+C2→C2→[1021000−21310218001]−2R1+R3→R3,−2C1+C3→C3→[1001000−21310014−201]R3+0.5R2→R3,C3+0.5C2→C3→[1001000−203100092−12121]⇒[100310−12121][1−32−37−52−58][100310−12121]T=[1000−200092] is diagonal⇒P=[13−120112001]⇒signature of A=(2,1)
解答:A=[⟨u1,u1⟩⟨u1,u2⟩⟨u1,u3⟩⟨u2,u1⟩⟨u2,u2⟩⟨u2,u3⟩⟨u3,u1⟩⟨u3,u2⟩⟨u3,u3⟩]=[2343142242235]

解答:P=[1/32/32/32/3ac2/3cb]⇒{(1/3,2/3,2/3)⋅(2/3,a,c)=0(1/3,2/3,2/3)⋅(2/3,c,b)=0(2/3,a,c)⋅(2/3,c,b)=0⇒{a=ba+c=−1/3ac=−2/9choose{a=−2/3b=−2/3c=1/3⇒P=[1/32/32/32/3−2/31/32/31/3−2/3]
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