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2024年3月1日 星期五

112年台大碩士班-線性代數C詳解

 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=[uT1ut2uT3]=[123238513345]rref(A)=[101012000000] basis of W:{(1,2,5,3),(2,3,1,4)} and dimension of W=2(b)B=[uT1ut2uT3eT1eT2eT3eT4]=[1231000238010051300103450001]rref(B)=[1010041711701200317517000102179170000111]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+R2R2[sin2αsin2βsin2γsin2α+cos2αsin2β+cos2βsin2γ+cos2γ111]=[sin2αsin2βsin2γ111111]det(A)=0,α,β,γA is not invertible,Q.E.D.

解答:f(λ)=|51λ12216040λ2857681λ|,f(λ)=048,24αf(λ)=λ3+bλ2+cλ+d=(λ+48)(λ24)(λα)λ2b4824α=α24(51λ)(40λ)(1λ)=(λ51)(λ+40)(λ1)λ2(51+401)=12α24=12α=36:,4848

解答:1251212130121141=17

解答:A=[1124]=PDP1=[1/2111][3002][2221]eA=[1/2111][e300e2][2221]=[e2(2e)e2(e1)2e2(1e)e2(2e1)]


解答:(AI)(A2I)=0mA(t)=(t2)(t1)(BI)(B2I)0mB(t)=(t1)(t1)2
解答:[AI]=[132100375010258001]3R1+R2R2,3C1+C2C2[102100021310218001]2R1+R3R3,2C1+C3C3[100100021310014201]R3+0.5R2R3,C3+0.5C2C3[100100020310009212121][10031012121][132375258][10031012121]T=[1000200092] is  diagonalP=[13120112001]signature of A=(2,1)
解答:A=[u1,u1u1,u2u1,u3u2,u1u2,u2u2,u3u3,u1u3,u2u3,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=1/3P=[1/32/32/32/32/31/32/31/32/3]

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解題僅供參考,其他歷年試題及詳解

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