Loading [MathJax]/jax/output/CommonHTML/jax.js

2024年11月4日 星期一

112年政大轉學考-線性代數詳解

國立政治大學112學年度轉學生招生考試

科目:線性代數
系所別:應用數學系三年級

解答: is a subspace of R2(0,0)0=m0+bb=0b=0={(x,y)R2:y=mx}Suppose {w1=(x1,y1)w2=(x2,y2){y1=mx1y2=mx2ay1+by2=(ax1+bx2)m(ax1+bx2,ay1+by2)=aw1+bw2 is a subspace of R2QED
解答:A=[211121112]det(AλI)=λ(λ3)2=0λ=0,3λ1=0(Aλ1I)v=0[211121112][x1x2x3]=0{x1=x3x2=x3v=x3[111] we choose v1=[111]λ2=3(Aλ3I)v=0[111111111][x1x2x3]=0x1+x2+x3=0v=[x2x3x2x3]=x2[110]+x3[101] we choose v2=[110],v3=[101]P=[v1v2v3],D=[λ1000λ2000λ3]A=[111110101][000030003][131313132313131323]
解答:A=[100020003]det(AλI)=(λ1)(λ2)(λ3)=0λ1=3(Aλ1I)v=0[200010000][x1x2x3]=0x1=x2=0v=x3[001],we choose v1=[001]λ2=2(Aλ2I)v=0[100000001][x1x2x3x4]=0x1=x3=0v=x2[010],we choose v2=[010]λ3=1(Aλ3I)v=0[000010002][x1x2x3]=0x2=x3=0v=x1[100],we choose v3=[100]P=[v1v2v2]Yes, they are similar. P=[001010100]
解答:(a)T:RnR[T] is 1 by nrank([T])=1dim(N(T))=nrank([T])=n1QED(b)Suppose avi+bw=0,i=1(n1), then T(avi+bw)=0aT(vi)+bT(w)=0bT(w)=0b=0(wN(T)T(w)0)avi+bw=avi=0a=0vi and w are linearly independent B is a basis of RnQED(c)From (b), we have uRnuspan(B)u=a1v1+a2v2++an1vn1+bw We can choose v=a1v1+a2v2++an1vn1span(B)vN(T) ,and choose b=T(u)T(w), where T(w)0. Then u=v+T(u)T(w)wQED
解答:(a) [AI]=[121100254010110001]R22R1R2,R3R1R3[121100012210011101]R12R2R2,R3+R2R3[103520012210001311]R1+3R3R1,R22R3R2[100413010412001311]A1=[413412311](b) rref(A)=I3Yes, they are independent.(c) A1b=[332]b=3A1+3A22A3
解答:A=[2101031001100103]det(AλI)=(λ2)3(λ3)=0λ1=2(Aλ1I)v=0[0101011001100101][x1x2x3x4]=0{x2=x4x2=x3v=a[1000]+b[0111],we choose v1=[1000],v2=[0111]λ2=3(Aλ2I)v=0[1101001001200100][x1x2x3x4]=0{x1=x4x2=x3=0v=a[1001],we choose v3=[1001]A is 4 by 4 and has 3 eigenvectors, then J=[2100020000200003]

======================= END ======================

解題僅供參考,轉學考歷年試題及詳解

沒有留言:

張貼留言