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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]R1/3R1[113132313000132301002133001012120001]R1+R2R2,2R1+R3R3,R1+R4R4[113132313000083537313100013735323010053234313001]3R2/8R2[113132313000015878183800013735323010053234313001]R2/3+R3R3,5R2/3+R4R4[11313231300001587818380000178118581810003818185801]8R3/17R3[11313231300001587818380000111175171178170003818185801]3R3/8+R4R4[1131323130000158781838000011117517117817000021741711173171]17R4/2R4[113132313000015878183800001111751711781700001211232172]2R4/3+R1R4,7R4/8+R2R2,11R4/17+R3R3[1131305311311730158013883162116119160010172121120001211232172]R3/3+R1R1,5R3/8+R2R2[11300435256236010013140010172121120001211232172]R2/3+R1R1[10001321252010013140010172121120001211232172]A1=[13/21/25/2131417/21/211/2211/23/217/2]x=A1b=[13/21/25/2131417/21/211/2211/23/217/2][1432]=[1/225/25/2]{x1=1/2x2=2x3=5/2x4=5/2
解答:|0121257303622542|R2R1|2573012103622542|R1+R4R4|2573012103620031|3R2+R3R3|2573012100050031|R3R4|2573012100310005|=2(1)(3)(5)=30


解答:(a)A=[12348765910111216151413]rref(A)=[1012012300000000]Ax=0{x1x32x4=0x2+2x3+3x4=0[x1x2x3x4]=s[1032]+t[0121]A basis is {[1032],[0121]},and dim(N(A))=2 (b)rref(A)=[1012012300000000]Col(A)={a[18916]+b[271015]a,bR}rank(A)=2 (c)Ax=b,(A|b),,rank(A)=rank(A|b)=rank(A)=rank(A|b)<rank(A)<rank(A|b)


解答:{a1=x3a2=x2a3=x1,Gram–Schmidt Processe1=a1

解答:A=\begin{bmatrix} 3&-2 & 4\\ -2& 6& 2\\ 4& 2& 3\end{bmatrix} \Rightarrow \det(A-\lambda I) = 0 \Rightarrow -(\lambda+2)(\lambda-7)^2=0\\ \text{Thus, the eigenvalues of }A \text{ are }\lambda=-2,7,7, \text{the corresponding eigenvectors are }\\ v_1=\begin{bmatrix} -1\\ -1/2\\ 1\end{bmatrix}, v_2=\begin{bmatrix} -1/2\\ 1\\ 0\end{bmatrix}, v_3=\begin{bmatrix} 1\\ 0 \\ 1 \end{bmatrix},\text{so }P=[v_1,v_2,v_3],\text{ but } v_2 \not \bot v_3\\ \text{We choose} \cases{u_2=-v_3-v_2 \\u_3= v_3/2-v_2} \Rightarrow u_2=\begin{bmatrix} -1/2\\ -1\\ -1 \end{bmatrix}, u_3=\begin{bmatrix} 1\\ -1\\ 1/2\end{bmatrix}\\ Q=[{v_1\over \Vert v_1 \Vert}, {u_2 \over \Vert u_2\Vert},{u_3 \over \Vert u_3\Vert }]= \begin{bmatrix}-2/3 & -1/3& 2/3\\ -1/3 & -2/3 & -2/3\\ 2/3&-2/3 & 1/3 \end{bmatrix} \Rightarrow Q^{-1}=Q^T=Q\\ \Rightarrow \bbox[red, 2pt]{ QAQ^{-1}= \begin{bmatrix}-2 & 0 &0 \\ 0 & 7 & 0\\ 0 & 0 & 7 \end{bmatrix}}

解答:A=\left[ \begin{matrix}0 & 1 \\1 & 1 \\1 & 0\end{matrix}\right] \Rightarrow B=A^TA= \begin{bmatrix} 2& 1\\ 1& 2\end{bmatrix} \Rightarrow \text{the eigenvalues of A are }1,3, \\ \text{and the corresponding eigenvectros are }u_1=  \begin{bmatrix} -1\\ 1 \end{bmatrix}, u_2= \begin{bmatrix} 1\\ 1 \end{bmatrix} \\ \Rightarrow {u_1 \over \Vert u_1\Vert} =\begin{bmatrix} -1/\sqrt 2\\ 1/\sqrt 2 \end{bmatrix}, {u_2 \over \Vert u_2 \Vert} =\begin{bmatrix} 1/\sqrt 2\\ 1 /\sqrt 2 \end{bmatrix} \\ \Rightarrow \sum=\begin{bmatrix} 1& 0 \\0& \sqrt 3 \end{bmatrix} ,V^T=\begin{bmatrix} -1/\sqrt 2& 1/\sqrt 2\\1/\sqrt 2& 1/\sqrt 2 \end{bmatrix} \Rightarrow A=U\sum V^T \\ \Rightarrow AV=\left[ \begin{matrix}0 & 1 \\1 & 1 \\1 & 0\end{matrix}\right] \begin{bmatrix} -1/\sqrt 2& 1/\sqrt 2\\1/\sqrt 2& 1/\sqrt 2 \end{bmatrix}= \begin{bmatrix} 1/\sqrt 2& 1/\sqrt 2\\ 0& 2/\sqrt 2\\ -1/\sqrt 2 & 1/\sqrt 2 \end{bmatrix} =\begin{bmatrix} 1/\sqrt 2& 1/\sqrt 6\\ 0& 2/\sqrt 6\\ -1/\sqrt 2 & 1/\sqrt 6 \end{bmatrix}\\ \Rightarrow U=AV\sum =\begin{bmatrix} 1/\sqrt 2& 1/\sqrt 6\\ 0& 2/\sqrt 6\\ -1/\sqrt 2 & 1/\sqrt 6 \end{bmatrix} \begin{bmatrix} 1& 0 \\0& \sqrt 3 \end{bmatrix} =\begin{bmatrix}1/\sqrt 2& 1/\sqrt 2\\ 0 & 2/\sqrt 2\\ -1/\sqrt 2& 1/\sqrt 2 \end{bmatrix} =\begin{bmatrix}1/\sqrt 2& 1/\sqrt 6\\ 0 & 2/\sqrt 6\\ -1/\sqrt 2& 1/\sqrt 6 \end{bmatrix}\\ \Rightarrow A=U\sum V^T=  \bbox[red, 2pt]{\begin{bmatrix} 1/\sqrt 2& 1/ \sqrt 6\\ 0 & 2/\sqrt 6\\ -1/\sqrt 2& 1/\sqrt 6 \end{bmatrix} \begin{bmatrix} 1& 0 \\0& \sqrt 3 \end{bmatrix} \begin{bmatrix} -1/\sqrt 2& 1/\sqrt 2\\1/\sqrt 2& 1/\sqrt 2 \end{bmatrix}}

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