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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

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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