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:$$\cases{3x_1+ x_2+x_3+2x_4 =1\\ x_1+ 3x_2+2x_3 +3x_4=4\\ 2x_1+ x_2+ 3x_3+3x_4=3\\ x_1+2x_2+ x_3+2x_4=2} \Rightarrow \begin{bmatrix}3 & 1& 1& 2\\ 1 & 3& 2& 3\\ 2& 1& 3& 3\\ 1& 2& 1& 2 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} =\begin{bmatrix} 1\\4\\3\\2\end{bmatrix} \equiv A\mathbf x=b\\ \left[\begin{array}{rrrr|rrrr}3 & 1& 1& 2 & 1& 0 & 0 & 0\\ 1 & 3& 2& 3 & 0 & 1& 0 & 0\\ 2& 1& 3& 3 & 0 & 0 & 1& 0\\ 1& 2& 1& 2 & 0 & 0 & 0 & 1\end{array} \right] \xrightarrow{R_1/3\to R_1} \left[ \begin{array}{rrrr|rrrr} 1 & \frac{1}{3} & \frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 \\1 & 3 & 2 & 3 & 0 & 1 & 0 & 0 \\2 & 1 & 3 & 3 & 0 & 0 & 1 & 0 \\1 & 2 & 1 & 2 & 0 & 0 & 0 & 1 \end{array} \right] \\ \xrightarrow{-R_1+R_2 \to R_2, -2R_1+R_3\to R_3, -R_1+R_4 \to R_4} \left[ \begin{array}{rrrr|rrrr}1 & \frac{1}{3} & \frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 \\0 & \frac{8}{3} & \frac{5}{3} & \frac{7}{3} & \frac{-1}{3} & 1 & 0 & 0 \\0 & \frac{1}{3} & \frac{7}{3} & \frac{5}{3} & \frac{-2}{3} & 0 & 1 & 0 \\0 & \frac{5}{3} & \frac{2}{3} & \frac{4}{3} & \frac{-1}{3} & 0 & 0 & 1
\end{array} \right] \\ \xrightarrow{3R_2/8 \to R_2} \left[ \begin{array}{rrrr|rrrr}1 & \frac{1}{3} & \frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 \\0 & 1 & \frac{5}{8} & \frac{7}{8} & \frac{-1}{8} & \frac{3}{8} & 0 & 0 \\0 & \frac{1}{3} & \frac{7}{3} & \frac{5}{3} & \frac{-2}{3} & 0 & 1 & 0 \\0 & \frac{5}{3} & \frac{2}{3} & \frac{4}{3} & \frac{-1}{3} & 0 & 0 & 1 \end{array} \right] \xrightarrow{-R_2/3+R_3\to R_3,-5R_2/3+R_4\to R_4} \\ \left[ \begin{array}{rrrr|rrrr} 1 & \frac{1}{3} & \frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 \\0 & 1 & \frac{5}{8} & \frac{7}{8} & \frac{-1}{8} & \frac{3}{8} & 0 & 0 \\0 & 0 & \frac{17}{8} & \frac{11}{8} & \frac{-5}{8} & \frac{-1}{8} & 1 & 0 \\0 & 0 & \frac{-3}{8} & \frac{-1}{8} & \frac{-1}{8} & \frac{-5}{8} & 0 & 1\end{array} \right] \xrightarrow{8R_3/17 \to R_3} \left[ \begin{array}{rrrr|rrrr}1 & \frac{1}{3} & \frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 \\0 & 1 & \frac{5}{8} & \frac{7}{8} & \frac{-1}{8} & \frac{3}{8} & 0 & 0 \\0 & 0 & 1 & \frac{11}{17} & \frac{-5}{17} & \frac{-1}{17} & \frac{8}{17} & 0 \\0 & 0 & \frac{-3}{8} & \frac{-1}{8} & \frac{-1}{8} & \frac{-5}{8} & 0 & 1 \end{array} \right] \\\xrightarrow{3R_3/8+R_4 \to R_4} \left[ \begin{array}{rrrr|rrrr} 1 & \frac{1}{3} & \frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 \\0 & 1 & \frac{5}{8} & \frac{7}{8} & \frac{-1}{8} & \frac{3}{8} & 0 & 0 \\0 & 0 & 1 & \frac{11}{17} & \frac{-5}{17} & \frac{-1}{17} & \frac{8}{17} & 0 \\0 & 0 & 0 & \frac{2}{17} & \frac{-4}{17} & \frac{-11}{17} & \frac{3}{17} & 1\end{array} \right] \xrightarrow{17R_4/2 \to R_4} \\\left[ \begin{array}{rrrr|rrrr} 1 & \frac{1}{3} & \frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 & 0 & 0 \\0 & 1 & \frac{5}{8} & \frac{7}{8} & \frac{-1}{8} & \frac{3}{8} & 0 & 0 \\0 & 0 & 1 & \frac{11}{17} & \frac{-5}{17} & \frac{-1}{17} & \frac{8}{17} & 0 \\0 & 0 & 0 & 1 & -2 & \frac{-11}{2} & \frac{3}{2} & \frac{17}{2} \end{array} \right] \xrightarrow{-2R_4/3+R_1 \to R_4, -7R_4/8+R_2 \to R_2, -11R_4/17+R_3 \to R_3}  \\\left[ \begin{array}{rrrr|rrrr} 1 & \frac{1}{3} & \frac{1}{3} & 0 & \frac{5}{3} & \frac{11}{3} & -1 & \frac{-17}{3} \\0 & 1 & \frac{5}{8} & 0 & \frac{13}{8} & \frac{83}{16} & \frac{-21}{16} & \frac{-119}{16} \\0 & 0 & 1 & 0 & 1 & \frac{7}{2} & \frac{-1}{2} & \frac{-11}{2} \\0 & 0 & 0 & 1 & -2 & \frac{-11}{2} & \frac{3}{2} & \frac{17}{2}\end{array} \right] \xrightarrow{-R_3/3+R_1 \to R_1, -5R_3/8+R_2 \to R_2}  \\\left[ \begin{array}{rrrr|rrrr} 1 & \frac{1}{3} & 0 & 0 & \frac{4}{3} & \frac{5}{2} & \frac{-5}{6} & \frac{-23}{6} \\0 & 1 & 0 & 0 & 1 & 3 & -1 & -4 \\0 & 0 & 1 & 0 & 1 & \frac{7}{2} & \frac{-1}{2} & \frac{-11}{2} \\0 & 0 & 0 & 1 & -2 & \frac{-11}{2} & \frac{3}{2} & \frac{17}{2} \end{array} \right] \xrightarrow{-R_2/3+R_1 \to R_1}  \left[ \begin{array}{rrrr|rrrr} 1 & 0 & 0 & 0 & 1 & \frac{3}{2} & \frac{-1}{2} & \frac{-5}{2} \\0 & 1 & 0 & 0 & 1 & 3 & -1 & -4 \\0 & 0 & 1 & 0 & 1 & \frac{7}{2} & \frac{-1}{2} & \frac{-11}{2} \\0 & 0 & 0 & 1 & -2 & \frac{-11}{2} & \frac{3}{2} & \frac{17}{2}\end{array} \right]  \\\Rightarrow A^{-1}= \begin{bmatrix}1& 3/2& -1/2& -5/2\\ 1& 3& -1& -4\\ 1& 7/2& -1/2& -11/2\\ -2& -11/2& 3/2& 17/2 \end{bmatrix} \\ \Rightarrow \mathbf x=A^{-1} b=\begin{bmatrix}1& 3/2& -1/2& -5/2\\ 1& 3& -1& -4\\ 1& 7/2& -1/2& -11/2\\ -2& -11/2& 3/2& 17/2 \end{bmatrix} \begin{bmatrix}1\\ 4\\ 3\\2  \end{bmatrix} =\begin{bmatrix}1/2\\ 2\\ 5/2\\-5/2  \end{bmatrix}  \Rightarrow \bbox[red,2pt]{\cases{x_1=1/2\\ x_2= 2\\ x_3=5/2\\ x_4=-5/2}}$$

解答:$$\left|\begin{matrix}0 & 1 & 2 & -1 \\2 & 5 & -7 & 3 \\0 & 3 & 6 & 2 \\-2 & -5 & 4 & -2\end{matrix}\right| \xrightarrow{R_2 \leftrightarrow -R_1} \left| \begin{matrix} 2 & 5 & -7 & 3 \\0 & -1 & -2 & 1 \\0 & 3 & 6 & 2 \\-2 & -5 & 4 & -2\end{matrix} \right| \xrightarrow {R_1+R_4\to R_4} \left| \begin{matrix} 2 & 5 & -7 & 3 \\0 & -1 & -2 & 1 \\0 & 3 & 6 & 2 \\0 & 0 & -3 & 1\end{matrix} \right| \\ \xrightarrow{3R_2+R_3\to R_3} \left| \begin{matrix} 2 & 5 & -7 & 3 \\0 & -1 & -2 & 1 \\0 & 0 & 0 & 5 \\0 & 0 & -3 & 1\end{matrix} \right| \xrightarrow{-R_3 \leftrightarrow R_4} \left| \begin{matrix}2 & 5 & -7 & 3 \\0 & -1 & -2 & 1 \\0 & 0 & -3 & 1 \\0 & 0 & 0 & -5\end{matrix} \right| \\=2\cdot(-1)\cdot(-3) \cdot (-5)= \bbox[red, 2pt]{-30}$$

解答:$$\mathbf{(a)}\;A=\begin{bmatrix}1& 2& 3& 4\\ 8 & 7 & 6 & 5 \\ 9 & 10& 11 & 12\\ 16 & 15 & 14 & 13\end{bmatrix} \Rightarrow rref(A)=\begin{bmatrix}1& 0& -1& -2\\ 0 & 1 & 2 & 3 \\ 0 & 0& 0 & 0\\ 0 & 0 & 0 & 0\end{bmatrix}\\ Ax=0 \Rightarrow \cases{x_1-x_3-2x_4=0\\ x_2+2x_3+3x_4=0} \Rightarrow \begin{bmatrix} x_1 \\ x_2 \\x_3\\ x_4\end{bmatrix} =s\begin{bmatrix} 1 \\0 \\ -3\\ 2 \end{bmatrix} +t\begin{bmatrix} 0 \\1 \\ -2\\ 1 \end{bmatrix} \\ \Rightarrow \text{A basis is } \bbox[red, 2pt]{\left\{ \begin{bmatrix} 1 \\0 \\ -3\\ 2 \end{bmatrix}, \begin{bmatrix} 0 \\1 \\ -2\\ 1 \end{bmatrix} \right\}}, \text{and }\dim(N(A))=\bbox[red, 2pt]2$$ $$\mathbf{(b)}\; rref(A)=\begin{bmatrix}1& 0& -1& -2\\ 0 & 1 & 2 & 3 \\ 0 & 0& 0 & 0\\ 0 & 0 & 0 & 0\end{bmatrix} \Rightarrow Col(A)= \bbox[red, 2pt]{\left\{ a\begin{bmatrix}1 \\ 8\\9 \\16 \end{bmatrix}+ b \begin{bmatrix}2 \\ 7\\10 \\15 \end{bmatrix} \mid a,b \in \mathbb R \right\}} \\ \Rightarrow rank(A)= \bbox[red, 2pt]2$$ $$\mathbf{(c)}\; Ax=b,假設擴增矩陣為(A|b), 基底代表線性獨立的個數,因此\\ 若rank(A)= rank(A|b)=未知數的個數 \Rightarrow 有唯一解\\ 若rank(A)=rank(A|b) <未知數的個數 \Rightarrow 有無窮多解\\ 若rank(A)\lt rank(A|b) \Rightarrow 無解$$

解答:$$令\cases{a_1=x_3\\ a_2=x_2\\ a_3=x_1} ,藉由\text{Gram–Schmidt Process}求正交基底\\ e_1={a_1\over \Vert a_1 \Vert} ={1\over \sqrt 2} \begin{bmatrix} 1\\ 0 \\ 0 \\ 1\end{bmatrix} \\ b_2= a_2-(a_2 \cdot e_1)e_1 =\begin{bmatrix} 1\\ 2 \\ 0 \\ 0 \end{bmatrix} -{1\over   2}\begin{bmatrix} 1\\ 0 \\ 0 \\ 1\end{bmatrix} =\begin{bmatrix} 1/2\\ 2 \\ 0 \\ -1/2\end{bmatrix}\\ \Rightarrow e_2= {b_2\over \Vert b_2\Vert} = \begin{bmatrix} \sqrt 2/6\\2\sqrt 2/3 \\ 0 \\ -\sqrt 2/6\end{bmatrix} \\ b_3=a_3-(a_3 \cdot e_1)e_1-(a_3\cdot e_2)e_2 =\begin{bmatrix} 1\\ 2 \\ 3 \\ 0\end{bmatrix} -{1\over   2}\begin{bmatrix} 1\\ 0 \\ 0 \\ 1\end{bmatrix} -\begin{bmatrix} 1/2\\ 2 \\ 0 \\ -1/2\end{bmatrix}= \begin{bmatrix} 0\\ 0 \\ 3 \\ 0\end{bmatrix} \\ \Rightarrow e_3= {b_3\over \Vert b_3 \Vert } =\begin{bmatrix} 0\\ 0\\ 1 \\ 0\end{bmatrix} \\ \Rightarrow \text{orthogonal basis }=\{e_1,e_2,e_3\} = \bbox[red, 2pt]{\left\{ \begin{bmatrix} 1/\sqrt 2\\ 0 \\ 0 \\ 1/\sqrt 2\end{bmatrix}, \begin{bmatrix} \sqrt 2/6\\2\sqrt 2/3 \\ 0 \\ -\sqrt 2/6 \end{bmatrix}, \begin{bmatrix} 0\\ 0\\ 1 \\ 0\end{bmatrix}  \right\}}$$

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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