115年成功大學工程科學碩士班-線性代數詳解

國立成功大學115學年度碩士班招生考試試題

系所:工程科學系
科目:線性代數

解答:$$\textbf{(a) }\bbox[red, 2pt]T: f(x)= \det(xI-A)=x^2+5x+4 \Rightarrow A^2+5A+4I=0 \Rightarrow A(A^2+5A+4I) =0  \\\qquad \Rightarrow A^3+5A^2+4A=0 \Rightarrow A^3+5A^2+4A+I=I \\ \textbf{(b) }\bbox[red, 2pt]T: [A\mid b] = \begin{bmatrix}2&-1& 3\\ -6& 3& 1 \end{bmatrix} \Rightarrow \cases{2x-y=3 \Rightarrow -6x+3y=-9\\ -6x+3=1} \Rightarrow -9\ne -1 \Rightarrow \text{ no solution} \\\textbf{(c) }\bbox[red, 2pt]T:u\cdot v=2+1-3=0 \Rightarrow u\bot v \\ \textbf{(d) }\bbox[red, 2pt] F: \det \left( \begin{bmatrix}1& 1& 4\\0& 2& 4\\ 1& -1& 0 \end{bmatrix} \right) =4+4-8=0 \Rightarrow \text{ linearly dependent} \\\textbf{(e) }\bbox[red, 2pt]T: \text{ by definition} \\\textbf{(f) }\bbox[red, 2pt]F: L(0)= \begin{bmatrix}2\\ 0  \end{bmatrix} \ne 0 $$

解答:$$\textbf{(a) }S=40 \Rightarrow f_d(x)=40+7x+x^2 \Rightarrow \cases{f_d(1) =48\\ f_d(2) = 58\\ f_d(3) = 70\\ f_d(4) = 84\\ f_d(5) =100 } \Rightarrow \bbox[red, 2pt]{\cases{\text{Eldest son: 48} \\ \text{Second daughter: 58} \\\text{Third son: 70} \\\text{Fourth daug}hter: 84\\ \text{Youngest son: 100}}}  \\\textbf{(b) }\text{Two points: (3,70) and (4,84) cannot determine the unknown coefficients }(a,b,c) \\\qquad \text{ for }f(x)=a+bx+cx^2 \Rightarrow \bbox[red, 2pt]{\text{nondeterministic}} \\\textbf{(c) }\cases{(1,48) \Rightarrow a+b+c=48\\ (2,58)\Rightarrow a+2b+4c=58\\ (5,100) \Rightarrow a+5b+25c=100} \Rightarrow \cases{a=40\\ b=7 \\c=1} \Rightarrow \text{ the wealthy man has }\bbox[red, 2pt]{40} \text{ million}$$

解答:$$\textbf{(a) }R_2-R_1\to R_2 \Rightarrow \bbox[red, 2pt]{E_1= \begin{bmatrix}1& 0& 0\\ -1& 1&0 \\0&0& 1\end{bmatrix} }\Rightarrow E_1A= \begin{bmatrix}1& 0& 0\\0& 1& 1 \\ 1& 1& 2 \end{bmatrix} \\ R_3-R_1\to R_3 \Rightarrow \bbox[red, 2pt]{E_2 = \begin{bmatrix}1& 0& 0\\ 0& 1& 0\\ -1& 0& 1 \end{bmatrix} }\Rightarrow E_2E_1 A= \begin{bmatrix}1& 0& 0\\ 0& 1& 1\\ 0& 1& 2 \end{bmatrix} \\R_3-R_2\to R_3 \Rightarrow \bbox[red, 2pt]{E_3 = \begin{bmatrix}1& 0& 0\\ 0& 1& 0\\ 0& -1&1 \end{bmatrix} }\Rightarrow E_3E_2E_1 A= \begin{bmatrix}1&0& 0\\ 0& 1& 1\\ 0& 0& 1 \end{bmatrix} \\ R_2-R_3\to R_2 \Rightarrow \bbox[red, 2pt]{E_4 = \begin{bmatrix}1& 0& 0\\ 0& 1& -1\\ 0& 0& 1 \end{bmatrix} }\Rightarrow E_4E_3E_2 E_1 A= \begin{bmatrix}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix} =I \\\textbf{(b) } \text{From (a), we have } E_3E_2E_1 A= \begin{bmatrix}1&0& 0\\ 0& 1& 1\\ 0& 0& 1 \end{bmatrix} =U \Rightarrow A=E_1^{-1} E_2^{-1}E_3^{-1} U \\\qquad \Rightarrow L= E_1^{-1} E_2^{-1}E_3^{-1} = \begin{bmatrix} 1 & 0 & 0 \\1 & 1 & 0 \\0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0 \\1 & 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0 \\0 & 1 & 1\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\1 & 1 & 0 \\1 & 1 & 1\end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{L= \begin{bmatrix} 1 & 0 & 0 \\1 & 1 & 0 \\1 & 1 & 1\end{bmatrix}, U=\begin{bmatrix}1&0& 0\\ 0& 1& 1\\ 0& 0& 1 \end{bmatrix} }$$

解答:$$\textbf{(a) }L(x) = \begin{bmatrix}x_1+2x_2+ 3x_3\\ x_1+x_3\\ x_1+x_2+2x_3 \end{bmatrix} = \begin{bmatrix}1& 2& 3\\ 1&0& 1\\ 1& 1& 2 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{A= \begin{bmatrix}1& 2& 3\\ 1&0& 1\\ 1& 1& 2 \end{bmatrix} } \\\textbf{(b) } \det(A-\lambda I) =0 \Rightarrow -\lambda(\lambda-4)(\lambda+1)=0 \\\Rightarrow \text{eigenvalues: }\cases{\lambda_1=0\\ \lambda_2=4\\ \lambda_3=-1} \Rightarrow \text{eigenvectors: }\cases{v_1= (1,1,-1)^T\\ v_2=(7,3,5)^T\\ v_3= (1,-1,0)^T} \Rightarrow P=[v_1\; v_2\; v_3] \\ \Rightarrow \bbox[red, 2pt]{P= \begin{bmatrix}1&7& 1\\ 1&3& -1\\ -1&5& 0 \end{bmatrix} },D= \begin{bmatrix}\lambda_1&0&0\\ 0&\lambda_2& 0\\ 0&0& \lambda_3 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{D= \begin{bmatrix}0&0&0\\ 0& 4&0\\ 0&0& -1 \end{bmatrix}} \\ \textbf{(c) }A=PDP^{-1} = \begin{bmatrix}1&7& 1\\ 1&3& -1\\ -1&5& 0 \end{bmatrix} \begin{bmatrix}0&0&0\\ 0& 4&0\\ 0&0& -1 \end{bmatrix} \begin{bmatrix}1&7& 1\\ 1&3& -1\\ -1&5& 0 \end{bmatrix}^{-1} \\ \qquad \Rightarrow e^A =\begin{bmatrix}1&7& 1\\ 1&3& -1\\ -1&5& 0 \end{bmatrix} \begin{bmatrix}e^0& 0&0\\ 0& e^4&0\\ 0&0& e^{-1} \end{bmatrix} \begin{bmatrix}1&7& 1\\ 1&3& -1\\ -1&5& 0 \end{bmatrix}^{-1}\\\qquad \Rightarrow \bbox[red, 2pt]{e^A ={1\over 20}  \begin{bmatrix}5+7e^4+8e^{-1} & 5+7e^4-12e^{-1} & -10+14e^4-4e^{-1} \\5+3e^4-8e^{-1} & 5+3e^{4}+12 e^{-1} & -10+6e^4+4e^{-1} \\ -5+5e^4& -5+5e^4& 10+10e^4 \end{bmatrix}}$$

解答:$$A = \begin{bmatrix}1& 1& 2\\ 0& 1& 1\\ 1& 0& 1 \end{bmatrix} \xrightarrow{R_3-R_1\to R_3}  \begin{bmatrix}1& 1& 2\\ 0& 1& 1\\ 0& -1& -1 \end{bmatrix} \xrightarrow{R_2+R_3 \to R_3} \begin{bmatrix}1& 1& 2\\ 0& 1& 1\\ 0& 0& 0 \end{bmatrix} \xrightarrow{R_1-R_2\to R_1} \begin{bmatrix}1& 0& 1\\ 0& 1& 1\\ 0& 0& 0 \end{bmatrix} \\ \Rightarrow \text{RREF}(A) = \begin{bmatrix}1& 0& 1\\ 0& 1& 1\\ 0& 0& 0 \end{bmatrix} \Rightarrow \text{rank}(A)=2\\ \mathbf{R(A^T) }: \text{basis: }\{(1,0,1), (0,1,1)\} \Rightarrow R(A^T) =\{a(1,0,1) +b(0,1,1) \mid a,b\in \mathbb R\}  \\ \qquad \Rightarrow \bbox[red, 2pt]{R(A^T)= \{(a,b,a+b) \mid a,b \in \mathbb R\}} \\ \mathbf {N(A)}: \begin{bmatrix}1& 0& 1\\ 0& 1& 1\\ 0& 0& 0 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1+x_3 =0\\ x_2+x_3 =0} \Rightarrow \bbox[red, 2pt]{N(A) =\{(t,t,-t) \mid t\in \mathbb R\}} \\ \mathbf {R(A)}: \text{basis:} \{(1,0,1), (1,1,0)\} \Rightarrow R(A)=\{a(1,0,1)+b(1,1,0) \mid a,b\in \mathbb R\}  \\ \qquad \Rightarrow \bbox[red, 2pt]{R(A)=\{(a+b, b,a) \mid a,b\in \mathbb R\}} \\ \mathbf{N(A^T)}: \begin{bmatrix}1& 0& 1\\ 1& 1& 0\\ 2& 1& 1 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1+ x_3=0 \\ x_1+x_2=0\\ 2x_1+x_2+x_3=0} \Rightarrow x_2=x_3= -x_1 \\ \qquad \Rightarrow \bbox[red, 2pt]{N(A^T) =\{(t,-t,-t) \mid t\in \mathbb R\}}$$

解答:$$\textbf{(a) }\text{Let }\cases{u_0={1\over \sqrt 2} \\ u_n=\cos(nx), n=1,2,3} \Rightarrow \langle u_0,u_n\rangle = {1\over \pi}\int_{-\pi}^\pi {1\over \sqrt 2} \cos(nx)\,dx ={1\over \sqrt 2 \pi} \left. \left[ {1\over n}\sin(nx) \right] \right|_{-\pi}^\pi=0 \\\qquad \Rightarrow \langle u_0, u_n\rangle =0, \text{ for }n=1,2,3 \Rightarrow \langle u_m,u_n \rangle ={1\over \pi} \int_{-\pi}^\pi \cos(mx) \cos(nx)\,dx=0, \text{ for }m\ne n \\ \qquad \Rightarrow \langle u_i,u_j \rangle =0, \text{ for }i\ne j\text{ and }0\le i,j\le 3\\ \qquad \text{Additionally, }\langle u_0,u_0 \rangle ={1\over \pi}\int_{-\pi}^\pi{1\over 2}\,dx ={2\pi\over 2\pi} =1 \\ \qquad \langle u_n, u_n \rangle ={1\over \pi} \int_{-\pi}^\pi\cos^2(nx)\,dx ={1\over 2\pi}  \int_{-\pi}^\pi(1+\cos(2nx))\,dx=1, \text{ for }n=1,2,3 \\ \qquad \Rightarrow \bbox[red, 2pt]{\text{Yes, }S \text{ is an orthonormal set}} \\\textbf{(b) }\sin^4 x= \left( {1-\cos(2x)\over 2} \right)^2 ={1\over 4}(1-2\cos(2x)+ \cos^2(2x)) ={1\over 4}(1-2\cos(2x)+{1\over 2} (1+\cos(4x)) ) \\ \qquad \Rightarrow \sin^4(x)={3\over 8}-{1\over 2}\cos(2x)+{1\over 8}\cos(4x) \not \in S\\\Rightarrow \bbox[red, 2pt]{\sin^4 x \text{ cannot be a linear combination of elements in }S} \\\textbf{(c) } \int_{-\pi}^\pi \sin^4(x)\,dx =  \int_{-\pi}^\pi  \left( {3\over 8}-{1\over 2}\cos(2x)+{1\over 8}\cos(4x) \right)\,dx ={3\over 8}\cdot 2\pi =\bbox[red, 2pt]{3\pi\over 4}$$

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解題僅供參考，碩士班歷年試題及詳解