113年調查三等-工程數學詳解

113年法務部調查局調查人員考試試題

考 試 別：調查人員
等 別：三等考試
類 科 組：電子科學組
科 目：工程數學

解答: $$\textbf{(一)}\; A=\begin{bmatrix} 2 & 1 & -2 \\3 & -2 & 0 \\3 & 1 & -3\end{bmatrix} \Rightarrow \det(A-\lambda I) = \begin{vmatrix} 2-\lambda & 1 & -2 \\3 & -2-\lambda & 0 \\3 & 1 & -3-\lambda \end{vmatrix}=0 \Rightarrow -(\lambda-1)(\lambda+3)( \lambda+1)=0\\ \qquad \Rightarrow 特徵值\lambda= \bbox[red, 2pt]{1,-3,-1} \\\textbf{(二)}\; \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow  \begin{bmatrix} 1 & 1 & -2 \\3 & -3 & 0 \\3 & 1 & -4\end{bmatrix} \begin{bmatrix} x_1 \\x_2 \\x_3\end{bmatrix} =0 \Rightarrow \cases{x_1=x_3\\ x_2 =x_3} \Rightarrow v= x_3\begin{pmatrix} 1 \\1 \\ 1\end{pmatrix}\\ \qquad \text{ choose }v_1= \begin{pmatrix} 1 \\1 \\ 1\end{pmatrix} \\ \lambda_2=-3 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow  \begin{bmatrix} 5 & 1 & -2 \\3 & 1 & 0 \\ 3 & 1 & 0\end{bmatrix} \begin{bmatrix} x_1 \\x_2 \\x_3\end{bmatrix} =0 \Rightarrow \cases{x_1=x_3\\ x_2+3x_3=0} \Rightarrow v=x_3\begin{pmatrix} 1 \\-3 \\ 1\end{pmatrix}\\ \qquad \text{ choose }v_2= \begin{pmatrix} 1 \\-3 \\ 1\end{pmatrix}\\ \lambda_3=-1 \Rightarrow (A-\lambda_3 I)v=0 \Rightarrow  \begin{bmatrix} 3 & 1 & -2 \\3 & -1 & 0 \\3 & 1 & -2\end{bmatrix} \begin{bmatrix} x_1 \\x_2 \\x_3\end{bmatrix} =0 \Rightarrow \cases{3x_1=x_3\\ x_2=x_3}  \Rightarrow v=x_3\begin{pmatrix} 1/3 \\1 \\ 1\end{pmatrix}\\\qquad \text{ choose }v_3= \begin{pmatrix} 1/3 \\1 \\ 1\end{pmatrix}\\ \Rightarrow \cases{D= \begin{bmatrix} \lambda_1 & 0 & 0\\ 0 &\lambda_2 & 0\\ 0 & 0 & \lambda_3\end{bmatrix} =\begin{bmatrix} 1& 0 & 0\\ 0 & -3& 0\\ 0 & 0 & -1\end{bmatrix}\\[1ex] P= [v_1 v_2 v_3] =\begin{bmatrix}1& 1& {1\over 3}\\ 1 & -3 & 1\\ 1& 1& 1 \end{bmatrix}} \Rightarrow \bbox[red, 2pt]{\cases{D= \begin{bmatrix} 1& 0 & 0\\ 0 & -3& 0\\ 0 & 0 & -1\end{bmatrix}\\ P= \begin{bmatrix}1& 1& {1\over 3}\\ 1 & -3 & 1\\ 1& 1& 1 \end{bmatrix}}} \\\textbf{(三)} \; \det(A-\lambda)=0 \Rightarrow (\lambda-1)(\lambda+3)(\lambda+1) =\lambda^3+ 3 \lambda^2-\lambda -3=0 \Rightarrow A^3+3A^2-A-3I=0 \\ \qquad \Rightarrow A^4+3A^3-A^2-5A+2I= A(A^3+3A^2-A-3I)-2A+2I =-2A+2I \\ \qquad =-2 \begin{bmatrix} 2 & 1 & -2 \\3 & -2 & 0 \\3 & 1 & -3\end{bmatrix}+2\begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix} -2 & -2 & 4 \\-6 & 6 & 0 \\-6 & -2 & 8\end{bmatrix}}$$

解答: $$A= \begin{bmatrix}1 & -4 \\1 & 5\end{bmatrix}= \begin{bmatrix}-2 & -4 \\1 & 2\end{bmatrix} +3\begin{bmatrix}1 & 0 \\0 & 1\end{bmatrix} =B+3I \Rightarrow B^2=\begin{bmatrix}-2 & -4 \\1 & 2\end{bmatrix} ^2=0 \Rightarrow B^n=0, n\ge 2\\ \Rightarrow e^{Bt}=I+Bt+{(Bt)^2\over 2!} +{(Bt)^3\over 3!}+ \cdots = I+Bt= \begin{bmatrix}-2t+1 & -4t \\t & 2t+1\end{bmatrix}  \\ \Rightarrow e^{At}= e^{(Bt+3tI)} =e^{Bt} \cdot e^{3tI} =  \begin{bmatrix}-2t+1 & -4t \\t & 2t+1\end{bmatrix} \begin{bmatrix}e^{3t} & 0 \\0 & e^{3t}\end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix} (-2t+1) e^{3t} & -4te^{3t} \\te^{3t} & (2t+1)e^{3t}\end{bmatrix} }$$

解答: $$\textbf{(一)}\; f(z)={2z^2+ 9z+5 \over (z+2)^2(z-3)}= \sum_{n=0}^\infty a_n(z-1)^n =a_0+ a_1(z-1) + a_2(z-1)^2 +a_3(z-1)^3 + \cdots\\ \Rightarrow f'(z)= -{2(z^3+7z^2+6z+ 17) \over (z-3)^2(z +2)^3}=a_1+ 2a_2(z-1)+ 3a_3(z-1)^2 + \cdots\\ \Rightarrow f''(z)={2(2z^4 +19z^3+ 21z^2+ 145z-49) \over (z-3)^3 (z+2)^4} =2a_2+6a_3(z-1)+ \cdots \\  \Rightarrow \cases{a_0=f(1) = {2+9+5\over 3^2(1-3)} =-{8\over 9} \\ a_1=f'(1) = -{31\over 54}\\2a_2= f''(1) =-{23\over 1728}} \Rightarrow \bbox[red, 2pt]{\cases{a_0= -{8\over 9} \\ a_1=  = -{31\over 54}\\ a_2 =-{23\over 3456}} } \\ \textbf{(二)}離原點最近的不可解析的點z=-2 \Rightarrow \bbox[red, 2pt]{R=2}$$

解答: $$L\{y(t)\} +2L\{y(t)\}L\{ \cos 2t\} =L\{ e^{-t}\} \Rightarrow Y(s)+2 Y(s)\cdot {s\over s^2+4} ={1\over s+1} \\ \Rightarrow Y(s)= {1\over s+1} \cdot {s^2+4\over s^2+2s+4} ={-2s-8\over 3(s^2+2s+4)} +{5\over 3(s+1)} \\=-{2\over 3}\cdot {s+1\over (s+1)^2+3}-2\cdot {1\over (s+1)^2+3} +{5\over 3}\cdot {1\over s+1} \\ \Rightarrow y(t) =L^{-1}\{Y(s)\} =-{2\over 3}L^{-1} \left\{  {s+1\over (s+1)^2+3} \right\}-2 L^{-1} \left\{  {1\over (s+1)^2+3} \right\}+{5\over 3} L^{-1}\left\{  {1\over s+1} \right\} \\\Rightarrow \bbox[red, 2pt]{y(t) =-{2\over 3}e^{-t} \cos (\sqrt 3t)-{2\over \sqrt 3}e^{-t} \sin(\sqrt 3 t)+{5\over 3}e^{-t}}$$

解答: $$\textbf{(一)}\;\iint f_{X,Y}\,dxdy =1 \Rightarrow \int_0^{10} \int_0^\infty be^{-(x+y)}\,dy\,dx =\int_0^{10} \left. \left[ -be^{-(x+y)}\right] \right|_0^\infty \,dx =\int_0^{10} be^{-x} \,dx \\= \left. \left[ -be^{-x} \right] \right|_0^{10} =b(1-e^{-10}) =1 \Rightarrow b=\bbox[red, 2pt]{1\over 1-e^{-10}} \\ \textbf{(二)}\; f_{X}(x) =\int_0^\infty f_{X,Y}(x,y)\,dy =\left. \left[ -be^{-(x+y)}\right] \right|_0^\infty =be^{-x} \Rightarrow \bbox[red, 2pt]{f_X(x) =\begin{cases} {1\over 1-e^{-10}} e^{-x},&0 \lt x\lt 10 \\0,& 其他\end{cases}} \\\textbf{(三)}f_Y(y) =\int_0^{10} f_{X,Y}(x,y) \,dx=\left. \left[ -be^{-(x+y)}\right] \right|_0^{10} =e^{-y } \Rightarrow  \bbox[red, 2pt]{f_Y{(y)} =\begin{cases}   e^{-y},&0 \lt y\lt\infty \\0,& 其他\end{cases}}$$
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