111年北科大自動化碩士班-工程數學詳解

 國立臺北科技大學111學年度碩士班招生考試

系所組別: 自動化科技研究所
科目:工程數學

解答: $$\textbf{(1)} \cases{\begin{bmatrix}1 & 2 \\3 & 4\end{bmatrix}^{-1} =\begin{bmatrix}-2 & 1 \\3/2 & -1/2 \end{bmatrix} \\[1ex] \begin{bmatrix}1 & 0 \\0 & 4\end{bmatrix}^{-1} =\begin{bmatrix}1 & 0 \\ 0 & 1/4 \end{bmatrix} \\[1ex]} \Rightarrow \begin{bmatrix}1 & 2 \\3 & 4\end{bmatrix}+\begin{bmatrix}1 & 0 \\0 & 4\end{bmatrix} =\begin{bmatrix}2 & 2 \\3 & 8\end{bmatrix}\\\qquad  \Rightarrow \cases{\begin{bmatrix}2 & 2 \\3 & 8\end{bmatrix}^{-1} =\begin{bmatrix}4/5 & -1/5 \\-3/10 & 1/5\end{bmatrix} \\[1ex] \begin{bmatrix}-2 & 1 \\3/2 & -1/2 \end{bmatrix}+ \begin{bmatrix}1 & 0 \\ 0 & 1/4 \end{bmatrix}= \begin{bmatrix}-1 & 1 \\ 3/2 & -1/4 \end{bmatrix}}\\ 由上可知\cases{A =\begin{bmatrix}1 & 2 \\3 & 4\end{bmatrix}\\[1ex] B=\begin{bmatrix}1 & 0 \\0 & 4\end{bmatrix}} \Rightarrow A^{-1} +B^{-1} \ne (A+B)^{-1} \Rightarrow \bbox[red, 2pt]{\text{False}}\\ \textbf{(2)}\; A\text{ is orthogonal} \Rightarrow AA^T=I \Rightarrow (\det(A))^2=1 \Rightarrow \det(A)\ne 0 \Rightarrow rank(A)=n \Rightarrow \bbox[red, 2pt]{\text{True}}\\ \textbf{(3)}\; rank(A)=n  \Rightarrow n\le m \text{ and }b \text{ is }n\times 1 \Rightarrow b \in  \text{column space of }A \Rightarrow \text{ there is a solution} \Rightarrow \bbox[red,2pt]{True} \\ \; \textbf{(4)} \;2\begin{pmatrix} 4\\2\\3\end{pmatrix} -3\begin{pmatrix} 2\\3 \\1 \end{pmatrix} =\begin{pmatrix} 2\\-5\\ 3\end{pmatrix} \Rightarrow \text{linearly dependent} \Rightarrow \bbox[red, 2pt]{\textbf{True}}$$

2. (20%) In a certain town 30 percent of the married women get divorced each year and 20 percent of the single women get married each year. There are 8000 married women and 2000 single women and the total population remains constant. Find the number of married women and single women after 5 years. What will be the long-range prospects if these percentages of marriages and divorces continue indefinitely into the future?

解答: $$令\cases{x:已婚婦女數\\ y:單身婦女數} \Rightarrow \left[\matrix{x_{i+1}\\y_{i+1}}\right] =\left[\matrix{70\% & 20\%\\30\% & 80\%}\right] \left[\matrix{x_{i}\\y_{i}}\right] ,其中\cases{x_0=8000\\ y_0=2000} \\ 令A=\left[\matrix{70\% & 20\%\\30\% & 80\%}\right] =\left[\begin{matrix}-1 & \frac{2}{3} \\1 & 1\end{matrix} \right] \left[\begin{matrix} \frac{1}{2} & 0 \\0 & 1\end{matrix} \right] \left[\begin{matrix}\frac{-3}{5} & \frac{2}{5} \\\frac{3}{5} & \frac{3}{5} \end{matrix} \right]\\ \Rightarrow \left[\matrix{x_{5}\\y_{5}}\right] =A^5 \left[ \matrix{x_{0} \\y_{0}} \right] =\left[\begin{matrix}-1 & \frac{2}{3} \\1 & 1\end{matrix} \right] \left[\begin{matrix} (\frac{1}{2})^5 & 0 \\0 & 1^5\end{matrix} \right] \left[ \begin{matrix} \frac{-3}{5} & \frac{2}{5} \\\frac{3}{5} & \frac{3}{5} \end{matrix} \right]\left[\matrix{8000 \\2000} \right] =\left[ \begin{matrix}\frac{67}{160} & \frac{31}{80} \\\frac{93}{160} & \frac{49}{80} \end{matrix} \right] \left[ \matrix{8000 \\2000} \right] =\left[\matrix{4125 \\5875} \right]\\ \Rightarrow \left[\matrix{x_{\infty}\\y_{\infty}}\right] =A^\infty \left[ \matrix{x_{0} \\y_{0}} \right] =\left[\begin{matrix}-1 & \frac{2}{3} \\1 & 1\end{matrix} \right] \left[\begin{matrix} (\frac{1}{2})^\infty & 0 \\0 & 1^\infty\end{matrix} \right] \left[ \begin{matrix} \frac{-3}{5} & \frac{2}{5} \\\frac{3}{5} & \frac{3}{5} \end{matrix} \right]\left[\matrix{8000 \\2000} \right] = \left[ \begin{matrix}\frac{2}{5} & \frac{2}{5} \\\frac{3}{5} & \frac{3}{5} \end{matrix} \right] \left[ \matrix{8000 \\2000} \right]  =\left[\matrix{4000 \\6000} \right]\\ \Rightarrow 五年後\bbox[red, 2pt]{\cases{已婚婦女4125人\\ 單身婦女5875人}},長期展望\bbox[red, 2pt]{\cases{已婚婦女4000人\\ 單身婦女6000人}}$$

解答: $$\textbf{(1)}A=\begin{bmatrix}3 & 0 \\8 & -1\end{bmatrix} \Rightarrow \det(A-\lambda I)=\lambda^2-2\lambda -3 =(\lambda+1)(\lambda-3) =0 \Rightarrow \lambda=-1,3\\ \lambda_1=-1 \Rightarrow (A-\lambda_1 I)v =0 \Rightarrow \begin{bmatrix}4 & 0 \\8 & 0\end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} =0 \Rightarrow x_1=0 \Rightarrow v=\begin{bmatrix}0 \\x_2 \end{bmatrix},取v_1= \begin{bmatrix}0 \\1 \end{bmatrix} \\ \lambda_2=3 \Rightarrow (A-\lambda_2 I)v =0 \Rightarrow \begin{bmatrix}0 & 0 \\8 & -4\end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} =0 \Rightarrow 2x_1=x_2 \Rightarrow v=\begin{bmatrix}x_1 \\2x_1 \end{bmatrix}, 取v_2=\begin{bmatrix}1 \\2 \end{bmatrix} \\ \Rightarrow \text{the characteristic equation: }\bbox[red,2pt]{\lambda^2-2\lambda -3=0}, \text{eigenvalues: }\bbox[red,2pt]{-1,3} \\\text{ ,and associated eigenvectors: }\bbox[red, 2pt]{\begin{bmatrix}0 \\1 \end{bmatrix}, \begin{bmatrix}1 \\2 \end{bmatrix}} \\\textbf{(2)}\; A=\begin{bmatrix}-2 & -1 \\5 & 2\end{bmatrix} \Rightarrow \det(A-\lambda I)=\lambda^2+1 =0 \Rightarrow \lambda=\pm i \\ \lambda_1=i \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}-2-i & -1 \\5 & 2-i\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix}=0 \Rightarrow 5x_1+(2-i)x_2=0 \\\qquad \Rightarrow v=\begin{bmatrix}(i-2)x_2/5 \\ x_2\end{bmatrix} ,取 v_1=\begin{bmatrix}i-2 \\ 5\end{bmatrix} \\\lambda_2=-i \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}-2+i & -1 \\5 & 2+i\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix}=0 \Rightarrow 5x_1+(2+i)x_2=0 \\\qquad \Rightarrow v=\begin{bmatrix}(-i-2)x_2/5 \\ x_2\end{bmatrix} ,取 v_2= \begin{bmatrix}-i-2 \\ 5\end{bmatrix} \\ \Rightarrow \text{the characteristic equation: }\bbox[red,2pt]{\lambda^2+1=0}, \text{eigenvalues: }\bbox[red,2pt]{i,-i} \\\text{ ,and associated eigenvectors: }\bbox[red, 2pt]{\begin{bmatrix}i-2 \\5 \end{bmatrix}, \begin{bmatrix}-i-2 \\5 \end{bmatrix}}$$

解答: $$先求齊次解,y''-4y'+3y=0 \Rightarrow \lambda^2-4\lambda+3=0 \Rightarrow \lambda=1,3 \Rightarrow y_h= c_1e^x +c_2 e^{3x}\\ 再求特解y_p=Axe^{3x} \Rightarrow y_p'=Ae^{3x}+ 3Axe^{3x} \Rightarrow y_p''=6Ae^{3x} +9Axe^{3x} \\ \Rightarrow y_p''-4y_p'+ 3y_p=2Ae^{3x} =4e^{3x} \Rightarrow A=2\\ y=y_h+y_p = c_1e^x+ c_2e^{3x} +2xe^{3x} \Rightarrow y'=c_1e^x+ (3c_2+2)e^{3x} + 6xe^{3x} \\ 初始值\cases{y(0)=c_1+c_2= -1\\ y'(0)=c_1+3c_2+2=3} \Rightarrow \cases{c_1=-2\\c_2=1} \Rightarrow \bbox[red, 2pt]{y= -2e^x+ e^{3x} +2xe^{3x}}$$

解答: $$L\{ y''\}+ L\{y\} =L\{ \delta(x-1)\} \Rightarrow s^2Y(s)-sy(0)-y'(0)+ Y(s)= e^{-s} \\ \Rightarrow (s^2+1)Y(s)=e^{-s}+y'(0) \Rightarrow Y(s)={e^{-s} \over s^2+1} +{y'(0)\over s^2+1}\\ \Rightarrow y(x)=L^{-1}\{ {e^{-s} \over s^2+1}\} +L^{-1}\{ {y'(0)\over s^2+1}\} =u(x-1)\sin(x-1)+ y'(0)\sin(x) \\ \Rightarrow y(2)=\sin 1+y'(0)\sin 2=0 \Rightarrow y'(0)=-{\sin 1\over \sin 2}\\ \Rightarrow \bbox[red, 2pt]{y= u(x-1)\sin(x-1)-{\sin 1\over \sin 2}\sin x}$$

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