114年台北科大自動化碩士班-工程數學詳解

國立臺北科技大學114學年度碩士班招生考試

系所組別:1501、1502 自動化科技研究所
第一節 工程數學 試題

解答:$$x{dy\over dx}=y+x^3+3x^2-2x \Rightarrow {1\over x}{dy\over dx}-{y \over x^2}= x+3-{2\over x} \Rightarrow \left( {y\over x}\right)'=x+3-{2\over x} \\ \Rightarrow {y\over x}= \int \left( x+3-{2\over x} \right)\,dx ={1\over 2}x^2+3x-2\ln x +c_1 \Rightarrow \bbox[red, 2pt]{y={1\over 2}x^3 +3x^2-2x\ln x+c_1x}$$

解答:$$y'={2y\over x}+x^2e^x \Rightarrow {1\over x^2}y'-{2\over x^3}y =e^x \Rightarrow \left( {y\over x^2} \right)'= e^x \Rightarrow {y\over x^2} =\int e^x\,dx = e^x+c_1 \\ \Rightarrow  y= x^2e^x +c_1x^2  \Rightarrow y(2)=4e^2+c_1 4 =0 \Rightarrow c_1=-e^2\Rightarrow  \bbox[red, 2pt]{y= x^2e^x -e^2x^2}$$

解答:$$\cases{L\{ y_1'\}+ 2L\{y_1\}-3L\{y_2\}=0\\ L\{y_2'\}-4L\{y_1\}+L\{y_2\}=0} \Rightarrow \cases{sY_1(s)-4+2Y_1(s)- 3Y_2(s) =0\\ sY_2(s)-3-4Y_1(s) +Y_2(s) =0} \\ \Rightarrow \cases{(s+ 2)Y_1(s) -3Y_2(s)= 4\\ (s+1)Y_2(s)-4Y_1(s)=3  }  \Rightarrow Y_2(s)={(s+2)Y_1(s)-4 \over 3}\\ \Rightarrow {(s+1)(s+2)Y_1(s)-4(s+1) \over 3}-4Y_1(s)=3 \Rightarrow Y_1(s)={4s+13\over s^2+3s-10}\\ \Rightarrow Y_2(s)={s+2\over 3}\cdot {4s+13\over s^2+3s-10}-{4\over 3}={3s+22\over s^2+3s-10}\\ \Rightarrow \cases{y_1(t)= L^{-1}\{Y_1(s) =L^{-1}\left\{ {3\over s-2}+{1\over s+5} \right\} \\ y_2(t) =L^{-1} \{Y_2(S)\} =L^{-1}\left\{ {4\over s-2}-{1\over s+5}\right\}} \Rightarrow \bbox[red, 2pt]{\cases{y_1(t)= 3e^{2t}+e^{-5t} \\y_2(t) =4e^{2t}-e^{-5t}}}$$

解答:$$[A\mid I] =\left[ \begin{array}{rr|rr} 1& 1& 1& 0\\ 2& -1& 0& 1\end{array} \right] \xrightarrow {R_2-2R_1\to R_2} \left[ \begin{array}{rr|rr} 1 & 1 & 1 & 0\\0 & -3 & -2 & 1\end{array} \right] \xrightarrow {R_2/(-3)\to R_2} \left[ \begin{array}{rr|rr} 1 & 1 & 1 & 0\\0 & 1 & \frac{2}{3} & - \frac{1}{3}\end{array} \right] \\ \xrightarrow{R_1-R_2\to R_1} \left[ \begin{array}{rr|rr} 1 & 0 & \frac{1}{3} & \frac{1}{3}\\0 & 1 & \frac{2}{3} & - \frac{1}{3}\end{array} \right] \Rightarrow A^{-1} = \bbox[red, 2pt]{\begin{bmatrix}  \frac{1}{3} & \frac{1}{3}\\  \frac{2}{3} & - \frac{1}{3} \end{bmatrix}}$$

解答:$$A^{-1} =  \begin{bmatrix}  1 & 2& -1\\ 3 & 4& 2\\ 0& 1& -2 \end{bmatrix} \Rightarrow A=  \begin{bmatrix}  10 & -3 & -8\\-6 & 2 & 5\\-3 & 1 & 2 \end{bmatrix} \Rightarrow AB^{-1 } = \begin{bmatrix}  10 & -3 & -8\\-6 & 2 & 5\\-3 & 1 & 2 \end{bmatrix} \begin{bmatrix}  0 & 1 & 1\\1 & 0 & 1\\-2 & 3 & 2 \end{bmatrix} \\= \bbox[red, 2pt]{\begin{bmatrix} 13 & -14 & -9\\-8 & 9 & 6\\-3 & 3 & 2\end{bmatrix}}$$

解答:$$A=\begin{bmatrix}1 & 4 \\ 2&3 \end{bmatrix}  \Rightarrow \det(A-\lambda I)=(\lambda+1) (\lambda-5) =0 \Rightarrow \lambda=-1,5 \\ \lambda_1=-1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}2 & 4 \\2 & 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2\end{bmatrix}=0 \Rightarrow x_1+2x_2=0\\\qquad  \Rightarrow v= x_2\begin{pmatrix} -2  \\ 1 \end{pmatrix}, \text{choose } v_1 =\begin{pmatrix} -2  \\ 1 \end{pmatrix} \\\lambda_2=5 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}-4 & 4 \\2 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} =0 \Rightarrow x_1=x_2 \\\qquad  \Rightarrow v= x_1 \begin{pmatrix} 1  \\ 1 \end{pmatrix},\text{choose }v_1=\begin{pmatrix} 1  \\ 1 \end{pmatrix} \\ \Rightarrow P=[v_1\; v_2] =\begin{bmatrix} -2& 1 \\1 & 1\end{bmatrix} \Rightarrow P^{-1}AP= \begin{bmatrix} -1& 0 \\0 & 5\end{bmatrix} \\ \textbf{(a) }\text{ eigenvalues: }\bbox[red, 2pt]{-1,5} \text{ and  eigenvectros: } \bbox[red, 2pt] {\begin{pmatrix} -2  \\ 1 \end{pmatrix}, \begin{pmatrix} 1  \\ 1 \end{pmatrix}} \\\textbf{(b) }P= \bbox[red, 2pt] {\begin{bmatrix} -2& 1 \\1 & 1\end{bmatrix}}$$

解答:$$A=\begin{bmatrix}1/\sqrt 2& 1/\sqrt 2& 0\\ 1/\sqrt 2& -1/\sqrt 2 & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}1& 0 &0\\ 0 & 4& 0\\ 0& 0&   9 \end{bmatrix} \begin{bmatrix}1/\sqrt 2& 1/\sqrt 2& 0\\ 1/\sqrt 2& -1/\sqrt 2 & 0\\ 0 & 0 & 1 \end{bmatrix} ^{-1} \\= \begin{bmatrix}\frac{\sqrt{2}}{2} & 2 \sqrt{2} & 0 \\\frac{\sqrt{2}}{2} & -2 \sqrt{2} & 0 \\0 & 0 & 9 \end{bmatrix} \begin{bmatrix} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\\frac{\sqrt{2}}{2} & \frac{-\sqrt{2}}{2} & 0 \\0 & 0 & 1\end{bmatrix}=  \begin{bmatrix}\frac{5}{2} & \frac{-3}{2} & 0 \\\frac{-3}{2} & \frac{5}{2} & 0 \\
0 & 0 & 9\end{bmatrix} =  \begin{bmatrix}a & f & e \\f & b & d \\e & d & c\end{bmatrix} \Rightarrow \cases{a= 5/2\\ b=5/2\\ c=9\\ d=0\\ e=0\\ f=-3/2}\\ 故選\bbox[red, 2pt]{(BCD)}$$

解答:$$(A)\times:\det(A)= \begin{vmatrix} 1+\alpha_1 & \alpha_2 & \cdots & \alpha_n \\ \alpha_1 & 1+\alpha_2 & \cdots &\alpha_n \\\vdots & \vdots & \ddots & \vdots \\\alpha_1 & \alpha_2 & \cdots & 1+\alpha_n \end{vmatrix} \xrightarrow{ \sum_{k=1}^n C_k \to C_1} \begin{vmatrix} 1+ \sum_{k=1}^n \alpha_k & \alpha_2 & \cdots & \alpha_n \\ 1+ \sum_{k=1}^n \alpha_k & 1+\alpha_2 & \cdots &\alpha_n \\\vdots & \vdots & \ddots & \vdots \\ 1+ \sum_{k=1}^n \alpha_k & \alpha_2 & \cdots & 1+\alpha_n \end{vmatrix} \\= (1+ \sum_{k=1}^n \alpha_k) \begin{vmatrix} 1 & \alpha_2 & \cdots & \alpha_n \\ 1 & 1 +\alpha_2 & \cdots &\alpha_n \\\vdots & \vdots & \ddots & \vdots \\ 1 & \alpha_2 & \cdots & 1+\alpha_n \end{vmatrix} \xrightarrow{\sum_{k=2}^n (C_k-\alpha_kC_1 \to C_k) } (1+ \sum_{k= 1}^n \alpha_k) \begin{vmatrix} 1 & 0 & \cdots & 0 \\ 1 & 1 & \cdots &0 \\\vdots & \vdots & \ddots & \vdots \\ 1 & 0 & \cdots & 1 \end{vmatrix} \\=1+ \sum_{k=1}^n \alpha_k \ne 1+ n\sum_{k =1}^n \alpha_k\\(C)\times: Q \text{ is an orthogonal projection matrix} \Rightarrow Q^2=Q \\ \Rightarrow (I-uu^T)(I-uu^T) =I-2uu^T+ u(u^Tu)u^T =I-2uu^T+ ||u||^2uu^T =I-(2-||u||^2)uu^T \\ \Rightarrow I-uu^T=I-(2-||u||^2)uu^T \Rightarrow 2-||u||^2=1 \Rightarrow ||u||^2=1 \Rightarrow u \text{ is a unit vector} \\ Qx=(I-uu^T)x=x-u(u^Tx)=0 \Rightarrow nullity(Q)=1 \Rightarrow rank(Q)=4-1=3\\ Qu=(I-uu^T)u=u-u=0 =0\cdot u \Rightarrow u \text{   is an eigenvector with an eigenvalue of 0.} \\\text{Let }v \bot u \Rightarrow Qv=(I-uu^T)v=v-0= 1\cdot v \Rightarrow u \text{ is an eigenvector with an eigenvalue of 1.} \\ \Rightarrow \text{eigenvalues of }Q \text{ are }\{1,1,1,0\} \Rightarrow \det(Q)=0 \Rightarrow rank(Q)+\det(Q)=3 \ne 4 \\(D)\times: A=\begin{bmatrix}a & z_1 &z_2  \\\bar z_1 & b & z_3 \\\bar z_2 & \bar z_3 &c \end{bmatrix} \text{ is a 3x3 Hermitian matrix, where }a,b ,c \in \mathbb R \\\qquad \Rightarrow \det(A) =abc+ \bar z_1 z_2\bar z_3+ z_1\bar z_2z_3-az_3\bar z_3-bz_2\bar z_2-cz_1\bar z_1 \in \mathbb R\\  \qquad \because\cases{\bar z_1 z_2\bar z_3+ z_1\bar z_2z_3=\bar z_1 z_2\bar z_3+ \overline{z_1 z_2\bar z_3} \in \mathbb R \\ z_1\bar z_1, z_2\bar z_2,z_3\bar z_3 \in \mathbb R }\\ 故選\bbox[red, 2pt]{(B)}$$ 註: (B) 是對的, 但證明很長.....

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解題僅供參考，其他碩士班試題及詳解