114年中山大學光電碩士班-工程數學詳解

 國立中山大學114學年度碩士班考試入學招生考試試題

科目名稱:工程數學【光電系碩士班】

解答:$$\det(A)=3 \ne 0 \Rightarrow A^{-1} \text{ exists }\Rightarrow Ax=0 \Rightarrow A^{-1}Ax=A^{-1}0 \Rightarrow x=0 \\ \Rightarrow \bbox[red, 2pt]{\text{ the system has exactly one solution}}$$

解答:$$P \text{ diagonalizes A} \Rightarrow A=PDP^{-1}, \text{ where }D= \begin{bmatrix}\lambda_1 & 0& \dots& 0\\ 0& \lambda_2 & \dots& 0\\\vdots& \vdots & \ddots & \vdots\\0& 0& \dots& \lambda_n \end{bmatrix} \\ \Rightarrow A^k = \underbrace{(PDP^{-1})(PDP^{-1}) \cdots(PDP^{-1})}_{k \text{ times}} =PD(P^{-1}P)D(P^{-1}P) \cdots (P^{-1}P)DP^{-1} \\=P(D\cdot D\cdots D)P^{-1} =PD^kP^{-1} \Rightarrow P \text{ diagonalizes }A^k \; \bbox[red, 2pt]{QED.} \\ A^k= PD^kP^{-1} =P\underbrace{\begin{bmatrix}\lambda_1 & 0& \dots& 0\\ 0& \lambda_2 & \dots& 0\\\vdots& \vdots & \ddots & \vdots\\0& 0& \dots& \lambda_n \end{bmatrix} \cdots \begin{bmatrix}\lambda_1 & 0& \dots& 0\\ 0& \lambda_2 & \dots& 0\\\vdots& \vdots & \ddots & \vdots\\0& 0& \dots& \lambda_n  \end{bmatrix}}_{k\text{ times}}P^{-1} \\=P \begin{bmatrix}\lambda_1^k & 0& \dots& 0\\ 0& \lambda_2^k & \dots& 0\\\vdots& \vdots & \ddots & \vdots\\0& 0& \dots& \lambda_n^k \end{bmatrix} P^{-1} \Rightarrow P^{-1} A^kP= \bbox[red, 2pt]{ \begin{bmatrix}\lambda_1^k & 0& \dots& 0\\ 0& \lambda_2^k & \dots& 0\\\vdots& \vdots & \ddots & \vdots\\0& 0& \dots& \lambda_n^k \end{bmatrix}}$$

解答:$$\textbf{(a) }\det(A) =-(k-1)(k-9)(2k-3) \\ \textbf{Case I }k\ne1,k\ne 9,k\ne {3\over 2} \Rightarrow \det(A)\ne 0 \Rightarrow \text{rank}(A)=3 \\\textbf{Case II }k=9 \Rightarrow A= \begin{bmatrix}-4& 4& 2\\ 4& -4& -2\\ -2&-2 & -15 \end{bmatrix} \Rightarrow \text{rref}(A)= \begin{bmatrix} 1 & 0 & \frac{7}{2}\\0 & 1 & 4\\0 & 0 & 0 \end{bmatrix} \Rightarrow \text{rank}(A)=2 \\ \textbf{Case III }k={3\over 2} \Rightarrow A = \begin{bmatrix}7/2& 4& 2\\ 4& 7/2 &-2\\ -2& -2& 0 \end{bmatrix} \Rightarrow \text{rref}(A) = \begin{bmatrix}1&0&-4\\ 0&1& 4\\ 0&0&0 \end{bmatrix} \Rightarrow \text{rank}(A)=2 \\ \textbf{Case IV }k=1 \Rightarrow A= \begin{bmatrix}4& 4& 2\\ 4& 4& -2\\ -2&-2& 1 \end{bmatrix} \Rightarrow \text{rref}(A) = \begin{bmatrix}1& 1& 0\\ 0&0& 1\\ 0& 0& 0 \end{bmatrix} \Rightarrow \text{rank}(A)=2 \\ \Rightarrow \bbox[red, 2pt]{\text{rank}(A) \begin{cases} 3,& k\ne1,k\ne 9,k\ne {3\over 2}\\2, & k=1,9,3/2\end{cases} } \\ \textbf{(b) }k=2 \Rightarrow [A\mid I]= \left[ \begin{array}{rrr|rrr} 3 & 4 & 2& 1& 0& 0\\4 & 3 & -2& 0& 1& 0\\-2 & -2 & -1 & 0& 0& 1\end{array} \right] \xrightarrow{(-1/2)R_3 \to R_3} \left[ \begin{array}{rrr|rrr} 3 & 4 & 2 & 1 & 0 & 0\\4 & 3 & -2 & 0 & 1 & 0\\1 & 1 & \frac{1}{2} & 0 & 0 & - \frac{1}{2}\end{array} \right] \\ \xrightarrow{R_1-3R_3\to R_1, R_2-4R_3\to R_2} \left[ \begin{array}{rrr|rrr} 0 & 1 & \frac{1}{2} & 1 & 0 & \frac{3}{2}\\0 & -1 & -4 & 0 & 1 & 2\\1 & 1 & \frac{1}{2} & 0 & 0 & - \frac{1}{2}\end{array} \right] \xrightarrow{R_2+R_1\to R_2, R_3-R_1\to R_3} \\\left[ \begin{array}{rrr|rrr} 0 & 1 & \frac{1}{2} & 1 & 0 & \frac{3}{2}\\0 & 0 & - \frac{7}{2} & 1 & 1 & \frac{7}{2}\\1 & 0 & 0 & -1 & 0 & -2\end{array} \right] \xrightarrow{-(2/7)R_2 \to R_2} \left[ \begin{array}{rrr|rrr} 0 & 1 & \frac{1}{2} & 1 & 0 & \frac{3}{2}\\0 & 0 & 1 & - \frac{2}{7} & - \frac{2}{7} & -1\\1 & 0 & 0 & -1 & 0 & -2\end{array} \right] \\ \xrightarrow{R_1-(1/2)R_2 \to R_1} \left[ \begin{array}{rrr|rrr} 0 & 1 & 0 & \frac{8}{7} & \frac{1}{7} & 2\\0 & 0 & 1 & - \frac{2}{7} & - \frac{2}{7} & -1\\1 & 0 & 0 & -1 & 0 & -2\end{array} \right] \xrightarrow{R_1 \leftrightarrow R_3}\;\xrightarrow{R_2 \leftrightarrow R_3} \\\left[ \begin{array}{rrr|rrr} 1 & 0 & 0 & -1 & 0 & -2\\0 & 1 & 0 & \frac{8}{7} & \frac{1}{7} & 2\\0 & 0 & 1 & - \frac{2}{7} & - \frac{2}{7} & -1\end{array} \right] \Rightarrow \bbox[red, 2pt]{A^{-1} = \begin{bmatrix} -1 & 0 & -2\\ \frac{8}{7} & \frac{1}{7} & 2\\ - \frac{2}{7} & - \frac{2}{7} & -1 \end{bmatrix}}$$

解答:$$\det(A-\lambda I) =(1-\lambda)^2=0 \Rightarrow \bbox[red, 2pt]{\text{ eigenvalue: }\lambda =1} \\ (A-\lambda I) v=0 \Rightarrow  \begin{bmatrix}0& 1\\ 0&0 \end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix}=0 \Rightarrow x_2=0 \Rightarrow v= k \begin{bmatrix}1\\ 0 \end{bmatrix}, \text{ choosing } \bbox[red, 2pt]{\text{eigenvector: }\begin{bmatrix}1\\ 0 \end{bmatrix}} \\ \cases{\text{Algebraic multiplicity of }\lambda=1 \text{ is }2 \\ \text{Geometric multiplicity is 1}} \Rightarrow \text{ the geometric multiplicity }\lt  \text{ the algebraic multiplicity} \\ \Rightarrow \bbox[red, 2pt]{A \text{ cannot be diagonalized}}$$

解答:$$\textbf{Student 1: }\cases{b \text{ and }B\text{ are wrong} \\ a\text{ and }A\text{ are correct}} \Rightarrow y_A(x)=e^{-2x}( \cos 3x+ 2\sin 3x) \Rightarrow y_A(0)=1=A \\\qquad \text{the characteristic eq.: }r^2+ar+b=0 \Rightarrow r=-2\pm 3i \Rightarrow a=4 \\\textbf{Student 2: }\cases{b \text{ and }B\text{ are correct} \\ a\text{ and }A\text{ are wrong}} \Rightarrow y_B(x)=-3e^x+2e^{3x} \Rightarrow y'_B(x)= -3e^x+6e^{3x} \\ \qquad \Rightarrow y'_B(0)=3=B \Rightarrow r^2+ar+b=(r-1)(r-3) \Rightarrow b=3\\ \text{Correct constants:}\cases{A=1\\ B=3\\a=4\\ b=3} \Rightarrow y''+4y'+3y=0 \Rightarrow r^2+4r+3=0 \Rightarrow r=-3,-1 \\ \Rightarrow y=c_1e^{-x}+c_2e^{-3x} \Rightarrow y'=-c_1e^{-x}-3c_2e^{-3x} \Rightarrow \cases{y(0)=c_1+c_2= A=1\\ y'(0)=-c_1-3c_2=B=3} \\ \Rightarrow \cases{c_1=3\\ c_2=-2} \Rightarrow \bbox[red, 2pt]{y=3e^{-x}-2e^{-3x}}$$

解答:$$6x^2y+12xy+y^2 +(6x^2+2y)y'=0 \Rightarrow (6x^2y+y^2)+ (6x^2y+y^2)'=0 \\ \Rightarrow e^x(6x^2y+y^2)+ e^x (6x^2y+y^2)'=0 \Rightarrow (e^x(6x^2y+ y^2))'=0 \Rightarrow \bbox[red, 2pt]{e^x(6x^2y+ y^2)=C  }$$

解答:$$y_h= e^{-2t}+e^{-t} \Rightarrow \cases{r_1=-2\\ r_2=-1} \Rightarrow (r+2)(r+1) =r^2+3r+2=0 \Rightarrow y''+3y'+2y=0 \\ \Rightarrow \cases{A=3\\ B=2}; \quad  y_p={1\over 2}e^{-3t} \Rightarrow y'_p=-{3\over 2}e^{-3t} \Rightarrow y''_p = {9\over 2}e^{-3t} \Rightarrow y_p''+3y_p'+2y_p =e^{-3t}=C \\ \Rightarrow A*B*C = \bbox[red, 2pt]{6e^{-3t}}\\\bbox[cyan,2pt]{題目有疑義}，y''+Ay'+By=C應更正為y''+Ay'+By=Ce^{-3t},這樣C才是\text{coefficient}$$

解答:$$\vec F=(xy-1)\vec i+yz\vec j+xz\vec k \Rightarrow \iiint_V \nabla\cdot \vec F\; dV =\int_0^1 \int_0^1 \int_0^1 (x+y+z) \,dxdydz \\ =\int_0^1 \int_0^1 \left( {1\over 2}+y+z \right) \,dydz = \int_0^1 \left( {1\over 2}+{1\over 2}+z  \right)\,dz =\int_0^1 (1+z)\,dz ={3\over 2} \\ \begin{array}{l}\text{Face} & \text{Plane}& \text{normal }\vec n& \vec F\cdot \vec n & \text{Flux}\\\hline \text{Front}& x=1& (1,0,0)& y-1 & \int_0^1 \int_0^1 (y-1)\,dydz=-{1\over 2}\\ \text{Back}& x=0& (-1,0,0)& 1 & \int_0^1 \int_0^1 1\,dydz=1\\ \text{Right}& y=1&(0,1,0) &z & \int_0^1 \int_0^1 z\,dxdz ={1\over 2}\\ \text{Left}& y=0 &(0,-1,0) & 0& 0\\ \text{Top} & z=1& (0,0,1)& x & \int_0^1 \int_0^1 x\,dxdy ={1\over 2}\\ \text{Bottom}& z=0& (0,0,-1)& 0  &0\\\hline\end{array} \\ \Rightarrow \text{total flux:} -{1\over 2}+1+{1\over 2}+{1\over 2}={3\over 2} \\ \Rightarrow \iint_S \vec F\cdot \vec n\,dS=  \iiint_V (\nabla\cdot \vec F)\,dV ={3\over 2}\Rightarrow \text{ The Divergence Theorem is verified.}$$

解答:$$z(x,y)=3000-2x^2-9y^2 \Rightarrow \nabla z=(z_x,z_y) =(-4x,-18y) \Rightarrow \nabla z(4,1)= \bbox[red, 2pt]{(-16,-18)}$$

解答:$$ \cases{x(t)=2\cos t\\ y(t)=1\\ z(t)=2\sin t} \Rightarrow \cases{x'(t)=-2\sin t\\ y'(t)=0\\ z'(t)=2\cos t}, 0\le t\le \pi \\\Rightarrow \int_C \vec F\cdot d\vec r =\int_0^\pi (-6\cos t,0,2)\cdot (-2\sin t,0,2\cos t)\,dt =\int_0^\pi (12\sin t \cos t+4\cos t)\,dt \\=\int_0^\pi (6\sin(2t)+4\cos t)\,dt =\bbox[red, 2pt]0$$

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解題僅供參考，碩士班歷年試題及詳解