114年中山大學光電碩士班-工程數學詳解

 國立中山大學114學年碩士班考試入學招生考試

科目名稱：工程數學【光電系碩士班】

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

解答:$$\text{diagonal matrix }D=[d_{ij}] \Rightarrow D^k =[d_{ij}^k],  \text{where }\cases{d_{ii}= \lambda_i\\ d_{ij}=,0, i\ne j} \Rightarrow A=PD P^{-1} \\ \Rightarrow A^k= \overbrace{(PD P^{-1}) \cdots(PD P^{-1})}^{k \text{ times}} =PD^k P^{-1} \Rightarrow P^{-1}A^k=D^kP^{-1} \Rightarrow P^{-1}A^k P=D^k \\ \Rightarrow \bbox[red, 2pt]{P \text{ diagonalizes }A^k \text{ and }P^{-1}A^k P=D^k}$$

解答:$$\textbf{(a) }\det(A) = -2k^3+23k^2-48k+27 =-(k-1)(k-9)(2k-3) =0 \Rightarrow k=1,9,{3\over 2} \\\quad \textbf{Case I }k=1 \Rightarrow A= \begin{bmatrix} 4 & 4 & 2\\4 & 4 & -2\\-2 & -2 & 1\end{bmatrix} \Rightarrow rref(A) = \begin{bmatrix} 1 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0 \end{bmatrix} \Rightarrow rank(A)=2 \\\textbf{Case II }k=9 \Rightarrow A= \begin{bmatrix} -4 & 4 & 2\\4 & -4 & -2\\-2 & -2 & -15\end{bmatrix} \Rightarrow rref(A) = \begin{bmatrix} 1 & 0 & \frac{7}{2}\\0 & 1 & 4\\0 & 0 & 0 \end{bmatrix} \Rightarrow rank(A)=2 \\\textbf{Case III }k=3/2 \Rightarrow A= \begin{bmatrix} \frac{7}{2} & 4 & 2\\4 & \frac{7}{2} & -2\\-2 & -2 & 0\end{bmatrix} \Rightarrow rref(A) = \begin{bmatrix} 1 & 0 & -4\\0 & 1 & 4\\0 & 0 & 0 \end{bmatrix} \Rightarrow rank(A)=2 \\\quad \Rightarrow \bbox[red, 2pt]{\text{rank}(A) =f(k) =\begin{cases}2& k=1,9,3/2\\ 3& \text{otherwise} \end{cases}} \\\textbf{(b) }k=2 \Rightarrow A= \begin{bmatrix} 3 & 4 & 2\\4 & 3 & -2\\-2 & -2 & -1\end{bmatrix} \Rightarrow \bbox[red ,2pt]{A^{-1} = \begin{bmatrix} -1 & 0 & -2\\\frac{8}{7} & \frac{1}{7} & 2\\- \frac{2}{7} & - \frac{2}{7} & -1 \end{bmatrix}}$$

解答:$$A= \begin{bmatrix} 1&1 \\0& 1\end{bmatrix} \Rightarrow \det(A- \lambda I) =(\lambda-1)^2=0 \\ \Rightarrow \bbox[red, 2pt]{\text{ eigenvalue of }A=1 }\\ (A-I)v=0 \Rightarrow \begin{bmatrix} 0&1\\ 0& 0\end{bmatrix} \begin{bmatrix} x_1\\ x_2\end{bmatrix} =0 \Rightarrow x_2=0 \Rightarrow \bbox[red, 2pt]{\text{eigenvector of }A= \begin{pmatrix} 1\\ 0\end{pmatrix}} \\ \text{Matrix }A\text{ has only one eigenvector, it cannot be diagonalized}$$

解答:$$y_A(x) =e^{-2x}( \cos 3x+ 2\sin 3x) \Rightarrow \cases{\lambda=-2\pm 3i \\ y_A(0)= 1=A} \Rightarrow \lambda^2+4\lambda+13 \Rightarrow a=4 \\ y_B=-3e^x+2e^{3x} \Rightarrow y'_B=-3x^x+6e^{3x} \Rightarrow \cases{\lambda=1,3\\ y_B'(0)=3=B} \Rightarrow \lambda^2-4\lambda+3=0 \Rightarrow b=3 \\ \Rightarrow y''(x)+ 4y'(x)+ 3y(x)=0 \text{ with initial conditions }\cases{y(0)=1\\ y'(0)=3 } \\ \Rightarrow \lambda^2+4\lambda+3=0 \Rightarrow \lambda=-3,-1 \Rightarrow y(x)=c_1e^{-x}+ c_2e^{-3x} \Rightarrow y'(x)=-c_1e^{-x}-3c_2e^{-3x} \\ \Rightarrow \cases{y(0) =c_1+ c_2=1\\ y'(0)=-c_1-3c_2=3} \Rightarrow \cases{c_1=3\\ c_2=-2} \Rightarrow y(x)=3e^{-x}-2e^{-3x} \\ \Rightarrow \bbox[red, 2pt]{\cases{a=4,b=3\\ A=1,B=3 \\ y(x)=3e^{-x}-2e^{-3x}}}$$

解答:$$6x^2y+ 12xy+y^2+(6x^2+2y)y'=0 \Rightarrow (6x^2y+ 12xy+y^2)dx+ (6x^2+2y)dy =0 \\\cases{P(x,y) =6x^2y+ 12xy+y^2\\ Q(x,y) = 6x^2+2y} \Rightarrow \cases{P_y= 6x^2+12x+2y\\ Q_x= 12x } \Rightarrow P_x \ne Q_y \Rightarrow \text{Not Exact} \\ \Rightarrow {P_y-Q_x\over Q} = 1 \Rightarrow u'=u \Rightarrow \text{integrating factor } u=e^x \Rightarrow \cases{uP =(6x^2y+ 12xy+y^2)e^x\\ uQ  = (6x^2+2y)e^x}\\ \Rightarrow (uP)_y=( 6x^2+12x+ 2y )e^x =(uQ)_x   \Rightarrow \text{ Exact} \\ \Rightarrow \Phi(x,y)= \int (6x^2y+ 12xy+y^2)e^x\,dx = \int (6x^2+2y)e^x\,dy \\ \Rightarrow \Phi(x,y) = (6x^2y-12xy+ 12y)e^x+(-12xy+12y)e^x +y^2 e^x +\phi(y)=(6x^2y +y^2) e^x + \rho(x) \\ \Rightarrow \Phi(x,y)=\bbox[red, 2pt]{(6x^2y-24xy+24y+y^2)e^x+c_1 =0}$$

解答:$$y_h=e^{-2t}+e^{-t} \Rightarrow \lambda=-2,-1 \Rightarrow (\lambda+2)(\lambda+1) =\lambda^2+ 3\lambda+2 \Rightarrow \cases{A=3\\ B=2} \\ y_p={1\over 2}e^{-3t} \Rightarrow y_p'=-{3\over 2}e^{-3t} \Rightarrow y_p''={9\over 2}e^{-3t} \Rightarrow y_p''+Ay_p'+By_p =e^{-3t}({9\over 2}-{9\over 2}+1) \\=e^{-3t}\cdot 1 \Rightarrow C=1 \Rightarrow A\times B\times C=\bbox[red, 2pt]6 \\ 註:題目應該是y''+Ay'+By=Ce^{-3t}$$

解答:$$\vec F=(xy-1)\vec i+yz\vec j+xz\vec k \Rightarrow   \text{div} \vec  F=y+z+x \Rightarrow  \iiint_D (x+y+z) \,dV \\=\int_0^1 \int_0^1 \int_0^1 (x+y+z)\,dx\,dy\,dz ={3\over 2} \\ S_1: x=1 \Rightarrow \iint_{S_{1}} (xy-1,yz,xz) \cdot (1,0,0)dA =-{1\over 2} \\ S_2:x=0 \Rightarrow \iint_{S_{2}} (xy-1,yz,xz) \cdot (-1,0,0)dA =1 \\ S_3:y=1\Rightarrow \iint_{S_{3}} (xy-1,yz,xz) \cdot (0,1,0)dA ={1\over 2} \\S_4: y=0\Rightarrow \iint_{S_{4}} (xy-1,yz,xz) \cdot (0,-1,0)dA =0\\ S_5: z=1 \Rightarrow \iint_{S_{5}} (xy-1,yz,xz) \cdot (0,0,1)dA ={1\over 2} \\ S_6: z=0\Rightarrow \iint_{S_{6}} (xy-1,yz,xz) \cdot (0,0,-1)dA =0 \\ \Rightarrow \iint_S \vec F \cdot \vec n \,dA=-{1\over 2}+1+{1\over 2}+0+{1\over 2} ={3\over 2} \Rightarrow \iiint_D \text{div }\vec F\,dV= \iint_S \vec F\cdot \vec n\, dA\quad \bbox[red, 2pt]{QED}$$

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

解答:$$\cases{x(t) =2\cos t\\ y(t)=1\\ z(t) =2\sin t} \Rightarrow \cases{x'(t)=-2\sin t\\ y'(t)=0\\ z'(t)=2\cos t} \\ \Rightarrow \int_C \vec F\cdot d\vec r =\int_0^\pi (-6\cos t,0,2) \cdot (-2\sin t, 0, 2\cos t)\,dt =\int_0^\pi (6\sin 2t +4\cos t)\,dt \\= \left. \left[ -3\cos 2t +4\sin t\right] \right|_0^\pi = \bbox[red, 2pt]0$$

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解題僅供參考，碩士班歷年試題及詳解