114年中山大學電機碩士班-工程數學乙詳解

 國立中山大學114學年碩士班考試入學招生考試

科目名稱：工程數學乙【電機系碩士班乙組選考】

第1-10題為複選題,每題5分,總分50分。每題有5個選項,其中至少有1個是正確答案,答錯1個選項者,得3分;答錯2個選項者,得1分;答錯多於2個選項或未作答者,該題以零分計算。

解答:$$(E)\bigcirc: \mathbf a_1+2\mathbf a_2=3\mathbf a_3 \Rightarrow \mathbf a_1+2\mathbf a_2-3\mathbf a_3=0, \text{the coefficients (1,2,-3) are not all zero.} \\ \quad \text{By definition, the vectors }\mathbf a_1, \mathbf a_2,\mathbf a_3 \text{ are linearly dependent.} \\(C) \bigcirc: \text{ Matrix }A\text{ has 3 columns, and }\mathbf a_1, \mathbf a_2,\mathbf a_3 \text{ are linearly dependent.} \Rightarrow rank(A)\le 2 \\(A)\times:  \mathbf a_1+ \mathbf a_2+\mathbf a_3 =\mathbf b \Rightarrow A \begin{bmatrix}1\\1\\1 \end{bmatrix}=\mathbf b \Rightarrow \text{ at least one solution }[1,1,1]^T\\\quad \text{And }rank(A)\le 2 \Rightarrow \text{ there are infinitely many solutions.} \\(B) \bigcirc: \text{ the linear system is consistent }\Rightarrow rank(A)= rank([A,b]) \\(D)\times: [1,2,3]^T \text{ is a solution} \Rightarrow \mathbf a_1+2 \mathbf a_2+3\mathbf a_3 =\mathbf b =\mathbf a_1+\mathbf a_2+\mathbf a_3 \Rightarrow \mathbf a_2+2\mathbf a_3=0 \\ \quad \text{This contradicts the given condition }\mathbf a_2+2\mathbf a_3\ne 0 \Rightarrow [1,2,3]^T \text{ is NOT a solution} \\ \Rightarrow \text{The true statements: }\bbox[red, 2pt]{(BCE)}$$

解答:$$\textbf{(A) }\bigcirc: A= \begin{bmatrix} 2 & 0 & 0 & 0 & 2\\0 & 1 & 1 & 1 & 0\\0 & 1 & 1 & 1 & 0\\0 & 1 & 1 & 1 & 0\\2 & 0 & 0 & 0 & 2\end{bmatrix} \xrightarrow{R_5-R_1\to R_5, R_3-R_2\to R_3, R_4-R_2\to R_4} \begin{bmatrix}2 & 0 & 0 & 0 & 2\\0 & 1 & 1 & 1 & 0\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\end{bmatrix} \\\qquad \xrightarrow{R_1/2\to R_1} \begin{bmatrix}1 & 0 & 0 & 0 & 1\\0 & 1 & 1 & 1 & 0\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 \end{bmatrix} \Rightarrow rref(A)= \begin{bmatrix}1 & 0 & 0 & 0 & 1\\0 & 1 & 1 & 1 & 0\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 \end{bmatrix} \Rightarrow rank(A)=2 \\\textbf{(B)}\bigcirc: trace(A)=2+1+1+1+2= 7 \\\textbf{(C)} \times: rank(A)=2 \ne 5 \Rightarrow \det(A)=0 \\\textbf{(D)} \bigcirc: \det(A-\lambda I) =-\lambda^3(\lambda-3)(\lambda-4) =0 \Rightarrow 3 \text{ is an eigenvalue of }A \\\textbf{(E)}\bigcirc: \text{All eigenvalues (0,3,4) are non-negative} \\ \Rightarrow \text{The true statements are }\bbox[red, 2pt]{(ABDE)}$$

解答:$$Q \text{ is an orthogonal matrix} \Rightarrow Q^TQ=I \Rightarrow A^TA=(QR)^T(QR) =R^T(Q^TQ)R =R^TIR \\ \Rightarrow A^TA=R^TR = \begin{bmatrix}1& 0& 0\\ 2&4& 0\\ 3& 5& 6 \end{bmatrix} \begin{bmatrix}1& 2& 3\\ 0&4& 5\\ 0& 0& 6 \end{bmatrix} = \begin{bmatrix}1& 2& 3\\ 2& 20& 26\\ 3& 26& 70 \end{bmatrix} = (d_{i ,j})_{3\times 3} \\ (A) \bigcirc: \mathbf a_3^T \mathbf a_1 = d_{3,1} =3 \\(B)\times :\mathbf a_3^T \mathbf a_2 = d_{3,2}=26 \ne 24\\ (C) \bigcirc: \mathbf a_3^T \mathbf a_3 =  d_{3,3}=70\\ (D)\times \mathbf a_2^T \mathbf a_1 = d_{2,1}=2 \ne 3\\ (E) \times: \mathbf a_2^T \mathbf a_2 = d_{2,2}=20\ne 30 \\ \Rightarrow \text{The true statements: }\bbox[red, 2pt]{(AC)}$$

解答:$$C=[\mathbf a_1\; \mathbf a_2] = \begin{bmatrix}1& 2\\ 0& 1\\ 1& 0\\ 2& 2 \end{bmatrix} \Rightarrow P=C(C^TC)^{-1}C^T = \left[\begin{matrix}\frac{1}{2} & \frac{1}{3} & - \frac{1}{6} & \frac{1}{3}\\\frac{1}{3} & \frac{1}{3} & - \frac{1}{3} & 0\\- \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} & \frac{1}{3}\\\frac{1}{3} &0 & \frac{1}{3} & \frac{2}{3}\end{matrix}\right] \\ \Rightarrow P \mathbf a_3 = \left[\begin{matrix}\frac{1}{2} & \frac{1}{3} & - \frac{1}{6} & \frac{1}{3}\\\frac{1}{3} & \frac{1}{3} & - \frac{1}{3} & 0\\- \frac{1}{6} & - \frac{1}{3} & \frac{1}{2} & \frac{1}{3}\\\frac{1}{3} &0 & \frac{1}{3} & \frac{2}{3}\end{matrix} \right] \begin{bmatrix} 3\\4\\5\\6 \end{bmatrix} = \begin{bmatrix}4\\{2\over 3} \\{8\over 3} \\{20\over 3} \end{bmatrix} =\mathbf q_3 \Rightarrow \cases{q_{31} =4\\ q_{32} =2/3 \ne 4/3\\ q_{33} =8/3 \ne 10/3\\ q_{34} =20/3\ne 22/3} \\ \Rightarrow \mathbf a_1^T \mathbf q_3 = \begin{bmatrix} 1& 0& 1& 2 \end{bmatrix} \begin{bmatrix} 4 \\{2\over 3} \\{8\over 3} \\{20\over 3} \end{bmatrix} =20 \Rightarrow \text{ True statements: }\bbox[red, 2pt]{(AE)}$$

解答:$$(A)\times: A= \begin{bmatrix}0&-1\\ 1&0 \end{bmatrix} \Rightarrow AA^T= I, \text{ but trace}(A)=0\\ (B)\times: A= \begin{bmatrix} 1&0\\0&-1\end{bmatrix} \Rightarrow AA^T=I, \text{ but det}(A)=-1\ne 0 \\(C)\bigcirc: AA^T=I \Rightarrow A^{-1} =A^T \text{ exists  } \Rightarrow \text{rank}(A)=n \\(D)\bigcirc: A \text{ is an orthogonal matrix} \Rightarrow |\lambda_i| =1, \lambda_i \text{ is an eigenvalue of }A \Rightarrow \text{trace}(A) = \sum \lambda_i \le 3 \\(E)\times: A= \begin{bmatrix} 0&-1\\ 1&0 \end{bmatrix} \Rightarrow \det(A-\lambda I) =\lambda^2+1=0 \Rightarrow \lambda=\pm i \not \in \mathbb R \\ \Rightarrow \text{True statements: }\bbox[red, 2pt]{(CD)}$$

解答:$$(A)\times: A= \begin{bmatrix}1& 0\\0&1\\ 0& 0 \end{bmatrix} \Rightarrow \begin{bmatrix}1\\0\\0 \end{bmatrix} \text{ and }\begin{bmatrix}1\\0\\0 \end{bmatrix} \text{ are linearly independent, but }\\\qquad AA^T= \begin{bmatrix}1& 0& 0\\ 0& 1&0\\ 0&0&0 \end{bmatrix} \text{ is singular} \\(B) \bigcirc: A\text{ is }m\times n \Rightarrow A^T\text{ is }n\times m \Rightarrow \text{rank}(A^T)+ \dim(N(A^T)) =m \\(C)\times: \cases{AA^T\text{ is }m\times m\\ A^TA\text{ is }n\times n} \Rightarrow R(AA^T) \ne R(A^TA), \text{if }m\ne n \\(D)\times: A= \begin{bmatrix}1& 2 \end{bmatrix} \Rightarrow \cases{AA^T= [5] \\ A^TA = \begin{bmatrix}1&2\\3& 4 \end{bmatrix} } \Rightarrow \cases{\det(AA^T)=5\ne 0\\ \det(A^TA)=0} \\(E)\times: [3,4,5] \cdot [1,-1,2]^T=9 \ne 0 \\ \Rightarrow \text{True statement: }\bbox[red, 2pt]{(B)}$$

解答:$$(E)\times: rank(A^2)=rank(A)-1 \lt rank(A) \Rightarrow A \text{ is singular} \\(A)\times: rank(A)-rank(A^2)=1 \Rightarrow \text{ there is exactly one Jordan block for }\lambda=0\text{ with size }\ge 2\\\qquad \text{A matrix is diagonalizable iff all its Jordan blocks have size 1}\Rightarrow (A) \text{ is false} \\ \Rightarrow \text{The true statements: }\bbox[red, 2pt]{(BCD)}$$

解答:$$(D)\bigcirc: \text{The projection }\mathbf p \text{ of }\mathbf u \text{ onto }\mathbf v \text{ creates a right triangle where }\mathbf u \text{ is }\\\qquad \text{the hypotenuse, }\mathbf p\text{ is on leg, and }\mathbf u-\mathbf p\text{ is the other leg.} \\\qquad \Rightarrow ||\mathbf u||^2 =||\mathbf p||^2+||\mathbf u-\mathbf p||^2 \Rightarrow ||\mathbf u-\mathbf p||^2=||\mathbf u||^2-||\mathbf p||^2 \\(A)\bigcirc: ||\mathbf u||^2 =||\mathbf p||^2+||\mathbf u-\mathbf p||^2 \Rightarrow ||\mathbf u|| \ge ||\mathbf u-\mathbf p|| \\(E)\bigcirc: \mathbf p\text{ is the unique vector in the subspace spanned by }\mathbf v\text{ that is closest to }\mathbf u. \\\qquad \Rightarrow ||\mathbf u-\mathbf p|| \le ||\mathbf u-\alpha\mathbf v|| \\(B)\times:||\mathbf u||^2 =||\mathbf p||^2+||\mathbf u-\mathbf p||^2 \not \Rightarrow ||\mathbf p||\ge ||\mathbf u-\mathbf p|| \\(C)\times: \text{Orthogonality implies }\mathbf p=0, \text{ not }\mathbf p=\mathbf u\\\Rightarrow \text{The true statements: }\bbox[red, 2pt]{(ADE)}$$

解答:$$(A)\bigcirc: \langle u_3, u_3\rangle =u_3(-1)^2+ u_3(0)^2+ u_3(1)^2= {1\over 9}+ {4\over 9}+{1\over 9} ={2\over 3} \\(D)\times: p(x)=c_1u_1+c_2u_2+ c_3u_3 \Rightarrow c_i= {\langle p,u_i\rangle \over \langle u_i,u_i \rangle} \Rightarrow \cases{c_1=6/3=2\\ c_2=-6/2=-3\\ c_3=4/(2/3) =6} \\\qquad \Rightarrow p(x)=2u_1-3u_2+6u_3 \\(B)\bigcirc: \text{Projection }q(x) \text{ onto span}\{u_1,u_2\} \Rightarrow q(x) =c_1u_1+c_2u_2 =2u_1-3u_2 \\(C)\bigcirc: ||q(x)||^2 = \langle 2u_1-3u_2, 2u_1-3u_2\rangle =2^2\cdot 3+(-3)^2\cdot 2+0=30 \\(E)\times:||p(x)||^2 =\langle 2u_1-3u_2+6u_3, 2u_1-3u_2+6u_3 \rangle =54 \ne 572/3 \\ \Rightarrow \text{Correct statements: }\bbox[red, 2pt]{(ABC)}$$

解答:$$L(\mathbf x) = \begin{bmatrix}d_2x_3-d_3x_2\\ d_3x_1-d_1x_3\\ d_1x_2-d_2x_1 \end{bmatrix} =   \begin{bmatrix}0&-d_3& d_2 \\ d_3&0&-d_1\\ -d_2&d_1 &0\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix} \Rightarrow \cases{[a_{21}, a_{22},a_{23}] =[d_3,0,-d_1] \\ [a_{31}, a_{32}, a_{33}] =[-d_2,d_1,0]} \\ \Rightarrow \cases{a_{21}=d_3 \ne d_2\\ a_{22}=0 \ne d_1\\ a_{31}=-d_2 \ne -d_3\\ a_{32}=d_1\\ a_{33 }=0} \Rightarrow \text{Correct statements: }\bbox[red, 2pt]{(DE)}$$

第11-14題需要詳明推導計算過程。如推導計算過程錯誤，將酌扣分數或不給分

11.

解答:$${dM_z(t) \over dt}={M_0-M_z(t) \over T_1} \Rightarrow \int {dM_z\over M_0-M_z} =\int{1\over T_1}dt \Rightarrow -\ln(M_0-M_z)={t\over T_1}+c_1 \\ \Rightarrow M_0-M_z=c_2e^{-t/T_1} \Rightarrow M_z(t)=M_0-c_2e^{-t/T_1} \Rightarrow M_z(t)=0=M_0-c_2 \Rightarrow c_2=M_0 \\ \Rightarrow \bbox[red, 2pt]{M_z(t)=M_0 \left( 1-e^{-t/T_1} \right)}$$


解答:$$y''+y'=e^x+e^{-x} \Rightarrow y''e^x+y'e^x=e^{2x}+1 \Rightarrow (y'e^x)'=e^{2x}+1 \\\Rightarrow y'e^x =\int (e^{2x}+1) \,dx ={1\over 2}e^{2x}+x+c_1 \Rightarrow y'={1\over 2}e^x+xe^{-x}+c_1e^{-x} \\ \Rightarrow y= \int \left( {1\over 2}e^x+xe^{-x}+c_1e^{-x} \right)\,dx ={1\over 2}e^x-xe^{-x}-e^{-x}-c_1e^{-x}+c_2 \\ \Rightarrow \bbox[red, 2pt]{y={1\over 2}e^x-xe^{-x}+ c_3e^{-x}+c_2}$$


解答:$$\cases{x'=-6x+2y\\ y'=-3x+y} \Rightarrow  \begin{bmatrix}x'\\ y' \end{bmatrix} = \begin{bmatrix}-6& 2\\ -3& 1 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix}  \Rightarrow A= \begin{bmatrix}-6& 2\\ -3& 1 \end{bmatrix} \Rightarrow \det(A-\lambda I) =\lambda( \lambda +5) =0 \\ \Rightarrow \cases{ \lambda_1 =0 \Rightarrow \text{ eigenvector }v_1= \begin{bmatrix}1\\3 \end{bmatrix} \\\lambda_2 =-5 \Rightarrow \text{ eigenvector }v_2= \begin{bmatrix}2\\1 \end{bmatrix}} \Rightarrow \mathbf x(t)= c_1e^{\lambda_1 t}v_1+ c_2e^{\lambda_2 t}v_2 \\ \Rightarrow \begin{bmatrix}x(t)\\y(t) \end{bmatrix} =c_1 \begin{bmatrix}1\\3 \end{bmatrix}+ c_2e^{-5t} \begin{bmatrix}2\\1 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{\cases{x(t)=c_1+ 2c_2e^{-5t} \\y(t) =3c_1+c_2e^{-5t}}}$$

解答:$$\textbf{(1) }\text{Legendre equation: }(1-x^2) y''-2xy'+\ell(\ell+1)y=0\\ \text{Now, we have }(x^2-1)y''+2xy'-2y=0 \Rightarrow (1-x^2)y''-2xy'+2y=0 \Rightarrow \ell(\ell+1)=2 \\ \Rightarrow \ell=1 \;(\text{taking the non-negative integer }) \\ \text{The first solution }P_{\ell}(x) =P_1(x)=x\\ \text{The second solution }Q_{\ell}(x)=Q_{1} ={x\over 2} \ln \left|{1+x\over 1-x} \right|-1 \\ \cases{\ln(1+x)=x-x^2/2 +x^3/3- x^4/4+ \cdots \\ \ln(1-x)=-x-x^2/2-x^3/3-x^4/4-\cdots} \Rightarrow \ln {1+x\over 1-x}= 2 \left( x+{x^3\over 3}+{x^5\over 5}+\cdots \right) =2 \sum_{n=0}^\infty {x^{2n+1} \over 2n+1} \\ \Rightarrow Q_1(x)=-1+ \sum_{n=1}^\infty {x^{2n} \over 2n-1} \Rightarrow y=c_1P_1+ c_2Q_1 \Rightarrow \bbox[red, 2pt]{y=c_1x +c_2  \left( -1+ \sum_{n=1}^\infty {x^{2n} \over 2n-1} \right)}\\\textbf{(2) }1-x^2=0 \Rightarrow \text{ radius }R= \bbox[red, 2pt]1$$

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解題僅供參考，碩士班歷年試題及詳解