114年暨南大學電機碩士班-工程數學詳解

 國立暨南國際大學114學年度碩士班入學考試

科目:工程數學 適用:電機

解答: $$rref(A) =\left[ \begin{matrix}1 & 0 & -1\\0 & 1 & 2\\0 & 0 & 0\end{matrix} \right] \Rightarrow \bbox[red, 2pt]{rank(A)=2}, rref(B) =\left[ \begin{matrix}1 & 0 & 1\\0 & 1 & -1\\0 & 0 & 0\end{matrix} \right] \Rightarrow \bbox[red, 2pt]{rank(B)=2 }\\ AB= \left[\begin{matrix}-3 & 0 & -3\\-6 & 0 & -6\\-9 & 0 & -9\end{matrix}\right]  \Rightarrow rref(AB) = \left[ \begin{matrix}1 & 0 & 1\\0 & 0 & 0\\0 & 0 & 0\end{matrix} \right] \Rightarrow \bbox[red, 2pt]{rank(AB)=1}$$

解答: $$\textbf{(a) }\bbox[red, 2pt]{\text{true}}: A \text{ is not invertible} \Rightarrow \det(A)=0 \Rightarrow \det(AB) =\det(A) \det(B)=0 \Rightarrow AB \text{ is not invertible} \\\textbf{(b) } \bbox[red, 2pt]{\text{false}}: \det(A) =\pm \text{(product of its pivots)}  \\\textbf{(c) } \bbox[red, 2pt]{\text{false}}: \cases{A=\begin{bmatrix} 1& 0\\0& 1\end{bmatrix} \\[1ex] B=\begin{bmatrix} -1& 0\\0& -1\end{bmatrix}} \Rightarrow \cases{A-B=\begin{bmatrix} 2& 0\\0& 2\end{bmatrix} \\\det(A)=1\\ \det(B)=1} \Rightarrow \det(A-B)=4 \ne \det(A)-\det(B) =0 \\\textbf{(d) }\bbox[red, 2pt]{\text{true}}: \det(AB)= \det(A) \det(B) =\det(B)\det(A)= \det(BA)$$

解答: $$\textbf{(a) }\bbox[red, 2pt]{\text{true}}: \text{Suppose }A\text{ is invertible and has an eigenvalue of 0, then }\\\qquad Av=\lambda v =0v=0 \Rightarrow v=A^{-1}0=0, \text{ but }v\text{ is an eigenvector (not a zero vector)} \\\qquad \Rightarrow A \text{ has an eigenvalue of 0, then }A\text{ is not invertible.} \\\textbf{(b) }\bbox[red, 2pt]{\text{false}}: \begin{bmatrix}1& 2\\ 4& 3 \end{bmatrix} =\begin{bmatrix} -1& 1/2\\ 1& 1\end{bmatrix} \begin{bmatrix}-1& 0\\0& 5 \end{bmatrix} \begin{bmatrix} -1& 1/2\\ 1& 1\end{bmatrix}^{-1} \Rightarrow A= \begin{bmatrix}1& 2\\ 4& 3 \end{bmatrix} \text{ and } B=\begin{bmatrix}-1& 0\\0& 5 \end{bmatrix} \text{ are similar.} \\ \qquad \text{ but } \cases{\text{eigenvectors of }A=\begin{bmatrix}-1 \\1 \end{bmatrix}, \begin{bmatrix}1/2 \\1 \end{bmatrix} \\\text{eigenvectors of }B =\begin{bmatrix}1 \\0 \end{bmatrix}, \begin{bmatrix}0 \\1 \end{bmatrix}} \Rightarrow \text{ their eigenvectors are different} \\ \textbf{(c) }\bbox[red, 2pt]{\text{true}}: \det(A^T-\lambda I) =\det((A -\lambda I)^T) =\det(A-\lambda I) \Rightarrow A^T \text{ and }A \text{ have the same eigenvalues}\\ \textbf{(d) }\bbox[red, 2pt]{\text{false}}: A =\begin{bmatrix} 0& 0\\1& 0\end{bmatrix}  \Rightarrow \det(A-\lambda I) =\lambda^2=0 \Rightarrow \lambda =0, \text{ but }A\ne 0 \\\textbf{(e) }\bbox[red, 2pt]{\text{false}}: \cases{Av_1= \lambda_1 v_1\\ Av_2 = \lambda_2 v_2}, \lambda_1\ne \lambda_2 \text{ and suppose that }v_1+v_2 \text{ is also an eigenvector of }A \\\qquad \Rightarrow A(v_1+v_2) =\lambda_3(v_1+v_2) \Rightarrow Av_1+Av_2=\lambda_1 v_1+ \lambda_2 v_2 =\lambda_3 v_1+ \lambda_3 v_2 \\\qquad \Rightarrow (\lambda_1-\lambda_3)v_1=(\lambda_3-\lambda_2)v_2 \Rightarrow v_1\parallel v_2 \text{  contradiction}$$

解答: $$\textbf{(a) }z=xy \Rightarrow z'=y+xy' \Rightarrow \cases{y=z/x\\ y'=z'/x-z/x^2} \Rightarrow xe^z\left({z'\over x}-{z\over x^2} \right)+ e^z\cdot {z\over x}-3=0\\\qquad \Rightarrow \bbox[red, 2pt]{e^zz'-3=0} \\\textbf{(b) }z'e^z=3 \Rightarrow \int e^z\,dz=\int 3\,dx \Rightarrow e^z=3x+c_1 \Rightarrow z=xy=\ln(3x+c_1) \Rightarrow \bbox[red, 2pt]{y={1\over x}\ln(3x+c_1)}$$

解答: $$\textbf{(a) }y=x^{-4} \Rightarrow y'=-4x^{-5} \Rightarrow y''=20x^{-6} \Rightarrow 20x^{-6}=-\left( {9\over x}(-4x^{-5})+{A\over x^2}x^{-4}\right) =(36-A)x^{-6} \\\qquad \Rightarrow 20=36-A \Rightarrow A= \bbox[red, 2pt]{16} \\\textbf{(b) }y''=-\left({9\over x}y'+{16\over x^2}y \right) \Rightarrow x^2y''+9xy'+16y =0 \\ y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow x^2y''+9xy' +16y= (m^2 +8m+16)x^m=0 \\ \qquad \Rightarrow m^2+8m +16= (m+4)^2=0 \Rightarrow m=-4 \Rightarrow \bbox[red, 2pt]{y=c_1x^{-4}+ c_2x^{-4} \ln x}x$$

解答: $$\textbf{(a) }y= \sum_{n=0}^\infty a_n x^n \Rightarrow y' =\sum_{n=0}^\infty na_n x^{n-1} \Rightarrow y'' = \sum_{n=0}^\infty n(n-1)a_n x^{n-2}\\ \Rightarrow y''+3xy'= \sum_{n=0}^\infty (3na_n x^n +n(n-1)a_n x^{n-2}) = \sum_{n=0}^\infty (3na_n +(n+2)(n+1)a_{n+2})x^n\\ y''+3xy'= y \Rightarrow \sum_{n=0}^\infty (3na_n +(n+2)(n+1) a_{n+2})x^n =\sum_{n=0}^\infty a_n x^n \\ \Rightarrow 3na_n +(n+2)(n+1 )a_{n+2} =a_n \Rightarrow (3n-1)a_n + (n+2)(n+1)a_{n+2}=0,n=0,1,\dots \\ \Rightarrow a_{n+2} =-{3n-1\over (n+2)(n+1)}a_n   \Rightarrow \bbox[red, 2pt]{a_{n+2} =\left({4\over n+1}-{7\over n+2} \right)a_n,n=0,1,\dots} \\ \textbf{(b) }a_{n+2} = \left({4\over n+1}-{7\over n+2} \right)a_n,n=0,1,2,\dots \\ \Rightarrow a_2={1\over 2}a_0 \Rightarrow a_3=-{1\over 3}a_1 \Rightarrow a_4= -{5\over 12}a_2 =-{5\over 24}a_0 \\ \Rightarrow \bbox[red, 2pt]{y=a_0+a_1x+{1\over 2}a_0x^2 -{1\over 3}a_1x^3-{5\over 24}a_0x^4+\cdots}$$

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