112年台師大電機碩士班-工程數學詳解

 國立臺灣師範大學112學年度碩士班招生考試試題

科目:工程數學
適用系所:電機工程學系

解答: $$[A\mid I] =\left[\begin{matrix}-4 & 0 & 0 & 1 & 0 & 0\\0 & 8 & 13 & 0 & 1 & 0\\0 & 3 & 5 & 0 & 0 & 1\end{matrix}\right] \xrightarrow{R_1/(-4) \to R_1, R_2/8 \to R_2} \left[\begin{matrix}1 & 0 & 0 & - \frac{1}{4} & 0 & 0\\0 & 1 & \frac{13}{8} & 0 & \frac{1}{8} & 0\\0 & 3 & 5 & 0 & 0 & 1\end{matrix}\right]\\ \xrightarrow{R_3- 3R_2 \to R_3} \left[\begin{matrix}1 & 0 & 0 & - \frac{1}{4} & 0 & 0\\0 & 1 & \frac{13}{8} & 0 & \frac{1}{8} & 0\\0 & 0 & \frac{1}{8} & 0 & - \frac{3}{8} & 1\end{matrix}\right] \xrightarrow{R_2-13R_3\to R_2} \left[\begin{matrix}1 & 0 & 0 & - \frac{1}{4} & 0 & 0\\0 & 1 & 0 & 0 & 5 & -13\\0 & 0 & \frac{1}{8} & 0 & - \frac{3}{8} & 1\end{matrix}\right]\\ \xrightarrow{8R_3\to R_3} \left[ \begin{matrix} 1 & 0 & 0 & - \frac{1}{4} & 0 & 0\\0 & 1 & 0 & 0 & 5 & -13\\0 & 0 & 1 & 0 & -3 & 8\end{matrix} \right] \Rightarrow A^{-1}=\bbox[red, 2pt]{ \left[ \begin{matrix}  - \frac{1}{4} & 0 & 0\\  0 & 5 & -13\\  0 & -3 & 8\end{matrix}\right]}$$

解答: $$3y''-24y'+48y=0 \Rightarrow y''-8y'+16y=0 \Rightarrow \lambda^2-8\lambda +16=0 \Rightarrow (\lambda-4)^2=0 \\ \Rightarrow \lambda=4 \Rightarrow \bbox[red, 2pt]{y=c_1e^{4x}+ c_2xe^{4x}}$$

解答: $$\bbox[red, 2pt]{{d\over dt}T(t)= k(T(t)-R)} \Rightarrow {d\over dt}T(t)- kT(t)=-kR \Rightarrow e^{-kt}{d\over dt}T(t)- ke^{-kt}=-kRe^{-kt} \\ \Rightarrow \left( e^{-kt }T(t)\right)'=-kRe^{-kt} \Rightarrow e^{-kt }T(t)=Re^{-kt}+c_0 \Rightarrow T(t)=R+c_0e^{kt} \\ \Rightarrow T(0)=R+c_0 \Rightarrow c_0=T(0)-R \Rightarrow \bbox[red, 2pt]{T(t)=R+(T(0)-R)e^{kt}}$$

解答: $$y''+p(x)y'+q(x)y=0 \Rightarrow \text{ The Wronskian: }y_1y_2'-y_1'y_2= e^{-\int p(x)\,dx} \\ \text{ Let }y_2=v(x)y_1 \Rightarrow y_2'=v'y_1+ vy_1' \Rightarrow y_1(v'y_1+ vy_1') -y_1vy_1= e^{-\int p(x)\,dx} \\ \Rightarrow v'y_1^2= e^{-\int p(x)\,dx} \Rightarrow v=\int {e^{-\int p(x)\,dx}\over y_1^2}\,dx \Rightarrow y_2=\int {e^{-\int p(x)\,dx}\over y_1^2}\,dx\cdot y_1 \\ \Rightarrow y=c_1y_1+c_2 y_2 \Rightarrow y=c_1y_1+ c_2\int {e^{-\int p(x)\,dx}\over y_1^2}\,dx\cdot y_1\qquad \bbox[red, 2pt]{QED}$$

解答: $$L\{ e^{at}f(t)+ f''(t) +t\} =L\{e^{at}f(t)\} +L\{f''(t)\}+ L\{t\} = \bbox[red, 2pt]{F(s-a)+s^2F(s)-sf(0)-f'(0)+{1\over s^2}}$$

解答: $$1. \bbox[red, 2pt]{false}:A=0 \Rightarrow AB=BA=0 \\ 2.\bbox[red, 2pt]{false}: \{(1,1), (0,1) \}\text{ is a basis in }\mathbb R^2, \text{ but } (1,1)\not \bot (0,1) \\ 3. \bbox[red, 2pt]{false}: A= \begin{bmatrix} 2& 0\\ 0 & 2\end{bmatrix} \Rightarrow rref(A)= \begin{bmatrix} 1& 0\\ 0 & 1 \end{bmatrix} \Rightarrow \det(A) =4 \ne \det(rref(A))=1\\ 4.\bbox[red, 2pt]{true}: A\text{ is invertible and n by n} \Rightarrow rank(A)=n \Rightarrow rref(A)=I \\5. \bbox[red, 2pt]{true}: rank(A)= rank(A^T) \Rightarrow rank(AB)= rank((AB)^T) =rank(B^TA^T) \\6. \bbox[red, 2pt]{true}: A\text{ is m by n} \Rightarrow rank(A)\le \min(m,n)\\\qquad \Rightarrow \text{row vectors or column vectors are linearly dependent}$$

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解題僅供參考，碩士班歷年試題及詳解