114年台師電機碩士班-工程數學詳解

 國立臺灣師範大學 114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:$$\cases{M(x,y)= 2xy+3y\\ N(x,y) = 4y^3+x^2+3x+4} \Rightarrow M_y=2x+3=N_x \Rightarrow Exact \\ \Rightarrow \Psi(x,y) =\int (2xy+3y)dx = \int(4y^3+x^2+3x+4)\,dy \\\Rightarrow x^2y+3xy+ \phi(y) =y^4+x^2y+3xy+4y+ \rho(x) \Rightarrow \Psi(x,y)=y^4+x^2y+3xy+4y=C \\ y(0)=1 \Rightarrow 1+4=C \Rightarrow C=5 \Rightarrow \bbox[red, 2pt]{y^4+x^2y+3xy+4y=5}$$

解答:$$L\{y''\}-4L\{y'\}+4L\{y\}=L\{\delta(t-1)\} \Rightarrow s^2Y(s)-1-4sY(s)+4Y(s) =e^{-s} \\ \Rightarrow Y(s) ={1+e^{-s} \over (s-2)^2} = {1\over (s-2)^2} +{e^{-s} \over (s-2)^2} \Rightarrow y(t) =L^{-1}\{Y(s)\} \\=L^{-1} \left\{{1\over (s-2)^2} \right\}+ L^{-1} \left\{{e^{-s} \over (s-2)^2} \right\}  \Rightarrow \bbox[red, 2pt]{y(t)=te^{2t}+u(t-1)(t-1)e^{2(t-1)}}$$

解答:$$\begin{bmatrix}2 & 4 & 3 & 5 \\-4 & -7 & -5 & -8 \\6 & 8 & 2 & 9 \\ 4 & 9 & -2 & 14 \end{bmatrix} \xrightarrow{R_2+2R_1\to R_2} \begin{bmatrix}2 & 4 & 3 & 5 \\0 & 1 & 1 & 2 \\6 & 8 & 2 & 9 \\4 & 9 & -2 & 14 \end{bmatrix} \xrightarrow{R_3-3R_1\to R_3} \begin{bmatrix} 2 & 4 & 3 & 5 \\0 & 1 & 1 & 2 \\ 0 & -4 & -7 & -6 \\4 & 9 & -2 & 14\end{bmatrix} \\ \xrightarrow{R_4-2R_1 \to R_4} \begin{bmatrix} 2 & 4 & 3 & 5 \\0 & 1 & 1 & 2 \\0 & -4 & -7 & -6 \\ 0 & 1 & -8 & 4\end{bmatrix} \xrightarrow{R_3+4R_2 \to R_3} \begin{bmatrix} 2 & 4 & 3 & 5 \\0 & 1 & 1 & 2 \\0 & 0 & -3 & 2 \\ 0 & 1 & -8 & 4\end{bmatrix} \xrightarrow{R_4-R_2\to R_4} \begin{bmatrix}2 & 4 & 3 & 5 \\0 & 1 & 1 & 2 \\ 0 & 0 & -3 & 2 \\0 & 0 & -9 & 2 \end{bmatrix}  \\ \xrightarrow{R_4-3R_3 \to R_4} \begin{bmatrix} 2 & 4 & 3 & 5 \\0 & 1 & 1 & 2 \\0 & 0 & -3 & 2 \\ 0 & 0 & 0 & -4\end{bmatrix} \Rightarrow \bbox[red, 2pt]{L = \begin{bmatrix} 1 & 0 & 0 & 0 \\-2 & 1 & 0 & 0 \\3 & -4 & 1 & 0 \\ 2 & 1 & 3 & 1\end{bmatrix}, U=\begin{bmatrix} 2 & 4 & 3 & 5 \\0 & 1 & 1 & 2 \\0 & 0 & -3 & 2 \\ 0 & 0 & 0 & -4\end{bmatrix}}$$

解答:$$\textbf{1) }\bbox[red, 2pt]{true }:A^{-1} \text{ exists} \Rightarrow [A\mid I] \xrightarrow{\text{row operations}} [I\mid A^{-1}] \Rightarrow rref(A)=I \\\textbf{2) }\bbox[red, 2pt]{true }:\text{Since the inverse operation for any elementary row operation is itself }\\ \qquad \text{a single elementary row operation, the inverse of any elementary matrix is}\\ \qquad \text{ always another elementary matrix.} \\\textbf{3) }\bbox[red, 2pt]{false }:\text{A counterexample, } B= \begin{bmatrix}2&0\\0&1 \end{bmatrix} \Rightarrow rref(B)=I, \text{ but }\det(B)=2 \ne 1 \\\textbf{4) }\bbox[red, 2pt]{false }:\mathbb R^2 \text{ has a basis of two vectors, that is, }(1,0) \text{ and }(0,1). \text{Its dimension is 2} \\\textbf{5) }\bbox[red, 2pt]{true }: S\text{ can span }V \text{ but has extra vectors}, \text{it is a spanning set but not a basis.} \\\qquad \text{The extra vectors are redundant and can be combined by the basis. Therefore,} \\\qquad \text{ They are linearly dependent.} \\\textbf{6) }\bbox[red, 2pt]{true }: \text{Let }A\text{ be }m\times n\text{ matrix, then }rank(A)\le \min(m,n) \Rightarrow \text{columns or rows are l.d.} \\\textbf{7) }\bbox[red, 2pt]{false }: \det(A)=0 \Rightarrow rank(A)\lt n \\\textbf{8) }\bbox[red, 2pt]{false }:\text{A counterexample, }\cases{P=0\\ Q\ne 0} \Rightarrow PQ=QP=0$$

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解題僅供參考，碩士班歷年試題及詳解