台灣聯合大學系統 115 學年度碩士班招生考試
類組: 電機類 科目: 工程數學 C
Problem 1-5 為複選題，答錯不倒扣
解答:$$(A) \bigcirc: A = \begin{pmatrix} 1 & 0 & -1 \\ 2 & 5 & 1 \\ 0 & -4 & 3 \end{pmatrix} \Rightarrow \det(A)=27 \ne 0 \\(B)\times: B= \begin{pmatrix} 2 & -4 & 1 \\ 0 & 3 & -1 \\ 6 & 0 & -1 \end{pmatrix} \Rightarrow \det(B)=0\\ (C)\bigcirc: C = \begin{pmatrix} 1 & 2 & -1 \\ 1 & 0 & 2 \\ 2 & 1 & 1 \end{pmatrix} \Rightarrow \det(C) = 3 \ne 0 \\(D) \times: D = \begin{pmatrix} -1 & 3 & 1 \\ 2 & -4 & -3 \\ 1 & -1 & -2 \end{pmatrix} \Rightarrow \det(D)=0 \\(E) \times: E = \begin{pmatrix} 1 & -3 & -2 \\ -3 & 1 & 3 \\ -2 & -10 & -2 \end{pmatrix} \Rightarrow \det(E)=0 \\ \Rightarrow \text{Correct answers: }\bbox[red, 2pt]{(AC)}$$
解答:$$(A) \bigcirc: \text{If the null space only contains the zero vector, the transformation is injective.} \\(B) \times: \text{$T$ is onto iff $R(T) = W$. The statement incorrectly equates the range to the domain $V$.} \\(C)\times: rank(T)\le dim(W) \Rightarrow dim(V)\le dim(W) \\(D)\bigcirc: dim(V)= null(T)+dim(W) \Rightarrow dim(V)\ge dim(W) \\(E)\bigcirc: \text{This is the exact definition of the Rank-Nullity Theorem.} \\ \text{False answers: }\bbox[red, 2pt]{(BC)}$$
解答:$$(C)\times: \text{The columns of the transpose matrix $A^t$ are exactly the rows of the original matrix $A$.}\\ \text{Therefore, the column space of $A^t$ is the row space of $A$.}\\ \text{Because row rank equals column rank, it must always be true that $\text{rank}(A^t) = \text{rank}(A)$.} \\(E)\times: Ax=b\text{ is consistent iff the vector $\mathbf{b}$ is in the column space of $A$.}\\ \text{This occurs iff adding $\mathbf{b}$ as an extra column does not increase the dimension of the column space; } \\ \text{meaning $\text{rank}([A\vert{}\mathbf{b}]) = \text{rank}(A)$. If $\text{rank}([A\vert{}\mathbf{b}]) = \text{rank}(A) + 1$, the system is strictly inconsistent.} \\ \Rightarrow \text{Correct answers: }\bbox[red, 2pt]{(ABD)}$$
解答:$$ A = \begin{pmatrix} 1 & 2 & 3 & 1 & 1 \\ 1 & 4 & 0 & 1 & 2 \\ 1 & 2 & -3 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \end{pmatrix} \Rightarrow RREF(A) = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 3 & 1 & 1 \\ 0 & 0 & -6 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \Rightarrow rank(A)=3 \Rightarrow \bbox[red, 2pt]{(C)}$$
解答:$$$$
Problem 6-10 為計算題，計算題應詳列計算過程
解答:$$\text{By Orthogonal Decomposition Theorem, we have } \mathcal{V} = \mathcal{W} \oplus \mathcal{W}^\perp \\ \mathbf x\in \mathcal V \Rightarrow \mathbf x=\mathbf w+ \mathbf y, \text{where $\mathbf{w} \in \mathcal{W}$ and $\mathbf{y} \in \mathcal{W}^\perp$}\\ \text{We are given the condition that $\mathbf{x} \notin \mathcal{W}$.}\\ \text{If $\mathbf{y}$ were equal to the zero vector ($\mathbf{y} = \mathbf{0}$), then $\mathbf{x} = \mathbf{w} + \mathbf{0} = \mathbf{w}$, which would imply that $\mathbf{x} \in \mathcal{W}$.}\\ \text{ This contradicts our initial premise. Therefore, it must be true that: } \mathbf{y} \neq \mathbf{0} \\ \langle \mathbf{x}, \mathbf{y} \rangle = \langle \mathbf{w} + \mathbf{y}, \mathbf{y} \rangle = \langle \mathbf{w}, \mathbf{y} \rangle + \langle \mathbf{y}, \mathbf{y} \rangle = 0 + \langle \mathbf{y}, \mathbf{y} \rangle = \Vert{}\mathbf{y}\Vert{}^2 \ne \mathbf 0\quad \bbox[red, 2pt]{QED.}$$
解答:$$\textbf{(i) } M = \begin{bmatrix} 5 & 4 & 3 \\ -1 & 0 & -3 \\ 1 & -2 & 1 \end{bmatrix} \Rightarrow \det(M-\lambda I)= -(\lambda+2)(\lambda-4)^2=0 \Rightarrow \bbox[red, 2pt]{\text{ The eigenvalues are: }-2,4} \\ \lambda_1=-2 \Rightarrow (M-\lambda_1 I)v =0 \Rightarrow \begin{bmatrix} 7 & 4 & 3 \\ -1 & 2 & -3 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow \cases{y=z\\ x+z=0} \\\qquad \Rightarrow v= z \begin{bmatrix}-1\\1\\1 \end{bmatrix}, \text{ choosing }v_1 = \begin{bmatrix}-1\\1\\1 \end{bmatrix} \\ \lambda_2=4 \Rightarrow (M-\lambda_2 I)v =0 \Rightarrow \begin{bmatrix} 1 & 4 & 3 \\ -1 & -4 & -3 \\ 1 & -2 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow \cases{y+z=0\\ x=z} \\ \qquad \Rightarrow v = z \begin{bmatrix}1\\-1\\1 \end{bmatrix}, \text{ choosing }v_2= \begin{bmatrix}1\\-1\\1 \end{bmatrix}\\ \bbox[red, 2pt]{\text{The eigenvectors are: } \begin{bmatrix}-1\\1\\1 \end{bmatrix} , \begin{bmatrix}1\\-1\\1 \end{bmatrix}} \\\textbf{(ii) } \cases{\lambda_1=-2 \Rightarrow 1\times 1 \text{ block} \\ \lambda_2=4 \Rightarrow 2\times 2\text{ block}} \Rightarrow J = \begin{bmatrix} -2 & 0 & 0 \\ 0 & 4 & 1 \\ 0 & 0 & 4 \end{bmatrix} \\ \text{We need a generalized eigenvector $\mathbf{v}_3$ to complete the basis for $\lambda_2 = 4$,}\\\text{satisfying the chain condition $(M - 4I)\mathbf{v}_3 = \mathbf{v}_2$ , i,e.,} \begin{bmatrix} 1 & 4 & 3 \\ -1 & -4 & -3 \\ 1 & -2 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \\ \Rightarrow \cases{y+z=0\\ x-z=1} \Rightarrow \text{choosing }z=0 \Rightarrow \mathbf{v}_3 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \Rightarrow Q = [\mathbf{v}_1 \mid \mathbf{v}_2 \mid \mathbf{v}_3] \Rightarrow Q = \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \\ \text{Because the vectors forming $Q$ are linearly independent, $Q$ is invertible.}\\ \text{We have constructively demonstrated that there exist matrices $J$ and $Q$ satisfying $M Q = Q J$,}\\ \text{which definitively shows that $J = Q^{-1} M Q$.} \qquad \bbox[red, 2pt]{QED.}$$
解答:$$L\{y'(t)\}+4 L\left\{ \int_0^t y(u)\cos 2(t-u)\,du\right\}-L\{1\}=0\\ \Rightarrow sY(s)-y(0)+4 Y(s)L\{ \cos(2t)\}-{1\over s}=0 \Rightarrow sY(s) + 4 \left( Y(s) \frac{s}{s^2 + 4} \right) - \frac{1}{s} = 0 \\ \Rightarrow Y(s) = \frac{s^2 + 4}{s^2(s^2 + 8)} = \frac{1/2}{s^2} + \frac{1/2}{s^2 + 8} \\ \Rightarrow y(t) = \mathcal{L}^{-1} \left\{ \frac{1}{2} \cdot \frac{1}{s^2} \right\} + \mathcal{L}^{-1} \left\{ \frac{1}{2} \cdot \frac{1}{s^2 + 8} \right\} = \frac{1}{2}t + \frac{\sqrt{2}}{8}\sin(2\sqrt{2}t) \Rightarrow \bbox[red, 2pt]{(E)}$$
解答:$$\text{Let } x(t) = c_n t^n + c_{n-1} t^{n-1} + \dots + c_1 t + c_0, c_n\ne 0 \Rightarrow \frac{dx}{dt} = n c_n t^{n-1} + (n-1) c_{n-1} t^{n-2} + \dots + c_1 \\ t^2 \frac{dx}{dt} + (at+b)x = ct+d \\ LHS = t^2 (n c_n t^{n-1} + \dots) + (at+b)(c_n t^n + \dots) = (n+a)c_n t^{n+1} + \text{lower order terms} \\ RHS=ct+d \\ LHS=RHS \Rightarrow \cases{(n+a)c_n = 0 \Rightarrow n+a=0 \Rightarrow a=-n \\ t^0 \text{ term: } b c_0 = d \Rightarrow c_0 = \frac{d}{b} \\ t^1 \text{ term: } a c_0 + b c_1 = c } \Rightarrow -n \left(\frac{d}{b}\right) + b c_1 = c \\ \Rightarrow b c_1 = c + \frac{nd}{b} = \frac{bc+nd}{b} \Rightarrow c_1 = \frac{bc+nd}{b^2} \Rightarrow bc+nd \neq 0 \\ \Rightarrow \text{an $n$th-order polynomial solution exists iff: } a = -n \quad \text{and} \quad bc+nd \neq 0 \\ \Rightarrow \text{The logical negation of the existence condition for all $n \in \mathcal{N}$:} \\\text{For all } n \in \mathcal{N}: \quad a \neq -n \quad \text{or} \quad bc+nd = 0 \\ \Rightarrow -a \notin \mathcal{N} \quad \text{or} \quad bc = ad \Rightarrow \bbox[red, 2pt]{(E)}$$
解答:$$\textbf{(i) }w(x,t)= X(x)T(t) \Rightarrow xX'T+XT'=xt\\ \text{Choosing }X(x)=x \Rightarrow X'(x)=1 \Rightarrow xT+xT'=xt \Rightarrow T'+T=t \Rightarrow T'e^t+Te^t=te^t\\ \Rightarrow (Te^t)'=te^t \Rightarrow Te^t= \int te^t\,dt =te^t-e^t+C \Rightarrow T(t)=t-1+Ce^{-t} \\ \Rightarrow w(x,t)=xT(t) =x(t-1+Ce^{-t}) \Rightarrow w(x,0)=x(C-1)=0 \Rightarrow C=1\\ \Rightarrow \bbox[red, 2pt]{w(x,t)=x(t-c+e^{-t})} \\ \textbf{(ii) } \cases{\mathcal{L}\left\{\frac{\partial w}{\partial t}\right\} = sW(x,s) - w(x,0) = sW(x,s) \\ \mathcal{L}\left\{x\frac{\partial w}{\partial x}\right\} = x\frac{dW}{dx} \\\mathcal{L}\{xt\} = x\mathcal{L}\{t\} = \frac{x}{s^2}} \Rightarrow x\frac{dW}{dx} + sW = \frac{x}{s^2} \\ \Rightarrow x^s\frac{dW}{dx} + sx^{s-1}W = \frac{x^s}{s^2} \Rightarrow \left( x^s W \right)'={x^s \over s^2} \Rightarrow x^s W = \int \frac{x^s}{s^2} dx = \frac{1}{s^2} \frac{x^{s+1}}{s+1} + C(s) \\ \Rightarrow W(x,s) = \frac{x}{s^2(s+1)} + C(s)x^{-s} \\ w(0,t) =0\Rightarrow W(0,s)=0 \Rightarrow C(s)=0 \Rightarrow W(x,s) = \frac{x}{s^2(s+1)} = x \left( -\frac{1}{s} + \frac{1}{s^2} + \frac{1}{s+1} \right) \\ \Rightarrow w(x,t) = x \left( \mathcal{L}^{-1}\left\{-\frac{1}{s}\right\} + \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} + \mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\} \right) =x(-1+t+e^{-t})\\ \Rightarrow \bbox[red, 2pt]{ w(x,t) = x(t - 1 + e^{-t}) }$$
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解題僅供參考，碩士班歷年試題及詳解
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