115年暨南大學電機碩士班-工程數學詳解

 國立暨南國際大學115學年度碩士班入學考試

科目:工程數學(線性代數、微分方程)
適用:電機系

解答:$$\textbf{(a) } A = \begin{bmatrix} 1 & 3 & 0 & 1 \\ 2 & 4 & 1 & 0 \\ 0 & 1 & 2 & -1 \end{bmatrix} \Rightarrow \text{RREF}(A)= \begin{bmatrix} 1 & 0 & 0 & - \frac{4}{5}\\0 & 1 & 0 & \frac{3}{5}\\0 & 0 & 1 & - \frac{4}{5}\end{bmatrix} \Rightarrow \text{rank}(A)=\bbox[red, 2pt]3 \\ \textbf{(b) } \text{rank}(A) + \text{nullity}(A) = n \Rightarrow  3 + \text{nullity}(A) = 4 \Rightarrow \text{nullity}(A) = \bbox[red, 2pt]1 \\ \textbf{(c) }\text{From(a), we have a basis for Col$(A)$ consists of the first three original column vectors of }A. \\ v_1 = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \quad v_2 = \begin{bmatrix} 3 \\ 4 \\ 1 \end{bmatrix}, \quad v_3 = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \\ \text{To find an orthogonal basis $\{u_1, u_2, u_3\}$, we apply the Gram-Schmidt orthogonalization process:} \\ u_1=v_1= \begin{bmatrix}1\\2\\0 \end{bmatrix} \Rightarrow  u_2 = v_2 - \frac{v_2 \cdot u_1}{u_1 \cdot u_1} u_1 = \begin{bmatrix} 3 \\ 4 \\ 1 \end{bmatrix} - \frac{11}{5} \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}  = \begin{bmatrix} 4/5 \\ -2/5 \\ 1 \end{bmatrix} \\ \text{To avoid fractions, we set }u_2= \begin{bmatrix}4\\-2\\5 \end{bmatrix}. \Rightarrow u_3=  v_3 - \frac{v_3 \cdot u_1}{u_1 \cdot u_1} u_1 - \frac{v_3 \cdot u_2 }{u_2  \cdot u_2 } u_2 \\  = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} - \frac{2}{5} \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} - \frac{8}{45} \begin{bmatrix} 4 \\ -2 \\ 5 \end{bmatrix} = \begin{bmatrix} -10/9\\ 5/9\\10/9 \end{bmatrix}. \text{We set }u_3= \begin{bmatrix}-2\\1\\2 \end{bmatrix}\\ \Rightarrow \text{An orthhogonal basis for the column space of }A: \bbox[red, 2pt]{\left\{ \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 4 \\ -2 \\ 5 \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \\ 2 \end{bmatrix} \right\}}$$

解答:$$\textbf{(a) }\bbox[red, 2pt]{\text{true}}: \text{rank}(AB) \leq \min(\text{rank}(A), \text{rank}(B)) \Rightarrow  \text{rank}(AB) \leq \text{rank}(A) < n \\\textbf{(b) }\bbox[red, 2pt]{\text{false}}: \cases{A= I_3\\ B=-I_3} \Rightarrow \cases{\text{nullity}(A) =\text{nullity}(B) = 0 \\ A+B= 0_3 \Rightarrow \text{nullity}(A+B) = 3} \Rightarrow 0+0\ne 3 \\\textbf{(c) }\bbox[red, 2pt]{\text{false}}: A=-I_3 \Rightarrow A^TA= (-I_3)^T(-I_3) =(-I_3)(-I_3)=I_3^2= I_3 \Rightarrow A\text{ is orghogonal} \\ \qquad \text{tr}(A)=\text{tr}(-I_3)= -3 \ne 3 \\ \textbf{(d) }\bbox[red, 2pt]{\text{true}}: \text{This is a standard, fundamental theorem in linear algebra.}$$

解答:$$\textbf{(a) }\bbox[red, 2pt]{\text{true}}:  \det(A-\lambda I)=0 \Rightarrow \det(A-0I)=\det(A)=0 \Rightarrow \det(A^T)=0 \Rightarrow \det(A^T-0I)=0 \\   \textbf{(b) }\bbox[red, 2pt]{\text{false}}:  A = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \Rightarrow A \text{ has }\cases{\text{eigenvalue }\lambda_1=1  \text{ and eigenvector }v_1= \begin{bmatrix}1\\ 0 \end{bmatrix} \\\text{eigenvalue }\lambda_2=2  \text{ and eigenvector }v_2= \begin{bmatrix}0\\ 1 \end{bmatrix}} \\ \qquad \Rightarrow B = P^{-1}AP = \begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix}0& -2\\1& 3 \end{bmatrix} \Rightarrow (B-\lambda_1 I)v=0 \\\qquad \Rightarrow  \begin{bmatrix} -1 & -2 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \Rightarrow u_1= k \begin{bmatrix}-2\\1 \end{bmatrix} \Rightarrow u_1\ne v_1 ,v_2 \\\textbf{(c) }\bbox[red, 2pt]{\text{true}}:  (AB)v= \lambda v \Rightarrow B(AB)v= B(\lambda v) \Rightarrow BA(Bv)= \lambda(Bv) \\\textbf{(d) }\bbox[red, 2pt]{\text{false}}: \text{ The zero matrix does have an eigenvalue, and that eigenvalue is exactly $0$.}  \\\textbf{(e) }\bbox[red, 2pt]{\text{false}}: Av=\lambda v \Rightarrow (Av)^T=(\lambda v)^T \Rightarrow v^TA^T=\lambda v^T \\ \qquad \text{This is not equivalent to the given statement $v^T A = \lambda v^T$ unless $A = A^T$ .}\\\qquad \text{For a general matrix $A$, the statement is false.}$$

解答:$$\textbf{(a) } (\cos y + e^{-x})\frac{dy}{dx} + \sin y = 2xe^{-x} \Rightarrow (\sin y - 2xe^{-x})dx + (\cos y + e^{-x})dy = 0 \\ \quad \cases{M(x,y)= \sin y-2xe^{-x} \\ N(x,y) =\cos y+e^{-x}} \Rightarrow \cases{M_y=\cos y\\ N_x=-e^{-x}} \Rightarrow M_y\ne N_x \Rightarrow \bbox[red, 2pt]{\text{Not Exact}} \\ \textbf{(b) }{M_y-N_x\over N}=1 \Rightarrow \text{ integrating factor }\mu(x)=e^x \Rightarrow \cases{\mu M=e^x\sin y-2x\\ \mu N=e^x \cos y+1} \\\quad \Rightarrow (\mu M)_y=e^x\cos y= (\mu N)_x \Rightarrow \text{Exact} \Rightarrow f(x,y)= \int(e^x \sin y-2x)\,dy = \int(e^x \cos y+1)\,dx \\ \quad \Rightarrow f(x,y)=e^x\sin y+y-x^2 \Rightarrow \text{the general solution: }\bbox[red, 2pt]{e^x\sin y+y-x^2=C}$$

解答:$$\textbf{(a) }  y=x+\alpha \Rightarrow y'=1 \Rightarrow y''=0 \Rightarrow  (x^2 + 3x + 2)(0) - (x + 2)(1) + (x + \alpha) = 0  \\ \quad \Rightarrow -(x + 2) + x + \alpha = 0 \Rightarrow \alpha= \bbox[red, 2pt]2 \\ \textbf{(b) }y_1=x+\alpha=x+2 \Rightarrow y=v(x)y_1= v(x)(x+2) \Rightarrow y'=v'(x+2)+v \Rightarrow y''=v''(x+2)+2v'\\\quad \Rightarrow  (x + 1)(x + 2)[v''(x + 2) + 2v'] - (x + 2)[v'(x + 2) + v] + v(x + 2) = 0  \\\quad \Rightarrow    (x^2 + 3x + 2)v'' + xv' = 0 \Rightarrow (x^2 + 3x + 2)w' + xw = 0 \quad (\text{let }w=v') \\\quad \Rightarrow  \frac{1}{w}dw = \frac{-x}{(x + 1)(x + 2)}dx \Rightarrow \int \frac{1}{w}dw = \int \left( \frac{1}{x + 1} - \frac{2}{x + 2} \right) dx \Rightarrow  w = \frac{x + 1}{(x + 2)^2} \\\quad \Rightarrow  v = \int \frac{x + 1}{(x + 2)^2} dx   = \ln\vert{}x + 2\vert{} + \frac{1}{x + 2} \Rightarrow y_2(x) = \left( \ln\vert{}x + 2\vert{} + \frac{1}{x + 2} \right)(x + 2)\\\quad =(x+2)\ln|x+2|+1 \Rightarrow  y(x) = C_1 y_1(x) + C_2 y_2(x) \\\quad \Rightarrow \bbox[red, 2pt]{y(x) = C_1(x + 2) + C_2\left((x + 2)\ln\vert{}x + 2\vert{} + 1\right)}$$

解答:$$\textbf{(a) }  y=e^{2x} \Rightarrow y'=2e^{2x} \Rightarrow y''=4e^{2x} \Rightarrow 4e^{2x}+4e^{2x}-3e^{2x}=5e^{2x}=\beta e^{2x} \Rightarrow \beta=\bbox[red, 2pt]5 \\ \textbf{(b) }y'' + 2y' - 3y = 0 \Rightarrow r^2+2r-3=0 \Rightarrow (r+3)(r-1)=0 \Rightarrow r=-3,1 \Rightarrow  y_h(x) = C_1 e^{-3x} + C_2 e^x \\\quad y_{p1} = Ax e^x \Rightarrow  y_{p1}'  = Ae^x(1 + x) \Rightarrow y_{p1}'' =   Ae^x(2 + x) \Rightarrow Ae^x(2 + x) + 2[Ae^x(1 + x)] - 3[Axe^x] = 4e^x \\\quad \Rightarrow 4A=4 \Rightarrow A=1 \Rightarrow  y_{p1} = x e^x  \Rightarrow  y_p = y_{p1} + y_{p2} = xe^x + e^{2x} \Rightarrow y=y_h+ y_p= \\ \quad \Rightarrow \bbox[red, 2pt]{y = C_1 e^{-3x} + C_2 e^x + xe^x + e^{2x}}$$

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解題僅供參考，碩士班歷年試題及詳解