114年中正大學電機碩士班-線代與微方詳解

 國立中正大學114學年度碩士班招生考試試題

科目名稱:線性代數與微分方程
系所組別:電機工程學系-電磁晶片組、計算機工程組、電力與電能處理甲組

線性代數(每一題10分)

解答:$$\textbf{(1) }B= \begin{bmatrix} a& b\\ c& d\end{bmatrix} \Rightarrow BB= \begin{bmatrix} a^2+bc& ab+bd\\ ac+cd& bc+d^2\end{bmatrix} = \begin{bmatrix} 5& 5\\ 5& 10\end{bmatrix} \Rightarrow \cases{a^2+bc=5 \cdots(1)\\ b(a+d)=5 \cdots(2)\\ c(a+d)=5 \cdots(3)\\bc+d^2=10 \cdots(4)} \\ {(2) \over (3)} \Rightarrow \cases{a =2d/3\\ b=d/3\\ c=d/3} \Rightarrow d=\pm 3 \Rightarrow \bbox[red, 2pt]{B= \begin{bmatrix} 2& 1\\ 1& 3\end{bmatrix}, \begin{bmatrix} -2& -1\\ -1& -3 \end{bmatrix}} \\\textbf{(2) } \det(A)\lt 0 \Rightarrow \det(B)\det(B)\lt 0 \Rightarrow \det(B) \in \mathbb C \Rightarrow A\text{ has no real square root} \quad \bbox[red, 2pt]{QED}$$

解答:$$\text{Let }\cases{B_1=\begin{bmatrix} 3& 4\\ 3& -4\end{bmatrix}\\ B_2= \begin{bmatrix} 0& 1\\ 1& 0 \end{bmatrix} \\B_3=\begin{bmatrix} 0& -8\\ -12& -2\end{bmatrix}}; \text{Suppose } a,b,c \in \mathbb R, \text{ then }aB_1+bB_2+ cB_3=0 \\ \Rightarrow \begin{bmatrix} 3a& 4a+b-8c\\ 3a+b-12c& -4a -2c\end{bmatrix} =0 \Rightarrow \cases{a=0\\ b=0\\ c=0} \Rightarrow B_1,B_2,B_3 \text{ are linearly independent} \\ B=\begin{bmatrix} \alpha& \beta\\ \gamma& \delta\end{bmatrix} \in M_{22} \Rightarrow aB_1+bB_2+cB_3 =B  \Rightarrow \cases{3a=\alpha\\ 4a+b-8c=\beta\\ 3a+b-12c=\gamma\\ -4a-2c=\delta} \Rightarrow \cases{a=\alpha/3\\ b= -20\alpha/3+\beta-4\delta\\c=-2\alpha/3-\delta/2} \\ \Rightarrow B \text{ is a linear combination of }B_1,B_2,B_3 \Rightarrow \{B_1,B_2,B_3\} \text{ is a basis of }M_{22} \quad \bbox[red, 2pt]{QED}$$

解答:$$A=\begin{bmatrix} 1& 0& 0 \\ 0& r-2& 2\\ 0& s-1& r+2\\ 0& 0& 3\end{bmatrix} \\\text{rank 1 } : \text{the first row vector (1,0,0) and the last row vector (0,0,3) are linearly independent} \\\qquad rank(A)\ge 2 \Rightarrow \bbox[red, 2pt]{rank(A) \ne 1}\\ \text{rank 2: } \cases{r=2\\ s=1} \Rightarrow A= \begin{bmatrix} 1& 0& 0 \\ 0& 0& 2\\ 0& 0& 4\\ 0& 0& 3\end{bmatrix} \Rightarrow rref(A) =\begin{bmatrix} 1& 0& 0 \\ 0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{bmatrix} \Rightarrow rank(A)=2 \\\qquad \Rightarrow \bbox[red, 2pt]{r=2 \text{ and }s=1, \text{ then rank}(A)=2} \\\text{rank 3: }r\ne 2 \text{ or }s\ne 1 \text{ then the second column vector of }A\ne 0\Rightarrow rank(A)=3 \\ \qquad \Rightarrow \bbox[red, 2pt]{r\ne 2 \text{ or }s\ne 1  \Rightarrow rank(A)=3}\\ \text{rank 4: } A \text{ is 4x3} \Rightarrow rank(A) \le \min(4,3) \Rightarrow \bbox[red, 2pt]{rank(A) \ne 4}$$

解答:$$S= \begin{bmatrix}1 &a  & b \\0 & 2 & c \\0 & 0 & 3\end{bmatrix} \Rightarrow S^2 = \begin{bmatrix}1 & 3a  & ac+4b \\0 & 4 & 5c\\0 & 0 & 9\end{bmatrix} =\begin{bmatrix}1 &3  & 5 \\0 & 4 & 5 \\0 & 0 & 9 \end{bmatrix}  \Rightarrow \cases{3a=3\\ ac+4b=5\\ 5c=5} \\ \Rightarrow \cases{a=1\\ b=1\\ c=1} \Rightarrow \bbox[red, 2pt]{S =\begin{bmatrix}1 &1  & 1 \\0 & 2 & 1 \\0 & 0 & 3\end{bmatrix} }$$

解答:$$\cases{P(-1,t,-t) \\Q(2s,1+s,s)} \Rightarrow ||P-Q||^2 = f(s,t)=(2s+1)^2+(1+s-t)^2 +(s+t)^2 \\ \Rightarrow \cases{f_s=4(2s+1)+2(1+s-t)+2(s+t) =0\\ f_t =-2(1+s-t)+2(s+t)=0} \Rightarrow \cases{2s+1=0\\ 2t=1} \Rightarrow \bbox[red, 2pt]{\cases{s=-{1\over 2} \\t={1\over 2}}}$$

微分方程(每一題10分)

解答:$$L\{y'' \}+4L\{y' \}+13L\{y \} =L\{\delta(t-\pi) \}+ L\{ t-3\pi\} \\ \Rightarrow s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0)) +13Y(s) =e^{-\pi s}+e^{-3\pi s} \\ \Rightarrow (s^2+4s+13)Y(s)-s-4 =e^{-\pi s}+e^{-3\pi s} \Rightarrow Y(s) ={1\over s^2+4s+13} \left(s+4+ e^{-\pi s}+e^{-3\pi s}\right) \\ \Rightarrow y(t) =L^{-1}\{Y(s)\}   = -{2\over 3}e^{-2t} \sin(3t)+e^{-2t}\cos(3t) +{4\over 3}e^{-2t} \sin(3t) \\\qquad +u(t-\pi)e^{-2(t-\pi) }{1\over 3}\sin(3(t-\pi)) +u(t-3\pi) e^{-2(t-3\pi)} {1\over 3}\sin(3(t-3\pi)) \\ \Rightarrow \bbox[red, 2pt]{y(t)= {2\over 3}e^{-2t} \sin(3t) +e^{-2t}\cos(3t)+u(t-\pi)e^{-2(t-\pi) }{1\over 3}\sin(3(t-\pi))  }\\ \qquad \qquad \bbox[red, 2pt]{+u(t-3\pi) e^{-2(t-3\pi)} {1\over 3}\sin(3(t-3\pi))}$$

解答:$$\cases{y(0)=1 \\ y'(0)=2 \\y''+xy'+y=0} \Rightarrow \cases{x_0=0\\ y_0=1\\ y_0' =2} \Rightarrow y_0''=-x_0y_0'-y_0=-1 \Rightarrow \cases{x_0=0\\ y_0=1\\ y_0'=2\\ y_0''=-1}\\ x_1 =x_0+h=0.1 \Rightarrow y_1=y_0+h\cdot y_0'+{h^2\over 2}y_0'' =1+0.1\cdot 2+{0.01\over 2}\cdot (-1)= 1.195 \\ \Rightarrow y_1'= y_0'+ h\cdot y_0'' =2+0.1\cdot (-1) =1.9 \Rightarrow y_1'' =-x_1y_1'-y_1=-0.1\cdot 1.9-1.195=-1.385\\ \Rightarrow y(0.2)=y_2=y_1+h\cdot y_1'+{h^2\over 2}y_1'' =1.195+0.1\cdot 1.9+{0.01\over 2}(-1.385) = \bbox[red, 2pt]{1.378075}$$

解答:$$y''-4y'+4y=0 \Rightarrow \lambda^2-4\lambda+4=0 \Rightarrow (\lambda-2)^2=0 \Rightarrow \lambda=2 \Rightarrow y_h=c_1e^{2x }+c_2xe^{2x}\\ \cases{y_1=e^{2x} \\y_2 =xe^{2x}} \Rightarrow W = \begin{vmatrix}y_1 &y_2\\ y_1'& y_2' \end{vmatrix} =\begin{vmatrix}e^{2x} &xe^{2x}\\ 2e^{2x}& e^{2x} +2xe^{2x} \end{vmatrix} =e^{4x} \\ \Rightarrow y_p =-e^{2x} \int{ xe^{2x} \cdot (12x^2-6x)e^{2x} \over e^{4x}} \,dx +xe^{2x} \int{e^{2x} \cdot (12x^2-6x)e^{2x} \over e^{4x}} \,dx \\\qquad =-e^{2x} \int(12x^3-6x^2)\,dx +xe^{2x} \int (12x^2-6x)\,dx =-e^{2x}(3x^4-2x^3) +xe^{2x}(4x^3-3x^2) \\ \Rightarrow y_p=x^4e^{2x}-x^3e^{2x} \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y= c_1e^{2x }+c_2xe^{2x}+x^4e^{2x}-x^3e^{2x}}$$

解答:$$\mathbf X'=\begin{bmatrix} 6&-1\\ 5& 4\end{bmatrix} \mathbf X \Rightarrow \begin{bmatrix} x_1'\\ x_2' \end{bmatrix} =\begin{bmatrix} 6&-1\\ 5& 4\end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix}  \Rightarrow \cases{x_1'=6x_1-x_2 \cdots(1) \\ x_2'= 5x_1+4x_2 \cdots(2)}\\ (1) \Rightarrow x_2=6x_1-x_1' \Rightarrow x_2'=6x_1'-x_1'' 代入(2) \Rightarrow 6x_1'-x_1''=5x_1+24x_1-4x_1'\\ \Rightarrow x_1''-10x_1' +29x_1=0 \Rightarrow \lambda^2-10\lambda+29=0 \Rightarrow \lambda= 5\pm 2i \Rightarrow x_1= e^{5t}(c_1 \cos 2t+ c_2 \sin 2t) \\ \Rightarrow x_1'=  e^{5t}((5c_1+2c_2) \cos 2t+ (5c_2-2c_1) \sin t) \Rightarrow x_2= 6x_1-x_1'\\= e^{5t}((c_1-2c_2) \cos 2t+ (c_2+2c_1) \sin 2t) \\又 \mathbf X(0)=\begin{bmatrix} -2\\ 8 \end{bmatrix} \Rightarrow \cases{x_1(0)=-2\\ x_2(0)=8} \Rightarrow \cases{c_1=-2 \\ c_1-2c_2=8} \Rightarrow \cases{c_1=-2\\ c_2=-5}\\ \Rightarrow \cases{x_1(t)=e^{5t}(-2\cos 2t-5 \sin 2t) \\x_2(t)= e^{5t}(8\cos 2t-9 \sin 2t)}  \Rightarrow \bbox[red, 2pt]{\mathbf X=  \begin{bmatrix} -2&-5\\ 8& -9 \end{bmatrix} \begin{bmatrix} \cos 2t\\ \sin 2t\end{bmatrix} e^{5t}}$$

解答:$$L\{f(t)\} =\int_0^\infty f(t)e^{-st}\,dt =\int_0^\pi \cos(4t)e^{-st}\,dt = \left. \left[ {e^{-st} \over s^2+16}(4\sin(4t)-s\cos(4t)) \right] \right|_0^\pi ={s\over s^2+16}(1-e^{-\pi s}) \\ \Rightarrow L\{y''\} +16L\{y\}=L\{f(t)\} \Rightarrow s^2Y(s)-1+16Y(s) ={s\over s^2+16}(1-e^{-\pi s}) \\ \Rightarrow Y(s)= {s\over (s^2+16)^2}(1-e^{-\pi s}) +{1\over s^2+16} \Rightarrow y(t)=L^{-1}\{Y(s)\}  \\=L^{-1} \left\{{s\over (s^2+16)^2}(1-e^{-\pi s}) +{1\over s^2+16} \right\} = {t\sin(4t) \over 8}- {(t-\pi) \sin 4(t-\pi)\over 8}u(t-\pi)+ {1\over 4}\sin(4t) \\ \Rightarrow  \bbox[red, 2pt]{y(t)={1\over 8}(t\sin(4t)+2\sin(4t)-u(t-\pi) \cdot (t-\pi)\sin 4(t-\pi))}$$

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