國立暨南國際大學113學年度碩士班入學考試
科目:工程數學(線性代數、微分方程)
解答: $$\textbf{(a)}\; \det(A)=0+12+12-3-0+24= \bbox[red, 2pt]{45} \\\textbf{(b)} \; [A\mid I] = \left[ \begin{array} {rrr|rrr} -2 & 2 & -3 & 1 & 0 & 0\\2 & 1 & -6 & 0 & 1 & 0\\-1 & -2 & 0 & 0 & 0 & 1\end{array} \right] \xrightarrow{R_1-2R_3\to R_1, R_2+2R_3\to R_2} \left[ \begin{array} {rrr|rrr}0 & 6 & -3 & 1 & 0 & -2\\0 & -3 & -6 & 0 & 1 & 2\\-1 & -2 & 0 & 0 & 0 & 1\end{array} \right] \\ \xrightarrow{-R_3\to R_3, -R_2/3 \to R_2} \left[ \begin{array} {rrr|rrr}0 & 6 & -3 & 1 & 0 & -2\\0 & 1 & 2 & 0 & - \frac{1}{3} & - \frac{2}{3}\\1 & 2 & 0 & 0 & 0 & -1\end{array} \right] \xrightarrow{R_1-6R_2 \to R_1, R_2-2R_2\to R_3} \left[ \begin{array} {rrr|rrr}0 & 0 & -15 & 1 & 2 & 2\\0 & 1 & 2 & 0 & - \frac{1}{3} & - \frac{2}{3}\\1 & 0 & -4 & 0 & \frac{2}{3} & \frac{1}{3} \end{array} \right] \\ \xrightarrow{R_1\leftrightarrow R_3} \left[ \begin{array} {rrr|rrr}1 & 0 & -4 & 0 & \frac{2}{3} & \frac{1}{3}\\0 & 1 & 2 & 0 & - \frac{1}{3} & - \frac{2}{3}\\0 & 0 & -15 & 1 & 2 & 2 \end{array} \right] \xrightarrow{R_3/(-15) \to R_3} \left[ \begin{array} {rrr|rrr}1 & 0 & -4 & 0 & \frac{2}{3} & \frac{1}{3}\\0 & 1 & 2 & 0 & - \frac{1}{3} & - \frac{2}{3}\\0 & 0 & 1 & - \frac{1}{15} & - \frac{2}{15} & - \frac{2}{15} \end{array} \right] \\ \xrightarrow{R_1+4R_3\to R_1, R_2-2R_3\to R_2} \left[ \begin{array} {rrr|rrr}1 & 0 & 0 & - \frac{4}{15} & \frac{2}{15} & - \frac{1}{5}\\0 & 1 & 0 & \frac{2}{15} & - \frac{1}{15} & - \frac{2}{5}\\0 & 0 & 1 & - \frac{1}{15} & - \frac{2}{15} & - \frac{2}{15} \end{array} \right] \Rightarrow \bbox[red, 2pt]{A^{-1}= \left[ \begin{matrix}- \frac{4}{15} & \frac{2}{15} & - \frac{1}{5}\\\frac{2}{15} & - \frac{1}{15} & - \frac{2}{5}\\- \frac{1}{15} & - \frac{2}{15} & - \frac{2}{15}\end{matrix}\right]} \\\textbf{(c)}\; \det(A-\lambda I) =-(\lambda+3)^2(\lambda-5) =0 \Rightarrow \lambda=-3,5 \\\quad \lambda_1=-3 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}1 & 2 & -3 \\2 & 4 & -6 \\-1 & -2 & 3\end{bmatrix} \begin{bmatrix} x_1\\ x_2 \\x_3 \end{bmatrix} =0 \Rightarrow x_1+2x_2=3x_3 \\\quad \Rightarrow v= x_2 \begin{pmatrix}-2\\1\\ 0\end{pmatrix} +x_3 \begin{pmatrix} 3 \\0 \\1\end{pmatrix}, \text{choose }v_1=\begin{pmatrix}-2\\1\\ 0 \end{pmatrix}, v_2= \begin{pmatrix} 3 \\0 \\1\end{pmatrix} \\ \lambda_2=5 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}-7 & 2 & -3 \\2 & -4 & -6 \\-1 & -2 & -5\end{bmatrix} \begin{bmatrix} x_1\\ x_2 \\x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1+x_3=0\\ x_2+2x_3=0} \\\quad \Rightarrow v= x_3 \begin{pmatrix}-1\\ -2\\ 1\end{pmatrix}, \text{ choose }v_3= \begin{pmatrix}-1\\ -2\\ 1\end{pmatrix} \\ \Rightarrow \bbox[red, 2pt]{\text{eigenvalues: }-3,5 \text{ and eigenvectors: }\begin{pmatrix}-2\\1\\ 0\end{pmatrix}, \begin{pmatrix} 3 \\0 \\1\end{pmatrix}, \begin{pmatrix}-1\\ -2\\ 1\end{pmatrix} } \\\textbf{(d)} \; AB= \begin{bmatrix}-2 & 2 & -3 \\2 & 1 & -6 \\-1 & -2 & 0 \end{bmatrix} \begin{bmatrix} 5\\8\\2\end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix} 0 \\6 \\-21\end{bmatrix}} \\\textbf{(e)} \; rref(A)=\begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 &1 \end{bmatrix} \Rightarrow A\mathbf x=0 \Rightarrow \begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 &1 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix} =0 \Rightarrow \mathbf x= \bbox[red, 2pt]{\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}}$$
解答: $$\cases{kx_1+ x_2+x_3 =2k\\ x_1+kx_2+ x_3=3\\ x_1+ x_2 +kx_3=1} \Rightarrow \begin{bmatrix} k & 1 &1 \\1 & k & 1 \\1 & 1 & k\end{bmatrix}\begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix}= \begin{bmatrix} 2k\\3 \\1\end{bmatrix} \equiv A\mathbf x= \mathbf b \\\textbf{(a)} \; \det(A) =k^3-3k+2 =(k-1)^2(k+ 2) \ne 0 \Rightarrow \bbox[red, 2pt]{k\ne 1,k\ne -2} \\ \textbf{(b)}\; \bbox[red, 2pt] {k=1} \Rightarrow \cases{x_1+ x_2+x_3 =2\\ x_1+ x_2+ x_3= 3\\ x_1+ x_2 + x_3=1} ,\text{no solution } \\ \textbf{(c)}\; \bbox[red, 2pt]{k=-2} \Rightarrow A=\begin{bmatrix} -2 & 1 &1 \\1 & -2 & 1 \\1 & 1 & -2\end{bmatrix} \Rightarrow rref(A) =\begin{bmatrix} 1 & 0 &-1 \\0 & 1 & -1 \\0 & 0 & 0\end{bmatrix}, \text{infinitely many solutions} \\ \textbf{(d)} \;x_1= \cfrac{\begin{vmatrix}2k & 1 & 1 \\ 3& k& 1 \\1 & 1 & k\end{vmatrix}}{\det(A)} =\cfrac{2k^3-6k+4}{k^3-3k+2}=2, x_2= \cfrac{\begin{vmatrix}k & 2k & 1 \\ 1& 3& 1 \\1 & 1 & k\end{vmatrix}}{\det(A)} =\cfrac{k^2+k-2}{k^3-3k+2} ={1\over k-1} \\ \quad x_3= \cfrac{\begin{vmatrix}k & 1 & 2k \\ 1& k& 3 \\1 & 1 & 1 \end{vmatrix}}{\det(A)} =\cfrac{-k^2-k+2}{k^3-3k+2} =-\cfrac{ 1}{k-1} \\ \quad \Rightarrow \bbox[red, 2pt]{ \begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix} =\begin{bmatrix}2 \\1/(k-1) \\-1/(k-1)\end{bmatrix}} \\\textbf{(e)} \;k=0 \Rightarrow \begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix} =\begin{bmatrix}2 \\1/(k-1) \\-1/(k-1)\end{bmatrix} =\bbox[red, 2pt]{\begin{bmatrix}2 \\ -1 \\1\end{bmatrix}}$$
解答: $$\textbf{(a)}\; y_1= x^A \Rightarrow y_1'=Ax^{A-1} \Rightarrow y_1''=A(A-1)x^{A-2} \\\quad \Rightarrow (x^2-x)y_1'' +Bxy_1'+y_1= (A^2-A+AB+1)x^A-(A^2-A)x^{A-1}=0\\ \quad \Rightarrow \cases{A^2-A+AB+1=0 \\ A^2-A=0} \Rightarrow \bbox[red, 2pt]{ \cases{A=1 \\B=-1}} \\\textbf{(b)}\; y_2=uy_1 = ux \Rightarrow y_2'=u'x+ u \Rightarrow y_2''=2u'+u''x \\ \quad \Rightarrow (x^2-x)y_2''-xy_2'+y_2= (x^3-x^2)u''+ (x^2-2x)u'=0\\ \quad \text{Let }v=u', \text{ then }v'+{x-2\over x^2-x}v=0 \Rightarrow \int {1\over v}dv = \int -{x-2\over x^2-x} \,dx \\ \quad \Rightarrow \ln v=\ln {x-1\over x^2} +c_1 \Rightarrow v=c_2\cdot {x-1\over x^2} \Rightarrow u= \int c_2\cdot {x-1\over x^2}\,dx =c_2(\ln x+{1\over x})+c_3 \\ \quad \Rightarrow y_2=ux = c_2(x\ln x+1)+ c_3x \Rightarrow y=y_1+y_2 \Rightarrow \bbox[red, 2pt] {y= c_2(x\ln x+1)+ c_4x}$$
解答: $$\textbf{(a)}\; \cases{P(x,y)= e^{x+y}+ 2xye^y\\ Q(x,y)=x^2e^y-2e^{-y}} \Rightarrow \cases{P_y=e^{x+y}+2xe^y+2xye^y \\Q_x=2xe^y}\\\quad \Rightarrow -{P_y-Q_x\over P} =-{e^{x+y} +2xye^y \over e^{x+y}+2xye^y} =-1 \Rightarrow u'=-u \Rightarrow \text{integrating factor }\bbox[red, 2pt]{u(y)=e^{-y}} \\\textbf{(b)}\; \cases{uP = e^{x}+ 2xy\\ uQ=x^2-2e^{-2y}} \Rightarrow \cases{(uP)_y= 2x\\ (uQ)_x=2x} \Rightarrow \text{exact} \\\quad \Rightarrow \Phi(x,y)=\int (e^x+2xy)\,dx = \int (x^2-2e^{-2y})\,dy \\\quad \Rightarrow \Phi(x,y)= e^x+x^2y+ \phi(y) =x^2y+e^{-2y}+ \rho(x) \\ \Rightarrow \bbox[red, 2pt]{e^x+ x^2y +e^{-2y}=C}$$
解答: $$\textbf{(a)}\; y_1=e^{-2x} \Rightarrow y_1'=-2e^{-2x} \Rightarrow y_1''=4e^{-2x} \Rightarrow y''+y'+Ey=(2+E)e^{-2x}=0 \\ \quad \Rightarrow \bbox[red, 2pt]{E=-2} \\ \textbf{(b)}\; y''+y'-2y =0 \Rightarrow \lambda^2+\lambda-2=0 \Rightarrow (\lambda+2)( \lambda-1)=0 \Rightarrow \lambda=1,-2 \\\quad \Rightarrow \bbox[red, 2pt]{ y=c_1e^x +c_2e^{-2x}}$$
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解題僅供參考,其他歷年試題及詳解
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