國立中央大學112學年度碩士班考試
所別:環境工程研究所碩士班
科目:工程數學
解答:$$A= \left[\begin{matrix}3 & 2 & 1\\0 & -1 & 4\\0 & 0 & 2\end{matrix}\right] \Rightarrow [A\mid I] =\left[ \begin{array}{rrr|rrr} 3 & 2 & 1 & 1 & 0 & 0\\0 & -1 & 4 & 0 & 1 & 0\\0 & 0 & 2 & 0 & 0 & 1 \end{array} \right] \xrightarrow{R_1/3\to R_1, -R_2\to R_2, R_3/2 \to R_3} \\\left[ \begin{array}{rrr|rrr} 1 & \frac{2}{3} & \frac{1}{3} & \frac{1}{3} & 0 & 0\\0 & 1 & -4 & 0 & -1 & 0\\0 & 0 & 1 & 0 & 0 & \frac{1}{2} \end{array} \right] \xrightarrow{R_1-(2/3)R_2 \to R_1, R_2+4R_3 \to R_2}\left[ \begin{array}{rrr|rrr} 1 & 0 & 3 & \frac{1}{3} & \frac{2}{3} & 0\\0 & 1 & 0 & 0 & -1 & 2\\0 & 0 & 1 & 0 & 0 & \frac{1}{2}\end{array} \right] \\ \xrightarrow{R_1-3R_3\to R_1}\left[ \begin{array}{rrr|rrr} 1 & 0 & 0 & \frac{1}{3} & \frac{2}{3} & - \frac{3}{2}\\0 & 1 & 0 & 0 & -1 & 2\\0 & 0 & 1 & 0 & 0 & \frac{1}{2}\end{array} \right] \Rightarrow A^{-1} = \bbox[red, 2pt]{\begin{bmatrix} \frac{1}{3} & \frac{2}{3} & - \frac{3}{2}\\ 0 & -1 & 2\\ 0 & 0 & \frac{1}{2} \end{bmatrix}}$$
解答:$$\vec V=4x\vec i-8y\vec j+4z\vec k \Rightarrow \cases{v_1=4x\\ v_2=-8y\\ v_3=4z} \Rightarrow \cases{\frac{\partial v_1}{\partial x} =4 \\ \frac{\partial v_2}{\partial y} = -8 \\ \frac{\partial v_3}{\partial z} = 4 } \Rightarrow \text{div }\vec V=4-8+4=\bbox[red, 2pt]0$$
解答:$$\vec r(t)=2t\vec i+2t^2\vec j+2t^3\vec k =(2t,2t^2,2t^3) \Rightarrow \vec r'(t)=(2,4t,6t^2) \Rightarrow \vec r''(t) =(0,4,12t) \\ \Rightarrow \vec r' \times \vec r'' = (2,4t,6t^2) \times (0,4,12t) = (24t^2,-24t,8) \\ \Rightarrow \kappa(t) =\cfrac{\Vert \vec r' \times \vec r''\Vert}{\Vert \vec r' \Vert^3} = \cfrac{\Vert (24t^2,-24t,8)\Vert}{\Vert (2,4t,6t^2) \Vert^3} = \bbox[red, 2pt]{ \cfrac{ \sqrt{9t^4+ 9t^2+1}}{(9t^4+4t^2+1) \sqrt{ 9t^4+4t^2+1}}}$$
解答:$$A= \left[ \begin{matrix}5 & 7 & -5\\0 & 4 & -1\\2 & 8 & -3\end{matrix}\right] \Rightarrow \cases{A^2= \left[\begin{matrix}15 & 23 & -17\\-2 & 8 & -1\\4 & 22 & -9\end{matrix}\right]\\[1ex] 2A= \left[\begin{matrix}10 & 14 & -10\\0 & 8 & -2\\4 & 16 & -6\end{matrix} \right]} \Rightarrow B=A^2+2A= \left[\begin{matrix}25 & 37 & -27\\-2 & 16 & -3\\8 & 38 & -15\end{matrix}\right] \\ \Rightarrow \det(B-\lambda I) =-(\lambda-3)(\lambda-8)(\lambda-15) =0 \Rightarrow \bbox[red,2pt]{特徵值=3,8,15} \\ \lambda_1= 3 \Rightarrow (B-\lambda_1 I)v =0 \Rightarrow \begin{bmatrix}22 & 37 & -27 \\-2 & 13 & -3 \\8 & 38 & -18 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\ x_3\end{bmatrix} =0 \Rightarrow \cases{3x_1=2x_3\\ 3x_2=x_3} \\ \qquad \Rightarrow v= x_3 \begin{pmatrix}2/3 \\1/3\\1 \end{pmatrix} \Rightarrow \text{choose }v_1=\begin{pmatrix}2 \\1\\ 3 \end{pmatrix} \\ \lambda_2=8 \Rightarrow (B-\lambda_2 I)v=0 \Rightarrow \left[\begin{matrix}17 & 37 & -27 \\ -2 & 8 & -3 \\8 & 38 & -23 \end{matrix}\right] \begin{bmatrix}x_1 \\x_2 \\ x_3 \end{bmatrix} =0 \Rightarrow \cases{2x_1 =x_3\\ 2x_2=x_3} \\\qquad \Rightarrow v= x_3 \begin{pmatrix}1/2 \\1/2\\1 \end{pmatrix} \Rightarrow \text{ choose }v_2= \begin{pmatrix} 1 \\1 \\ 2 \end{pmatrix} \\ \lambda_3=15 \Rightarrow (B-\lambda_3 I)v=0 \Rightarrow \begin{bmatrix}10 & 37 & -27 \\-2 & 1 & -3 \\8 & 38 & -30 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\ x_3\end{bmatrix} =0 \Rightarrow \cases{x_1+x_3=0\\ x_2=x_3} \\ \qquad \Rightarrow v= x_3 \begin{pmatrix}-1 \\1\\1 \end{pmatrix} \Rightarrow \text{ choose }v_3= \begin{pmatrix}-1 \\1\\1 \end{pmatrix} \\ \Rightarrow \bbox[red, 2pt]{特徵向量= \begin{pmatrix} 2 \\1\\ 3 \end{pmatrix}, \begin{pmatrix} 1 \\1 \\ 2 \end{pmatrix}, \begin{pmatrix}-1 \\1\\1 \end{pmatrix}}$$
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