110年北科大自動化碩士班-工程數學詳解

 國立臺北科技大學110學年度碩士班招生考試

自動化科技研究所
工程數學

解答: $$\begin{bmatrix}1 & 2& 0 & 1 \\2 & 4& 1& 4\\3 &6 & 3& 9 \end{bmatrix} =\begin{bmatrix}1 & 0 & 0 \\2 & x & 0 \\ 3 & y & z \end{bmatrix} \begin{bmatrix}1 & 2 & 0 & 1 \\0 & u & v & w \\0 & 0 & 0 & 0 \end{bmatrix} =\begin{bmatrix}1 & 2 & 0 & 1 \\ 2 & ux+4 & vx & wx+2 \\3 & uy+6 & vy & wy+3 \end{bmatrix} \\ \Rightarrow \cases{ux+4=4\\ vx=1\\ wx+2=4\\ uy+6=6\\ vy=3\\ wy+3=9} \Rightarrow \cases{ux=0\\ vx=1\\ wx=2\\ uy=0\\ vy=3\\ wy=6} \Rightarrow \bbox[cyan,2pt]{z \text{ is don't care !!}}\\ \textbf{(b)}\;B=\left[\begin{matrix}-4 & 0 & 5\\-3 & 3 & 5\\-1 & 2 & 2\end{matrix}\right] \Rightarrow [B \mid I]= \left[\begin{array}{rrr|rrr}-4 & 0 & 5 & 1 & 0 & 0\\-3 & 3 & 5 & 0 & 1 & 0\\-1 & 2 & 2 & 0 & 0 & 1 \end{array} \right] \xrightarrow{R_1-4R_3 \to R_1, R_2-3R_3\to R_2} \\ \left[ \begin{array}{rrr|rrr}0 & -8 & -3 & 1 & 0 & -4\\0 & -3 & -1 & 0 & 1 & -3\\-1 & 2 & 2 & 0 & 0 & 1\end{array} \right] \xrightarrow{-R_3\to R_3,-R_1\to R_1, -R_2\to R_2} \left[ \begin{array}{rrr|rrr}0 & 8 & 3 & -1 & 0 & 4\\0 & 3 & 1 & 0 & -1 & 3\\1 & -2 & -2 & 0 & 0 & -1\end{array} \right] \\ \xrightarrow {R_1\leftrightarrow R_3, (1/3) R_2 \to R_2} \left[ \begin{array}{rrr|rrr}1 & -2 & -2 & 0 & 0 & -1\\0 & 1 & \frac{1}{3} & 0 & - \frac{1}{3} & 1\\0 & 8 & 3 & -1 & 0 & 4\end{array} \right] \xrightarrow{R_1+2R_2\to R_1, -8R_2 +R_3\to R_3} \\ \left[ \begin{array}{rrr|rrr}1 & 0 & - \frac{4}{3} & 0 & - \frac{2}{3} & 1\\0 & 1 & \frac{1}{3} & 0 & - \frac{1}{3} & 1\\0 & 0 & \frac{1}{3} & -1 & \frac{8}{3} & -4\end{array} \right] \xrightarrow{4 R_3+R_1 \to R_1,R_2-R_3\to R_2} \left[ \begin{array}{rrr|rrr}1 & 0 & 0 & -4 & 10 & -15\\0 & 1 &  0 & 1 & -3 & 5\\0 & 0 & \frac{1}{3} & -1 & \frac{8}{3} & -4\end{array} \right] \\ \xrightarrow{3R_3\to R_3} \left[ \begin{array}{rrr|rrr}1 & 0 & 0 & -4 & 10 & -15\\0 & 1 & 0 & 1 & -3 & 5\\0 & 0 & 1 & -3 & 8 & -12 \end{array} \right] \Rightarrow B^{-1} = \bbox[red, 2pt]{\left[ \begin{matrix}-4 & 10 & -15\\1 & -3 & 5\\-3 & 8 & -12\end{matrix} \right]}$$

解答: $$\textbf{(a)}\; A= \left[\begin{matrix}1 & 2\\2 & 4\\3 & 6\end{matrix}\right] \Rightarrow B=A^TA= \left[\begin{matrix}1 & 2 & 3\\2 & 4 & 6\end{matrix}\right] \left[\begin{matrix}1 & 2\\2 & 4\\3 & 6\end{matrix}\right]= \left[\begin{matrix}14 & 28\\28 & 56\end{matrix}\right]\\ \qquad \det(B-\lambda I)=(14-\lambda)(56-\lambda)-28^2 = \lambda(\lambda-70)=0 \Rightarrow \text{eigenvalues: }\bbox[red, 2pt]{0,70} \\ \textbf{(b)}\; C=AA^T= \left[\begin{matrix}5 & 10 & 15\\10 & 20 & 30\\15 & 30 & 45\end{matrix}\right] \Rightarrow \det(C-\lambda I)=\lambda^2(70-\lambda) =0 \Rightarrow \lambda=0,70 \\ \lambda_1=0 \Rightarrow (A-\lambda_1 I)\mathbf x=0 \Rightarrow \begin{bmatrix}5 & 10 & 15\\10 & 20 & 30\\15 & 30 & 45 \end{bmatrix} \begin{bmatrix}x_1 \\x_2\\ x_3 \end{bmatrix} =0 \Rightarrow x_1+2 x_2+3x_3=0 \\\qquad \Rightarrow \mathbf x=s \begin{pmatrix}-2 \\1\\ 0 \end{pmatrix} +t\begin{pmatrix}-3 \\0\\ 1 \end{pmatrix} ,s,t\in \mathbb R, \text{ choose }v_1 = \begin{pmatrix} -2 \\1\\ 0 \end{pmatrix},v_2=  \begin{pmatrix}-3 \\0\\ 1 \end{pmatrix}\\ \lambda_2 =70  \Rightarrow (A-\lambda_2 I)\mathbf x=0 \Rightarrow \begin{bmatrix}-65 & 10 & 15\\10 & -50 & 30\\15 & 30 & -25 \end{bmatrix} \begin{bmatrix}x_1 \\x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{3x_1 =x_3\\ 3x_2=2x_3} \\\qquad \Rightarrow \mathbf x= k \begin{pmatrix}1/3 \\2/3 \\ 1 \end{pmatrix}, k\in \mathbb R ,\text{choose }v_3 =\begin{pmatrix} 1 \\2 \\ 3 \end{pmatrix} \\ \text{Applying Gram-Schmidt process, }  e_1={v_1\over \Vert v_1\Vert } = \begin{pmatrix}-2/\sqrt 5 \\1/\sqrt 5 \\ 0 \end{pmatrix}, e_2={v_2-(v_2\cdot e_1)e_1 \over \Vert v_2-(v_2\cdot e_1)e_1 \Vert} \\ =\begin{pmatrix} -3/\sqrt{70} \\-6/\sqrt{70} \\ 5/\sqrt{70} \end{pmatrix} \Rightarrow e_3={v_3-(v_3\cdot e_2)e_2-(v_3\cdot e_1)e_1 \over \Vert v_3-(v_3\cdot e_2)e_2-(v_3\cdot e_1)e_1 \Vert } = \begin{pmatrix} 1/\sqrt{14} \\2 /\sqrt{14} \\ 3/\sqrt{14} \end{pmatrix} \\ \Rightarrow \text{orthonormal eigenvectors of }AA^T: \bbox[red, 2pt]{\begin{pmatrix}-2/\sqrt 5 \\1/\sqrt 5 \\ 0 \end{pmatrix},  \begin{pmatrix} -3/\sqrt{70} \\-6/\sqrt{70} \\ 5/\sqrt{70} \end{pmatrix}, \begin{pmatrix} 1/\sqrt{14} \\2 /\sqrt{14} \\ 3/\sqrt{14} \end{pmatrix}}$$

解答: $$\textbf{(a)}\; \vec v=(2,3) \text{ is perpendicular to the line, due to }(2,3) \cdot (-3,2)=0, \\\qquad \text{where (-3,2) is the directional vector of the line.} \\ \textbf{(b)}\; \text{There are n column vectors } \Rightarrow nullity(A)=\bbox[red, 2pt]{0}$$

解答: $$\textbf{(a)}\; \det(A)= \left| \begin{matrix} 5 & 4 & 0 & 0 & 0 \\6 & 7 & 0 & 0 & 0 \\3 & 4 & 5 & 6 & 7 \\2 & 1 & 0 & 1 & 2 \\2 & 1 & 0 & 0 & 1 \end{matrix} \right| =5\begin{vmatrix}7 & 0 & 0 & 0\\4 & 5 & 6 & 7\\1 & 0 & 1 & 2\\1 & 0 & 0 & 1 \end{vmatrix}-4 \begin{vmatrix}6 & 0 & 0 & 0\\3 & 5 & 6 & 7\\2 & 0 & 1 & 2\\2 & 0 & 0 & 1 \end{vmatrix} \\\quad =5\cdot 7 \begin{vmatrix} 5 & 6 & 7\\0 & 1 & 2\\0 & 0 & 1 \end{vmatrix}-4\cdot 6 \begin{vmatrix} 5 & 6 & 7\\0 & 1 & 2\\0 & 0 & 1\end{vmatrix} =5\cdot 7\cdot 5-4\cdot 6\cdot 5=\bbox[red, 2pt]{55} \\\textbf{(b)}\; \det(B)= \begin{vmatrix} 1& b& b^2\\ b& b^2 & b^3\\ b^2& b^3& b^4\end{vmatrix} =b\begin{vmatrix} 1& 1& b^2\\ b& b  & b^3\\ b^2& b^2& b^4\end{vmatrix} =\bbox[red, 2pt] 0$$

解答: $$\textbf{(a)}\; A=\begin{bmatrix}-\sqrt 3/2 & -1/2 \\1/2 & -\sqrt 3/2 \end{bmatrix} =\begin{bmatrix}\cos(5\pi/6) & -\sin(5\pi/6) \\ \sin(5\pi/6) & \cos(5\pi/6) \end{bmatrix} \\\Rightarrow T \text{ rotates points in the xy plane counterclockwise through an angle θ about the origin } \\\textbf{(b)}\; {5\pi \over 6}\times 2011=837\times(2\pi)+{11\pi\over 6} \Rightarrow A^{2011}=\begin{bmatrix}\cos(11\pi/6) & -\sin(11\pi/6) \\ \sin(11\pi/6) & \cos(11\pi/6) \end{bmatrix} \\\qquad = \bbox[red, 2pt]{\begin{bmatrix} \sqrt 3/2 & 1/2 \\ -1/2 & \sqrt 3/2 \end{bmatrix}}$$

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