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:$$\textbf{(a)}\; \det(A)= \left| \begin{matrix} 5 & 4 & 0 & 0 & 0 \\6 & 7 & 0 & 0 & 0 \\3 & 4 & 5 & 6 & 7 \\2 & 1 & 0 & 1 & 2 \\2 & 1 & 0 & 0 & 1 \end{matrix} \right| =5\begin{vmatrix}7 & 0 & 0 & 0\\4 & 5 & 6 & 7\\1 & 0 & 1 & 2\\1 & 0 & 0 & 1 \end{vmatrix}-4 \begin{vmatrix}6 & 0 & 0 & 0\\3 & 5 & 6 & 7\\2 & 0 & 1 & 2\\2 & 0 & 0 & 1 \end{vmatrix} \\\quad =5\cdot 7 \begin{vmatrix} 5 & 6 & 7\\0 & 1 & 2\\0 & 0 & 1 \end{vmatrix}-4\cdot 6 \begin{vmatrix} 5 & 6 & 7\\0 & 1 & 2\\0 & 0 & 1\end{vmatrix} =5\cdot 7\cdot 5-4\cdot 6\cdot 5=\bbox[red, 2pt]{55} \\\textbf{(b)}\; \det(B)= \begin{vmatrix} 1& b& b^2\\ b& b^2 & b^3\\ b^2& b^3& b^4\end{vmatrix} =b\begin{vmatrix} 1& 1& b^2\\ b& b & b^3\\ b^2& b^2& b^4\end{vmatrix} =\bbox[red, 2pt] 0$$
解答:$$\textbf{(a)}\; A=\begin{bmatrix}-\sqrt 3/2 & -1/2 \\1/2 & -\sqrt 3/2 \end{bmatrix} =\begin{bmatrix}\cos(5\pi/6) & -\sin(5\pi/6) \\ \sin(5\pi/6) & \cos(5\pi/6) \end{bmatrix} \\\Rightarrow T \text{ rotates points in the xy plane counterclockwise through an angle θ about the origin } \\\textbf{(b)}\; {5\pi \over 6}\times 2011=837\times(2\pi)+{11\pi\over 6} \Rightarrow A^{2011}=\begin{bmatrix}\cos(11\pi/6) & -\sin(11\pi/6) \\ \sin(11\pi/6) & \cos(11\pi/6) \end{bmatrix} \\\qquad = \bbox[red, 2pt]{\begin{bmatrix} \sqrt 3/2 & 1/2 \\ -1/2 & \sqrt 3/2 \end{bmatrix}}$$
第三題的(b):A 它的每一行向量都是線性獨立的那rank(A)=n,而nullity(A)=0吧?
回覆刪除寫太快了, 應該是0, 謝!
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