國立台北科技大學109學年度碩士班招生考試
系所組別: 2151電機工程系碩士班戊組
第一節 線性代數(選考)
解答:$$\left[\begin{array}{rrr|rrr}0 & 1 & 2 & 1 & 0 & 0\\1 & 0 & 3 & 0 & 1 & 0\\4 & -3 & 8 & 0 & 0 & 1\end{array} \right] \xrightarrow{-4R_2+R_3 \to R_3}\left[ \begin{array}{rrr|rrr} 0 & 1 & 2 & 1 & 0 & 0\\1 & 0 & 3 & 0 & 1 & 0\\0 & -3 & -4 & 0 & -4 & 1 \end{array} \right] \xrightarrow{3R_1 +R_3\to R_3} \\\left[ \begin{array}{rrr|rrr} 0 & 1 & 2 & 1 & 0 & 0\\1 & 0 & 3 & 0 & 1 & 0\\0 & 0 & 2 & 3 & -4 & 1 \end{array} \right] \xrightarrow{-R_3+R_1\to R_1,-3R_3/2+ R_2 \to R_2} \left[ \begin{array}{rrr|rrr} 0 & 1 & 0 & -2 & 4 & -1\\1 & 0 & 0 & - \frac{9}{2} & 7 & - \frac{3}{2}\\0 & 0 & 2 & 3 & -4 & 1 \end{array} \right] \\ \xrightarrow{R_1 \leftrightarrow R_2, R_3/2 \to R_3}\left[ \begin{array}{rrr|rrr} 1 & 0 & 0 & - \frac{9}{2} & 7 & - \frac{3}{2}\\0 & 1 & 0 & -2 & 4 & -1\\0 & 0 & 1 & \frac{3}{2} & -2 & \frac{1}{2} \end{array} \right] \Rightarrow A^{-1}= \bbox[red, 2pt]{ \left[\begin{matrix}- \frac{9}{2} & 7 & - \frac{3}{2}\\-2 & 4 & -1\\\frac{3}{2} & -2 & \frac{1}{2}\end{matrix} \right]}$$
解答:$$B=\begin{bmatrix}1 & 1 &1 & \cdots & 1 \\1 & 2 &2 & \cdots & 2\\ 1 & 2 & 3& \cdots & 3\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 2& 3& \cdots & n\end{bmatrix} \xrightarrow {-R_1+R_k \to R_k, k=2,\dots, n} \begin{bmatrix}1 & 1 &1 & \cdots & 1 \\0 & 1 &1 & \cdots & 1\\ 0 & 1 & 2& \cdots & 2\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 1& 2& \cdots & n-1\end{bmatrix} \\ \Rightarrow \det(B)=\begin{vmatrix} 1 &1 & \cdots & 1\\ 1 & 2& \cdots & 2\\ \vdots & \vdots & \ddots & \vdots\\ 1& 2& \cdots & n-1\end{vmatrix} \xrightarrow {-R_1+R_k \to R_k, k=2,\dots, n-1}\begin{vmatrix} 1 &1 & \cdots & 1\\ 0 & 1& \cdots & 1\\ \vdots & \vdots & \ddots & \vdots\\ 0& 1& \cdots & n-2\end{vmatrix} \\ \rightarrow \cdots \rightarrow \begin{vmatrix} 1& 1\\ 1& 2\end{vmatrix} =1 \Rightarrow \det(B)=\bbox[red, 2pt]1$$
解答:$$A=\left[\begin{matrix}\frac{3}{5} & \frac{3}{10}\\\frac{2}{5} & \frac{7}{10} \end{matrix} \right] =\left[ \begin{matrix} -1 & \frac{3}{4} \\1 & 1 \end{matrix} \right] \left[ \begin{matrix} \frac{3}{10} & 0 \\0 & 1\end{matrix} \right] \left[ \begin{matrix} \frac{-4}{7} & \frac{3}{7} \\\frac{4}{7} & \frac{4}{7} \end{matrix}\right] \\\Rightarrow A^\infty =\left[ \begin{matrix} -1 & \frac{3}{4} \\1 & 1\end{matrix} \right] \left[ \begin{matrix} (\frac{3}{10})^\infty & 0 \\0 & 1^\infty \end{matrix} \right] \left[ \begin{matrix} \frac{-4}{7} & \frac{3}{7} \\\frac{4}{7} & \frac{4}{7} \end{matrix}\right] =\left[ \begin{matrix} -1 & \frac{3}{4} \\1 & 1\end{matrix} \right] \left[ \begin{matrix} 0 & 0 \\0 & 1 \end{matrix} \right] \left[ \begin{matrix} \frac{-4}{7} & \frac{3}{7} \\ \frac{4}{7} & \frac{4}{7} \end{matrix}\right] =\left[ \begin{matrix} \frac{3}{7} & \frac{3}{7} \\\frac{4}{7} & \frac{4}{7} \end{matrix} \right] \\ x_{\infty} =A^\infty x_0 =\left[ \begin{matrix} \frac{3}{7} & \frac{3}{7} \\\frac{4}{7} & \frac{4}{7} \end{matrix} \right] \left[ \begin{matrix} \frac{1}{2} \\\frac{1}{2} \end{matrix} \right] = \bbox[red, 2pt]{\left[ \begin{matrix} \frac{3}{7} \\\frac{4}{7} \end{matrix} \right]}$$
解答:$$A=[b_1 \mid b_2\mid b_3]=\left[\begin{matrix}1 & 2 & 1\\0 & 1 & -1\\3 & 8 & 2\end{matrix}\right] \Rightarrow rref(A)= \left[ \begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix} \right] \\ \Rightarrow \text{rank}(A)=3 \Rightarrow \mathbf B\text{ is a basis of }\mathbb R^3 \\ ab_1+bb_2+ cb_3 = \mathbf x \Rightarrow \left[\begin{matrix}1 & 2 & 1\\0 & 1 & -1\\3 & 8 & 2\end{matrix} \right] \left[\begin{matrix}a\\b\\c\end{matrix} \right] = \left[\begin{matrix}3\\-5\\ 4\end{matrix} \right] \Rightarrow \left[\begin{matrix}a\\b\\c\end{matrix} \right] =A^{-1}\mathbf x \\ =\left[\begin{matrix}10 & 4 & -3\\-3 & -1 & 1\\-3 & -2 & 1\end{matrix} \right] \left[ \begin{matrix}3\\-5\\ 4\end{matrix} \right] = \left[ \begin{matrix}-2\\ 0\\ 5\end{matrix} \right] \Rightarrow [x]_B= \bbox[red, 2pt]{\left[ \begin{matrix}-2\\ 0\\ 5\end{matrix} \right]}$$
解答:$$A=\begin{bmatrix}1 & 2 & 3\\0& -1 & 1\\ 1& 1& 4 \end{bmatrix} \Rightarrow C(A)=\bbox[red, 2pt]{ \left\{a\begin{pmatrix}1\\ 0\\ 1 \end{pmatrix} +b \begin{pmatrix}2\\ -1\\ 1 \end{pmatrix} +c\begin{pmatrix}3 \\ 1 \\ 4 \end{pmatrix} \mid a,b,c \in \mathbb R\right\}} \\ rref(A)=\begin{bmatrix}1 & 0 & 5\\0& 1 & -1\\ 0& 0& 0 \end{bmatrix} \Rightarrow \begin{bmatrix}1 & 0 & 5\\0& 1 & -1\\ 0& 0& 0 \end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1+5x_3=0\\ x_2=x_3} \\ \Rightarrow ker(A)= \bbox[red, 2pt]{ \left\{ k \begin{pmatrix}-5\\ 1\\ 1 \end{pmatrix} \mid k\in \mathbb R\right\}}$$
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