109年北科大電機碩士班-線性代數詳解

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:$$A=\left[ \begin{matrix}0 & 3 & -6 & 6 & 4 & -5\\3 & -7 & 8 & -5 & 8 & 9\\3 & -9 & 12 & -9 & 6 & 15\end{matrix} \right] \xrightarrow{R_2-R_3\to R_2} \left[ \begin{matrix}0 & 3 & -6 & 6 & 4 & -5\\0 & 2 & -4 & 4 & 2 & -6\\3 & -9 & 12 & -9 & 6 & 15\end{matrix} \right] \\ \xrightarrow{R_2/2\to R_2, R_3/3 \to R_3} \left[ \begin{matrix}0 & 3 & -6 & 6 & 4 & -5\\0 & 1 & -2 & 2 & 1 & -3\\1 & -3 & 4 & -3 & 2 & 5\end{matrix} \right] \xrightarrow{-3R_2+R_1\to R_1,3R_2+R_3\to R_3} \left[\begin{matrix}0 & 0 & 0 & 0 & 1 & 4\\0 & 1 & -2 & 2 & 1 & -3\\1 & 0 & -2 & 3 & 5 & -4\end{matrix}\right] \\ \xrightarrow{R_1 \leftrightarrow R_3} \left[\begin{matrix}1 & 0 & -2 & 3 & 5 & -4\\0 & 1 & -2 & 2 & 1 & -3\\0 & 0 & 0 & 0 & 1 & 4\end{matrix}\right] \xrightarrow{-5R_3+R_1\to R_1, -R_3+R_2 \to R_2} \left[\begin{matrix}1 & 0 & -2 & 3 & 0 & -24\\0 & 1 & -2 & 2 & 0 & -7\\0 & 0 & 0 & 0 & 1 & 4\end{matrix}\right] \\ \Rightarrow rref(A)= \bbox[red, 2pt]{\left[\begin{matrix}1 & 0 & -2 & 3 & 0 & -24\\0 & 1 & -2 & 2 & 0 & -7\\0 & 0 & 0 & 0 & 1 & 4\end{matrix}\right]}$$

解答:$$B=[A \mid b] =\left[\begin{matrix}3 & 5 & -4 & 7\\-3 & -2 & 4 & -1\\6 & 1 & -8 & -4\end{matrix}\right] \Rightarrow rref(B)= \left[\begin{matrix}1 & 0 & - \frac{4}{3} & -1\\0 & 1 & 0 & 2\\0 & 0 & 0 & 0\end{matrix}\right] \\ \Rightarrow \cases{x_1-{4\over 3}x_3=-1\\ x_2=2} \Rightarrow x=\begin{pmatrix} {4\over 3}x_3-1\\ 2\\ x_3\end{pmatrix} = \bbox[red, 2pt]{ \left\{ \begin{pmatrix} {4\over 3}k-1\\ 2\\ k\end{pmatrix} \mid k \in \mathbb R\right\}}$$

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:$$A=\begin{bmatrix}1 & 2 & 3\\0& -1 & 1\\ 1& 1& 4 \end{bmatrix} \Rightarrow C(A)=\bbox[red, 2pt]{ \left\{a\begin{pmatrix}1\\ 0\\ 1 \end{pmatrix} +b \begin{pmatrix}2\\ -1\\ 1 \end{pmatrix} +c\begin{pmatrix}3 \\ 1 \\ 4 \end{pmatrix} \mid a,b,c \in \mathbb R\right\}} \\ rref(A)=\begin{bmatrix}1 & 0 & 5\\0& 1 & -1\\ 0& 0& 0 \end{bmatrix} \Rightarrow \begin{bmatrix}1 & 0 & 5\\0& 1 & -1\\ 0& 0& 0 \end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1+5x_3=0\\ x_2=x_3} \\ \Rightarrow ker(A)= \bbox[red, 2pt]{ \left\{ k \begin{pmatrix}-5\\ 1\\ 1 \end{pmatrix} \mid k\in \mathbb R\right\}}$$

解答:$$\text{Let }\cases{\vec u=\begin{pmatrix} a\\ b\\ c\end{pmatrix}\\ \vec v=\begin{pmatrix} 1\\ -2\\ -3\end{pmatrix} \\ \vec u\bot \vec v}  \Rightarrow \vec u\cdot \vec v=0 \Rightarrow a-2b-3c=0 \Rightarrow \vec u=a\begin{pmatrix} 1\\ 0\\ 1/3\end{pmatrix}+b \begin{pmatrix} 0\\ 1\\ -2/3\end{pmatrix} \\ \Rightarrow \left\{ \begin{pmatrix} 1\\ 0\\ 1/3\end{pmatrix}, \begin{pmatrix} 0\\ 1\\ -2/3\end{pmatrix}\right\} \text{ is a basis which is orthogonal to }\vec v \\ \vec a_1= \begin{pmatrix} 1\\ 0\\ 1/3\end{pmatrix} \Rightarrow \vec e_1= {\vec a_1\over \Vert \vec a_1\Vert} ={3\over \sqrt{10}}\begin{pmatrix} 1\\ 0\\ 1/3\end{pmatrix} \\ \vec a_2= \begin{pmatrix} 0\\ 1\\ -2/3\end{pmatrix} \Rightarrow  \vec e_2= {\vec a_2-(\vec a_2\cdot \vec e_1)\vec e_1 \over \Vert \vec a_2-(\vec a_2\cdot \vec e_1)\vec e_1 \Vert} =\sqrt{5\over 7}\begin{pmatrix} 1/5\\ 1\\ -3/5\end{pmatrix} \\ \Rightarrow \bbox[red, 2pt]{ \{ ({3\over \sqrt{10}},0, {1\over \sqrt{10}}), ({1\over \sqrt{35}}, {5\over \sqrt{35}},-{3\over \sqrt{35}})\}} \text{ is a set of orthnonomal basis} \\ \text{which is orthogonal to (1,-2,-3)}$$

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