114年台北科大電機碩士班-線性代數詳解

國立臺北科技大學114學年度碩士班招生考試

系所組別:2140 電機工程系碩士班丁組
第一節 線性代數 試題(選考)

解答: $$\textbf{a. }\cases{x+y+2z+w =3\\ -x+z+2w=1\\ 2x+2y+w=-2\\ x+y+2z+3w=5} \Rightarrow \begin{bmatrix}1& 1& 2& 1\\ -1&0 & 1& 2\\ 2& 2& 0& 1\\ 1& 1& 2& 3 \end{bmatrix} \begin{bmatrix} x\\ y\\ z\\ w\end{bmatrix} =\begin{bmatrix} 3\\ 1\\ -2\\ 5\end{bmatrix} \Rightarrow A= \bbox[red, 2pt]{\begin{bmatrix}1& 1& 2& 1\\ -1&0 & 1& 2\\ 2& 2& 0& 1\\ 1& 1& 2& 3 \end{bmatrix}} \\\textbf{b. }\det(A) = \begin{vmatrix}1& 1& 2& 1\\ -1&0 & 1& 2\\ 2& 2& 0& 1\\ 1& 1& 2& 3 \end{vmatrix} \xrightarrow{R_1+R_2 \to R_2,R_3-2R_1\to R_3, R_4-R_1\to R_4} \begin{vmatrix}1& 1& 2& 1\\ 0&1 & 3& 3\\ 0& 0& -4& -1\\ 0& 0& 0& 2 \end{vmatrix} \\\quad = \begin{vmatrix}   1 & 3& 3\\   0& -4& -1\\   0& 0& 2 \end{vmatrix} =\bbox[red, 2pt]{-8} \\\textbf{c. }\det(A)=-8 \ne 0 \Rightarrow \bbox[red, 2pt] {\text{Yes, it is consistent.}} \\\textbf{d. }\det(A)=-8 \ne 0 \Rightarrow rank(A)=4 \Rightarrow \bbox[red, 2pt]{\text{Yes, they are linearly independent.}}\\ \textbf{e. }rref(A) =  \left[\begin{matrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{matrix}\right] \Rightarrow A\mathbf x=0 \Rightarrow \mathbf x= \bbox[red, 2pt]{\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix} }\\\textbf{f. }[A \mid I] = \left[ \begin{array}{rrrr|rrrr}1 & 1 & 2 & 1 & 1 & 0 & 0 & 0\\-1 & 0 & 1 & 2 & 0 & 1 & 0 & 0\\2 & 2 & 0 & 1 & 0 & 0 & 1 & 0\\1 & 1 & 2 & 3 & 0 & 0 & 0 & 1 \end{array}\right] \xrightarrow{R_1+R_2 \to R_2,R_3-2R_1\to R_3, R_4-R_1\to R_4} \\  \left[ \begin{array}{rrrr|rrrr}1 & 1 & 2 & 1 & 1 & 0 & 0 & 0\\0 & 1 & 3 & 3 & 1 & 1 & 0 & 0\\0 & 0 & -4 & -1 & -2 & 0 & 1 & 0\\0 & 0 & 0 & 2 & -1 & 0 & 0 & 1 \end{array}\right] \xrightarrow{R_1-R_2 \to R_1, R_3/(-4) \to R_3, R_4/2 \to R_4} \\\left[ \begin{array}{rrrr|rrrr}1 & 0 & -1 & -2 & 0 & -1 & 0 & 0\\0 & 1 & 3 & 3 & 1 & 1 & 0 & 0\\0 & 0 & 1 & \frac{1}{4} & \frac{1}{2} & 0 & - \frac{1}{4} & 0\\0 & 0 & 0 & 1 & - \frac{1}{2} & 0 & 0 & \frac{1}{2} \end{array}\right] \xrightarrow{R_1+R_3\to R_1, R_2-3R_3 \to R_2} \\ \left[ \begin{array}{rrrr|rrrr}1 & 0 & 0 & - \frac{7}{4} & \frac{1}{2} & -1 & - \frac{1}{4} & 0\\0 & 1 & 0 & \frac{9}{4} & - \frac{1}{2} & 1 & \frac{3}{4} & 0\\0 & 0 & 1 & \frac{1}{4} & \frac{1}{2} & 0 & - \frac{1}{4} & 0\\0 & 0 & 0 & 1 & - \frac{1}{2} & 0 & 0 & \frac{1}{2} \end{array}\right] \xrightarrow{R_1+(7/4)R_4 \to R_1, R_2-(9/4)R_4\to R_2, R_3-(1/3)R_4 \to R_3}  \\ \left[ \begin{array}{rrrr|rrrr}1 & 0 & 0 & 0 & - \frac{3}{8} & -1 & - \frac{1}{4} & \frac{7}{8}\\0 & 1 & 0 & 0 & \frac{5}{8} & 1 & \frac{3}{4} & - \frac{9}{8}\\0 & 0 & 1 & 0 & \frac{5}{8} & 0 & - \frac{1}{4} & - \frac{1}{8}\\0 & 0 & 0 & 1 & - \frac{1}{2} & 0 & 0 & \frac{1}{2} \end{array} \right] \Rightarrow \bbox[red, 2pt]{A^{-1}= \begin{bmatrix}\frac{-3}{8} & -1 & \frac{-1}{4} & \frac{7}{8} \\\frac{5}{8} & 1 & \frac{3}{4} & \frac{-9}{8} \\\frac{5}{8} & 0 & \frac{-1}{4} & \frac{-1}{8} \\\frac{-1}{2} & 0 & 0 & \frac{1}{2} \end{bmatrix} }\\ \textbf{g. }  \mathbf x= A^{-1}\begin{bmatrix} 3\\ 1\\ -2\\ 5\end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix} \frac{11}{4} \\\frac{-17}{4} \\\frac{7}{4} \\1\end{bmatrix}}$$

解答: $$\textbf{a. }\bbox[red, 2pt]{ \text{false}}:  a\begin{bmatrix} 1& 0\\0&  0\end{bmatrix}+ b\begin{bmatrix} 0& 1\\0&  0\end{bmatrix} +c\begin{bmatrix} 0& 0\\1&  0\end{bmatrix} =\begin{bmatrix} a& b\\c&  0\end{bmatrix} \ne \begin{bmatrix} 0& 0\\0&  2\end{bmatrix}, \forall a,b,c \in \mathbb R \\ \textbf{b. } \bbox[red, 2pt]{\text{true}}:A\mathbf x=\mathbf b \equiv \begin{bmatrix} 1& 2& 4\\0&  1& 3 \\ 1& 0& -1\end{bmatrix} \begin{bmatrix} x_1\\x_2\\ x_3 \end{bmatrix} =\begin{bmatrix} b_1\\b_2\\ b_3\end{bmatrix} \Rightarrow A= \begin{bmatrix} 1& 2& 4\\0&  1& 3 \\ 1& 0& -1\end{bmatrix} \Rightarrow \det(A)=1 \Rightarrow A^{-1} \text{ exists. } \\\quad \Rightarrow \mathbf x=A^{-1}\mathbf b \text{ is a unique solution.}\\ \textbf{c. }\bbox[red, 2pt]{\text{false}}: aM_1+b M_2+ cM_3 =\begin{bmatrix} a& -a\\ b & c\end{bmatrix} =M_4= \begin{bmatrix} 2& -1\\ 1& 3\end{bmatrix} \Rightarrow \cases{a=2\\a=1\\ b=1\\c=3} \\\quad \Rightarrow M_4\text{ cannot be a combination of }M_1,M_2, \text{ and }M_3 \Rightarrow span(S) \ne span(T)$$

解答: $$\textbf{a. }B=\{1,x,x^2\} =\{\begin{bmatrix} 1\\0\\0\end{bmatrix}, \begin{bmatrix}0\\ 1\\ 0 \end{bmatrix}, \begin{bmatrix}0\\ 0\\ 1 \end{bmatrix}\}, B'=\{1,x+1,x^2+x+1\} =\{ \begin{bmatrix} 1\\0\\0\end{bmatrix}, \begin{bmatrix}1\\ 1\\ 0 \end{bmatrix}, \begin{bmatrix}1 \\ 1\\ 1 \end{bmatrix}\} \\\quad \Rightarrow \begin{bmatrix} 1\\0\\0\end{bmatrix} =1\cdot\begin{bmatrix} 1\\0\\0\end{bmatrix},\begin{bmatrix}0\\ 1\\ 0 \end{bmatrix}= -1\cdot \begin{bmatrix}1\\ 0\\ 0 \end{bmatrix} +1\cdot \begin{bmatrix}1\\ 1\\ 0 \end{bmatrix}, \begin{bmatrix}0\\ 0\\ 1 \end{bmatrix} =1 \cdot \begin{bmatrix}1\\ 1\\ 1 \end{bmatrix} -1\cdot \begin{bmatrix}1\\ 1\\ 0 \end{bmatrix} \\ \Rightarrow [I]_B^{B'} = \bbox[red, 2pt]{\begin{bmatrix} 1& -1& 0\\ 0& 1& -1\\ 0& 0& 1\end{bmatrix}} \\\textbf{b. }p(x)=3-x+2x^2 =\begin{bmatrix}3\\ -1\\ 2 \end{bmatrix}_B \Rightarrow [p(x)]_{B'} =\begin{bmatrix} 1& -1& 0\\ 0& 1& -1\\ 0& 0& 1\end{bmatrix} \begin{bmatrix}3\\ -1\\ 2 \end{bmatrix}= \bbox[red, 2pt]{\begin{bmatrix}4\\ -3\\ 2 \end{bmatrix}}\\ \textbf{c. } \left[ \begin{array}{rrr|rrr} 1& -1& 0 & 1& 0 & 0\\ 0& 1& -1 & 0& 1& 0\\ 0& 0& 1 &0& 0& 1\end{array} \right] \xrightarrow{R_1+R_2 \to R_1} \left[ \begin{array}{rrr|rrr} 1& 0& -1 & 1& 1 & 0\\ 0& 1& -1 & 0& 1& 0\\ 0& 0& 1 &0& 0& 1\end{array} \right] \xrightarrow{R_1+R_3 \to R_1, R_2+R_3\to R_2}\\ \left[ \begin{array}{rrr|rrr} 1& 0& 0 & 1& 1 & 1\\ 0& 1& 0 & 0& 1& 1\\ 0& 0& 1 &0& 0& 1\end{array} \right] \Rightarrow \text{ inverse of }[I]_B^{B'} = \bbox[red, 2pt]{\left[\begin{array}{rrr|rrr}   1& 1 & 1\\   0& 1& 1\\  0& 0& 1\end{array} \right] }$$

解答: $$\textbf{a. }  2x^2-4x+6=ax^2+(a-2b)x+b \Rightarrow \cases{a=2\\ a-2b=-4\\ b=6} \Rightarrow \text{ No solution }(a,b,c) \text{ exists.} \\\quad \Rightarrow  \bbox[red, 2pt]{p(x) \not \in R(T)} \\\textbf{b. }ax^2+(a-2b)x+b =\begin{bmatrix} b\\ a-2b\\ a\end{bmatrix} =a\begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}+ b\begin{bmatrix} 1\\ -2\\ 0\end{bmatrix} \Rightarrow \text{a basis of }R(T)= \bbox[red, 2pt]{\left\{ \begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ -2\\ 0\end{bmatrix}\right\}}$$

解答: $$\textbf{a. }  M=\begin{bmatrix} 0.75& 0.25& 0.2\\ 0.15& 0.45& 0.4\\ 0.1& 0.3& 0.4 \end{bmatrix} \Rightarrow M^2= \begin{bmatrix} \frac{31}{50} & \frac{9}{25} & \frac{33}{100} \\\frac{11}{50} & \frac{9}{25} & \frac{37}{100} \\\frac{4}{25} & \frac{7}{25} & \frac{3}{10} \end{bmatrix} \Rightarrow M^2 \begin{bmatrix}0.25\\ 0.3\\ 0.45 \end{bmatrix} = \begin{bmatrix} \frac{823}{2000}= 0.4115\\ \frac{659}{2000} =0.3295 \\\frac{259}{1000} =0.1295\end{bmatrix} \\ \quad \Rightarrow \bbox[red, 2pt]{\cases{\text{plan }A: 41.15\%\\ \text{plan }B: 32.95\%\\ \text{plan }C: 12.95\%}} \\\textbf{b. }M= \begin{bmatrix} \frac{21}{10} & \frac{\sqrt{5}-2}{2} & \frac{-\sqrt{5}-2}{2} \\\frac{13}{10} & \frac{-\sqrt{5}}{2} & \frac{\sqrt{5}}{2} \\1 & 1 & 1\end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\0 & \frac{-\sqrt{5}+3}{10} & 0 \\0 & 0 & \frac{\sqrt{5}+3}{10} \end{bmatrix} \begin{bmatrix} \frac{5}{22} & \frac{5}{22} & \frac{5}{22} \\\frac{13 \sqrt{5}-25}{220} & \frac{-31\sqrt{5}-25}{220} & \frac{13\sqrt{5}+85}{220} \\\frac{-13 \sqrt{5}-25}{220} & \frac{31\sqrt{5}-25}{220} & \frac{-13\sqrt{5}+85}{220} \end{bmatrix} \\ \Rightarrow M^\infty = \begin{bmatrix} \frac{21}{10} & \frac{\sqrt{5}-2}{2} & \frac{-\sqrt{5}-2}{2} \\\frac{13}{10} & \frac{-\sqrt{5}}{2} & \frac{\sqrt{5}}{2} \\1 & 1 & 1\end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\0 & 0 & 0 \\0 & 0 & 0\end{bmatrix} \begin{bmatrix} \frac{5}{22} & \frac{5}{22} & \frac{5}{22} \\\frac{13 \sqrt{5}- 25}{220} & \frac{-31\sqrt{5}-25}{220} & \frac{13\sqrt{5}+85}{220} \\\frac{-13 \sqrt{5}- 25}{220} & \frac{31\sqrt{5}-25}{220} & \frac{-13\sqrt{5}+85}{220}\end{bmatrix} \\\qquad = \begin{bmatrix} \frac{21}{44} & \frac{21}{44} & \frac{21}{44} \\\frac{13}{44} & \frac{13}{44} & \frac{13}{44} \\\frac{5}{22} & \frac{5}{22} & \frac{5}{22}\end{bmatrix} \Rightarrow M^\infty \begin{bmatrix}0.25\\ 0.3\\ 0.45 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}\frac{21}{44} \\\frac{13}{44} \\\frac{5}{22} \end{bmatrix} }$$

解答: $$A= \begin{bmatrix} 1& 1\\ 3& -3\end{bmatrix} \Rightarrow W=AA^T = \begin{bmatrix} 2& 0\\ 0& 18\end{bmatrix} \Rightarrow \text{ eigenvalues: }\lambda_1=2, \lambda_2=18 \\ (W-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} 0& 0\\ 0& 16\end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix}=0 \Rightarrow x_2=0 \Rightarrow v=x_1 \begin{bmatrix} 1\\ 0\end{bmatrix}, \text{ choose eigenvector }v_1 = \begin{bmatrix}1\\ 0 \end{bmatrix} \\ (W-\lambda_2)v=0 \Rightarrow \begin{bmatrix} -16 & 0 \\0 & 0\end{bmatrix} \begin{bmatrix} x_1\\ x_2\end{bmatrix}=0 \Rightarrow x_1=0 \Rightarrow v=x_2\begin{bmatrix} 0\\ 1\end{bmatrix}, \text{ choose eigenvector }v_2=\begin{bmatrix}0\\ 1 \end{bmatrix}\\ \cases{\sigma_1= \sqrt{\lambda_1} =\sqrt 2\\ \sigma_2 = \sqrt{\lambda_2} = 3\sqrt 2} \Rightarrow \sum= \begin{bmatrix} \sigma_1 & 0\\ 0& \sigma_2 \end{bmatrix} =\begin{bmatrix} \sqrt 2 & 0\\ 0& 3\sqrt 2\end{bmatrix}  \Rightarrow U= \begin{bmatrix} v_1 v_2\end{bmatrix} =\begin{bmatrix}1& 0\\ 0& 1 \end{bmatrix} \\ \Rightarrow \cases{u_1={1\over \sigma_1} A^T v_1=\begin{bmatrix} \sqrt 2/2\\ \sqrt 2/2 \end{bmatrix} \\ u_2 ={1\over \sigma_2} A^Tv_2 =\begin{bmatrix}\sqrt 2/2\\ -\sqrt 2/2 \end{bmatrix}} \Rightarrow V= [u_1 u_2] =\begin{bmatrix} {\sqrt 2\over 2} &{\sqrt 2\over 2} \\ {\sqrt 2\over 2}& -{\sqrt 2\over 2}\end{bmatrix} \\ \bbox[red, 2pt]{A=U \Sigma V^T =\begin{bmatrix}1& 0\\0& 1 \end{bmatrix} \begin{bmatrix}\sqrt 2& 0\\ 0& 3\sqrt 2 \end{bmatrix} \begin{bmatrix} {\sqrt 2\over 2} &{\sqrt 2\over 2} \\ {\sqrt 2\over 2}& -{\sqrt 2\over 2}\end{bmatrix}}$$

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解題僅供參考，其他碩士班試題及詳解