110年北科大電子工程碩士班-工程數學

國立臺北科技大學110學年度碩士班招生考試

系所組別:電子工程碩士班乙組
科目:工程數學 

解答: $$\textbf{(一)}\; \bbox[red, 2pt]{可能}, 關係為\bbox[red, 2pt]{\vec u_2= c\vec u_1}, c \in \mathbb R , 其中\vec v_1={\vec u_1 \over \Vert \vec u_1 \Vert } \\\textbf{(二)}\; \bbox[red, 2pt]{可能}, 關係可能為\bbox[red, 2pt]{\vec u_1 \bot \vec u_2}, 其中\vec v_1={\vec u_1 \over \Vert \vec u_1 \Vert }, \vec v_2={\vec u_2 \over \Vert \vec u_2\Vert } \\\textbf{(三)}\;\bbox[red, 2pt]{不可能},關係為\bbox[red, 2pt]{\vec u_2\ne  c\vec u_1}, c \in \mathbb R ,可取\vec v_1={\vec u_1 \over \Vert \vec u_1 \Vert }, \vec v_2 ={\vec u_2-(\vec u_2\cdot \vec v_1)\vec v_1 \over \Vert \vec u_2-(\vec u_2\cdot \vec v_1)\vec v_1 \Vert}$$

解答: $$滿足A\mathbf  v=\lambda \mathbf v的非零向量\mathbf v為特徵向量,而純量\lambda為特徵值$$

解答: $$\textbf{(一)}\; \text{vector component of }\vec u \text{ along } \vec v =\bbox[red, 2pt]{{\vec u\cdot \vec v\over |\vec v|^2}\vec v}\\\qquad \text{vector component of }\vec u \text{ orthogonal to }\vec v=  \bbox[red, 2pt]{\vec u-{\vec u\cdot \vec v\over |\vec v|^2}\vec v} \\ \textbf{(二)}\; \left({\vec u\cdot \vec v\over |\vec v|^2}\vec v \right) \cdot \left(\vec u-{\vec u\cdot \vec v\over |\vec v|^2}\vec v \right) ={(\vec u\cdot \vec v)^2\over |\vec v|^2}-{(\vec u\cdot \vec v)^2 (|\vec v|)^2\over |\vec v|^4} ={(\vec u\cdot \vec v)^2\over |\vec v|^2}-{(\vec u\cdot \vec v)^2\over |\vec v|^2}=0\\\qquad \Rightarrow \left({\vec u\cdot \vec v\over |\vec v|^2}\vec v \right) \bot \left(\vec u-{\vec u\cdot \vec v\over |\vec v|^2}\vec v \right) ,\bbox[red, 2pt]{Q.E.D.}$$

解答: $$A=\left[\begin{matrix}-2 & 5 & 7\\0 & 1 & -3\\-4 & 11 & 11\end{matrix}\right] \xrightarrow{-2R_1+R_3\to R_3} \left[\begin{matrix}-2 & 5 & 7\\0 & 1 & -3\\0 & 1 & -3\end{matrix}\right] \xrightarrow{-R_2+R_3\to R_3, -5R_2+R_1\to R_1} \left[\begin{matrix}-2 & 0 & 22\\0 & 1 & -3\\0 & 0 & 0\end{matrix} \right]  \\\xrightarrow{-R_1/2 \to R_1} \left[ \begin{matrix}1 & 0 & -11\\0 & 1 & -3\\0 & 0 & 0\end{matrix}\right] \Rightarrow \cases{\textbf{(一)}\;列空間維度=\bbox[red, 2pt] 2\\ \textbf{(二)}\; 列空間基底\bbox[red, 2pt]{\{(1,0,-11), (0,1,-3)\}}} \\ 又\left[ \begin{matrix}1 & 0 & -11\\0 & 1 & -3\\0 & 0 & 0\end{matrix} \right] \begin{bmatrix}x_1 \\x_2 \\ x_3\end{bmatrix} =0 \Rightarrow \cases{x_1=11x_3\\ x_2=3x_3} \Rightarrow \textbf{(三)}\;\vec x = \bbox[red,2pt]{\left\{ \begin{pmatrix}11k\\ 3k\\ k \end{pmatrix} \middle |k \in \mathbb R \right\}}$$

解答: $$\textbf{(一)}\; \vec u\times \vec v = \begin{vmatrix}\vec i & \vec j &\vec k\\ 1& 2& -2 \\3 & 0 & 1 \end{vmatrix} =2\vec i-6\vec j-6\vec k-\vec j =\bbox[red, 2pt]{(2,-7,-6)} \\\textbf{(二)}\; \cases{(\vec u\times \vec v)\cdot \vec u=(2,-7,-6)\cdot (1,2,-2)=2-14+12=0\\ (\vec u\times \vec v)\cdot \vec v =(2,-7,-6) \cdot (3,0,1)=6-6=0} \Rightarrow \cases{(\vec u\times \vec v)\bot \vec u \\(\vec u\times \vec v)\bot \vec v}, \bbox[red, 2pt]{Q.E.D.}$$

解答: $$\textbf{(一)}\; A,B互斥 \Rightarrow A\cap B=\varnothing \Rightarrow P(A\cup B)=P(A)+P(B)-P(A \cap B) =0.3+0.2-0= \bbox[red, 2pt]{0.5} \\\textbf{(二)}\; A,B 獨立 \Rightarrow P(A\cap B)=P(A)P(B) \Rightarrow P(A\cup B)=P(A)+P(B)-P(A\cap B) \\\qquad = 0.3+0.2-0.3\cdot 0.2=0.5-0.06= \bbox[red, 2pt]{0.44}$$

解答: $$\textbf{(一)}\;  {4\over 5+4+7}=\bbox[red, 2pt]{0.25} \\\textbf{(二)}\;\cases{前二球皆為白球機率={4\over 16}\cdot {3\over 15}={1\over 20} \\ 第1球不是白球,第2球是白球的機率={12\over 16}\cdot {4\over 15}={1\over 5}}\\\qquad \Rightarrow 第二球為白球的機率={1\over 20}+{1\over 5}= \bbox[red, 2pt]{0.25} \\ \textbf{(三)}\; {1/20\over 1/4}= {1\over 5}=\bbox[red, 2pt]{0.2}$$

解答: $$\textbf{(一)}\;\cases{P_X(X=-1)=1/6+1/6=1/3\\ P_X(X=0)=1/3\\ P_X(X=1)=1/6+1/6=1/3} \Rightarrow \bbox[red, 2pt]{ \cases{P_X(X=-1)=1/3\\ P_X(X=0)=1/3 \\ P_X(X=1)=1/3}} \\\textbf{(二)}\; P(X=0\mid Y=0) ={P(X=0\cap Y=0)\over P(Y=0)} ={1/3 \over 1/3} =\bbox[red, 2pt] 1$$

解答: $$\textbf{(一)}\; \cases{E(X_i)=1\\ Var(X_i)=2} ,i=1-100 \Rightarrow \cases{E(\bar X)=E((X_1 +X_2+ \cdots+ X_{100})/100) \\ Var(\bar X)=Var((X_1 +X_2+ \cdots+ X_{100})/100) } \\ \qquad \Rightarrow \cases{E(\bar X)=(E(X_1)+E(X_2)+ \cdots +E(X_{100}))/100= 1\\ Var(\bar X)=(Var(X_1)+ Var(X_2)+ \cdots +Var(X_{100}))/ 10000 =0.02} \\\qquad \Rightarrow \bbox[red, 2pt]{ \cases{E(\bar X)=1\\ Var(\bar X)=0.02}}  \\\textbf{(二)}\; \bbox[red, 2pt]{\bar X \sim N(1,0.02)}$$

解答: $$\textbf{(一)}\; \int f(x)\,dx =1 \Rightarrow \int_2^5 c(1+x)\,dx =1 \Rightarrow {27\over 2}c=1 \Rightarrow c=\bbox[red, 2pt]{2\over 27} \\\textbf{(二)}\; P(X\lt 4)= \int_2^4 c(1+x)\,dx =8c =8\cdot {2\over 27} =\bbox[red, 2pt]{16\over 27}$$

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