111年北科大電子工程碩士班-工程數學詳解

 國立臺北科技大學 l11學 年度碩 士班招生考試

系所組別 :2220 電子工程系碩士班乙組
第一節 工程數學 試題

 解答：$$\mathbf{(a)}\; \mu=E(X)=2\cdot {1\over 3} +3\cdot {1\over 2}+11\cdot {1\over 6}={1\over 6}(4+9+11)= \bbox[red, 2pt]4\\ E(X^2) =4\cdot {1\over 3}+9\cdot {1\over 2}+121\cdot {1\over 6}={1\over 6}(8+27 +121)=26 \\ \sigma^2=var(X)= E(X^2)-(E(X))^2 =26-16=\bbox[red, 2pt]{10} \\ \textbf{(b)}\; \mu=E(X)= 1\cdot 0.4+ 3\cdot 0.1+ 4\cdot 0.2+ 5\cdot 0.3=\bbox[red, 2pt]3 \\ E(X^2)=1\cdot 0.4+ 9\cdot 0.1+16\cdot 0.2 + 25\cdot 0.3=12\\ \Rightarrow \sigma^2 = var(X)= E(X^2)-(E(X))^2=12-3^2=\bbox[red, 2pt]3$$

解答：$$E(X)=\int xf(x)\,dx = \int_0^2 x\cdot {x\over 2}\,dx =\bbox[red, 2pt] {4\over 3} \\ E(X^2)= \int x^2f(x)\,dx =\int_0^2 x^2\cdot {x\over 2}\,dx =2\\ \Rightarrow var(X)=E(X^2)-(E(X))^2=2-{16\over 9}= \bbox[red, 2pt]{2\over 9}$$

解答：$$f_{XY}(x,y) \Rightarrow \cases{f_X(x)= \int_{-\infty}^\infty f_{XY}(x,y)\,dy \\ f_Y(y)= \int_{-\infty}^\infty f_{XY}(x,y)\,dx } \Rightarrow F_Z(z)= P(Z\le z) =P(XY\le z) \\ =P(XY \le z,X\ge 0)+P(XY\le z, X\le 0) =P(Y\le z/X,X\ge 0)+ P(Y\ge z, X\le 0) \\ =\int_0^\infty f_X(x)\int_{-\infty}^{z/x} f_Y(y) \,dydx+ \int_{-\infty}^0 f_X(x)\int_{-\infty}^{z/x} f_Y(y) \,dydx \\ \Rightarrow pdf\text{ of Z} =f_Z(z)= {d\over dz}F_Z(z) = \int_0^\infty f_X(x)f_Y(z/x){1\over |x|}\,dx + \int_{-\infty}^0 f_X(x)f_Y(z/x){1\over |x|}\,dx \\ = \bbox[red, 2pt]{\int_{-\infty}^\infty f_X(x)f_Y(z/x){1\over |x|}\,dx, \text{ where }\cases{f_X(x)= \int_{-\infty}^\infty f_{XY}(x,y)\,dy \\ f_Y(y)= \int_{-\infty}^\infty f_{XY}(x,y)\,dx }}$$

解答：$$\textbf{(a)}\;\cases{f_X(X=1)=0.1+0.2+0.2= 0.5\\ f_X(X=3)=0.3+0.1+0.1=0.5} \Rightarrow  \bbox[red, 2pt]{\cases{f_X(X=1)=0.5\\ f_X(X=3)= 0.5}} \\ \quad \cases{f_Y(Y=-3)=0.1+ 0.3=0.4 \\ f_Y(Y=2)=0.2+0.1= 0.3\\ f_Y(Y=4)= 0.2+0.1=0.3} \Rightarrow \bbox[red, 2pt]{\cases{f_Y(Y=-3)=0.4\\ f_Y(Y=2)= 0.3\\ f_Y(Y=4)= 0.3}} \\\textbf{(b)}\; \cases{\mu_x=E(X)=1\cdot 0.5+3\cdot 0.5=2 \\ \mu_y=E(Y)=(-3)\cdot 0.4+2\cdot 0.3+ 4\cdot 0.3=0.6}, \\\qquad 又E(XY)=1\cdot(-3) \cdot 0.1+ 1\cdot 2\cdot 0.2+1\cdot 4\cdot 0.2+ 3\cdot (-3)\cdot 0.3+ 3\cdot 2\cdot 0.1+ 3\cdot 4\cdot 0.1=0 \\ \qquad \Rightarrow Cov(X,Y=E(XY)-\mu_x\mu_y =0-2\cdot 0.6= \bbox[red, 2pt]{-1.2} \\\textbf{(c)}\;\cases{E(X^2)= 1\cdot 0.5+9\cdot 0.5 = 5\\ E(Y^2)= 9\cdot 0.4+ 4\cdot 0.3+ 16\cdot 0.3= 9.6} \Rightarrow \cases{Var(X)=E(X^2)-\mu_x^2= 5-2^2=1 \\ Var(Y)=E(Y^2)-\mu_y^2=9.6-0.6^2= 9.24} \\ \qquad \Rightarrow \rho ={Cov(X,Y)\over \sigma_x \cdot \sigma_y} ={-1.2\over \sqrt{1} \cdot \sqrt{9.24}}= \bbox[red, 2pt]{-0.395} \\\textbf{(d)}\; Cov(X,Y) \ne 0 \Rightarrow \bbox[red,2pt]{\text{NOT independent}}$$

解答：$$假設\text{moment generating function of }X:M(t)\\ \Rightarrow M^{[k]}(0) = E(X^k) =0.8 \Rightarrow M(t)=\bbox[red, 2pt]{0.8e^{t}}$$

解答：$$64+320 +72-128-120-96 =456-344=\bbox[red, 2pt]{112}$$

解答：$$\textbf{(a)}\; A=\begin{bmatrix} 20& 12\\ 16 & 8\end{bmatrix} \Rightarrow \det(A)=-32 \Rightarrow A^{-1}={1\over -32}\begin{bmatrix} 8& -12\\ -16 & 20\end{bmatrix} =\bbox[red, 2pt]{\begin{bmatrix} -1/4& 3/8\\ 1/2 & -5/8\end{bmatrix}}$$

解答：$$\textbf{(a)}\; A=\begin{bmatrix}6 &6 \\3 & 9\end{bmatrix} \Rightarrow \det(A-\lambda I)=(\lambda -3)(\lambda-12)=0 \Rightarrow \lambda=3,12\\ \lambda_1=3 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}3 &6 \\3 & 6\end{bmatrix} \begin{bmatrix} x_1 \\x_2\end{bmatrix}=0 \Rightarrow x_1+2x_2=0 \\\qquad\Rightarrow v=x_2 \begin{bmatrix} -2 \\1\end{bmatrix}, 取v_1=\begin{bmatrix}-2 \\1\end{bmatrix} \\ \lambda_2=12 \Rightarrow \lambda=3,12\\ \lambda_2=12 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}-6 &6 \\3 & -3\end{bmatrix} \begin{bmatrix}x_1 \\x_2\end{bmatrix}=0 \Rightarrow x_1=x_2 \\\qquad\Rightarrow v=x_2 \begin{bmatrix}1 \\1\end{bmatrix}, 取v_2=\begin{bmatrix} 1 \\1\end{bmatrix} \\ \Rightarrow \text{eigenvalues: }\bbox[red, 2pt]{3,12}\text{ and eigenvectors: } \bbox[red, 2pt]{ \begin{bmatrix}-2 \\1\end{bmatrix}, \begin{bmatrix} 1 \\1\end{bmatrix}} \\ \textbf{(b)}\; A=\begin{bmatrix}0 &0&-4 \\2&4 & 2 \\2& 0 & 6\end{bmatrix} \Rightarrow \det(A-\lambda I)=-(\lambda -2)(\lambda-4)^2=0 \Rightarrow \lambda=2, 4\\ \lambda_1=2 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}-2 &0&-4 \\2&2 & 2 \\2& 0 & 4\end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1+2x_3=0 \\ x_2=x_3} \\\qquad\Rightarrow v= x_3 \begin{bmatrix} -2 \\1 \\1 \end{bmatrix}, 取v_1= \begin{bmatrix} -2 \\1 \\1 \end{bmatrix} \\ \lambda_2=4 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix} -4 &0&-4 \\2&0 & 2 \\2& 0 & 2\end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3 \end{bmatrix}=0 \Rightarrow x_1=-x_3  \\\qquad\Rightarrow v=x_2 \begin{bmatrix} 0 \\1 \\0 \end{bmatrix}+ x_3 \begin{bmatrix} -1 \\0 \\1 \end{bmatrix}, 取v_2=\begin{bmatrix}0 \\1 \\0 \end{bmatrix},v_3= \begin{bmatrix} -1 \\0 \\1 \end{bmatrix}\\ \qquad \Rightarrow \text{eigenvalues: }\bbox[red, 2pt]{2,4} \text{ and eigenvectors: }\bbox[red, 2pt]{\begin{bmatrix} -2 \\1 \\1 \end{bmatrix}, \begin{bmatrix}0 \\1 \\0 \end{bmatrix}, \begin{bmatrix} -1 \\0 \\1 \end{bmatrix}}$$

解答：$$\textbf{(a)}\; \cos\theta ={u\cdot v\over \Vert u\Vert \Vert v\Vert} ={2-129-20+ 4 \over \sqrt{1+9+25+16}\cdot \sqrt{4+1849+16+1}} ={-143\over \sqrt{51} \cdot \sqrt{1870}} \\\qquad = \bbox[red, 2pt]{-143\over 17\sqrt{330}} \\ \textbf{(b)}\cases{A=\begin{bmatrix}9 & 8 &7\\6 & 5 &4\end{bmatrix} \\[1ex] B =\begin{bmatrix}1 & 2 &3\\4 & 5 &6\end{bmatrix}} \Rightarrow \cases{A\cdot B=tr(AB^T)=119 \\ \Vert A\Vert = \sqrt{tr(AA^T)} =\sqrt{271}\\ \Vert B\Vert =\sqrt{ tr{BB^T}} = \sqrt{91}} \Rightarrow \cos \theta ={A\cdot B \over \Vert A\Vert \Vert B\Vert} =\bbox[red, 2pt]{119\over \sqrt{24661}}$$

解答：$$\textbf{(a)}\; A= \left[\begin{matrix}1 & 1 & 0 & -1 \\1 & 2 & 1 & 3 \\1 & 1 & -9 & 2 \\16 & -13 & 1 & 3\end{matrix}\right] \xrightarrow{-R_1+R_2\to R_2, -R_1+R_3\to R_3, -16R_1+R_4\to R_4} \left[ \begin{matrix}1 & 1 & 0 & -1 \\0 & 1 & 1 & 4 \\0 & 0 & -9 & 3 \\0 & 0 & 30 & 135\end{matrix} \right] \\ \xrightarrow{10R_3/3+R)4\to R_4} \left[ \begin{matrix} 1 & 1 & 0 & -1 \\
0 & 1 & 1 & 4 \\0 & 0 & -9 & 3 \\0 & 0 & 0 & 145 \end{matrix} \right] \Rightarrow rank(A)=4\\ \qquad 又\cases{u_1\cdot u_2=1+2-3=0\\ u_1\cdot u_3=1+1-2=0\\ u_1\cdot u_4=16-13-3=0 \\ u_2\cdot u_3=1+2-9-6=0\\ u_2\cdot u_4= 16-26+1+9= 0\\ u_3\cdot u_4= 16-13-9+6=0} \Rightarrow S \text{ is orthogonal and a basis in }R^4, \bbox[red, 2pt] {\text{Q.E.D}}\\ \textbf{(b)}\; 4 \text{ coefficients: }\alpha,\beta,\gamma,\delta \Rightarrow \alpha u_1 +\beta u_2+ \gamma u_3+ \delta u_4=v\\ \quad \Rightarrow \begin{bmatrix}1 & 1& 1 & 16 \\1 & 2& 1& -13\\ 0& 1& -9 & 1\\ -1 &  3& 2& 3 \end{bmatrix} \begin{bmatrix}\alpha \\ \beta\\ \gamma\\\delta  \end{bmatrix}=\begin{bmatrix}a \\ b\\ c\\ d  \end{bmatrix} \Rightarrow \begin{bmatrix}\alpha \\ \beta\\ \gamma\\\delta  \end{bmatrix} =\begin{bmatrix}1 & 1& 1 & -1 \\1 & 2& 1& 3\\ 0& 1& -9 & 2\\ -1 &  3& 2& 3 \end{bmatrix}^{-1} \begin{bmatrix}a \\ b\\ c\\ d  \end{bmatrix} \\ =\left[ \begin{matrix}\frac{1}{3} & \frac{1}{3} & 0 & \frac{-1}{3} \\\frac{1}{15} & \frac{2}{15} & \frac{1}{15} & \frac{1}{5} \\\frac{1}{87} & \frac{1}{87} & \frac{-3}{29} & \frac{2}{87} \\\frac{16}{435} & \frac{-13}{435} & \frac{1}{435} & \frac{1}{145} \end{matrix}\right] \begin{bmatrix} a \\ b\\ c\\ d  \end{bmatrix} =\bbox[red, 2pt]{\left[ \begin{matrix} \frac{a+b-c}{3} \\\frac{a+2b+3c+d}{15} \\\frac{a+b+2c-9 d}{87} \\\frac{16a-13b+3c+d}{435} \end{matrix}\right]}$$

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