115年中正通訊碩士班-線代與機率詳解

國立中正大學115學年度碩士班招生考試

科目名稱:線性代數與機率
系所組別:通訊工程學系-通訊甲組

解答:$$\textbf{(a) } (a^2-9)x_3=a-2 \Rightarrow a^2-9\ne 0 \Rightarrow \bbox[red, 2pt]{a\ne 3 \text{ and }a\ne -3} \\ \textbf{(b) }a^2-9=0 \Rightarrow \cases{a=3 \Rightarrow (3^2-9)x_3=3-2 \Rightarrow 0=1 \Rightarrow \text{contradiction } \\a=-3 \Rightarrow ((-3)^2-9)x_3=-3-2 \Rightarrow 0=-5 \Rightarrow \text{contradiction }}  \\\quad \Rightarrow \bbox[red, 2pt]{a=3 \text{ or }a=-3} \\ \textbf{(c) } \cases{a^2-9=0\\ a-2=0} \Rightarrow \cases{a=\pm 3\\ a=2} \Rightarrow \bbox[red, 2pt]{\text{None}}$$

解答:$$\textbf{(a) }x+2y-3z=0 \Rightarrow x=-2y+3z \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -2y + 3z \\ y \\ z \end{bmatrix} = y \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + z \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix} \\\quad \Rightarrow \text{ a basis for }W: \bbox[red, 2pt]{\left\{ \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix} \right\} } \\ \textbf{(b) }\text{Let $\mathbf{n}$ be the normal vector of the plane $W$, given by the coefficients of the equation:} \\ \mathbf{n} = \begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix} \Rightarrow P = I - \frac{\mathbf{n} \mathbf{n}^T}{\mathbf{n}^T\mathbf{n}} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \frac{1}{14} \begin{bmatrix} 1 & 2 & -3 \\ 2 & 4 & -6 \\ -3 & -6 & 9 \end{bmatrix} = \bbox[red, 2pt] { \frac{1}{14} \begin{bmatrix} 13 & -2 & 3 \\ -2 & 10 & 6 \\ 3 & 6 & 5 \end{bmatrix} } \\ \textbf{(c) } \mathbf{x}_0 = \begin{bmatrix} x_0 \\ y_0 \\ z_0 \end{bmatrix} \Rightarrow \text{proj}_W( \mathbf{x}_0) = P\mathbf{x}_0 = \frac{1}{14} \begin{bmatrix} 13 & -2 & 3 \\ -2 & 10 & 6 \\ 3 & 6 & 5 \end{bmatrix} \begin{bmatrix} x_0 \\ y_0 \\ z_0 \end{bmatrix} = \bbox[red, 2pt]{\frac{1}{14} \begin{bmatrix} 13x_0 - 2y_0 + 3z_0 \\ -2x_0 + 10y_0 + 6z_0 \\ 3x_0 + 6y_0 + 5z_0 \end{bmatrix} } \\ \textbf{(d) } \text{ distance $d$ from $(-1,1,-1)$ to $x+2y-3z=0$}\\ \Rightarrow d= \frac{\vert{}1(-1) + 2(1) - 3(-1)\vert{}}{\sqrt{1^2 + 2^2 + (-3)^2}} = \frac{4}{\sqrt{14}} = \bbox[red, 2pt]{\frac{2\sqrt{14}}{7} }$$ 解答:$$\textbf{(a) } B' = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 0 \\ 2 & -2 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 2 & 0 & 2 \end{bmatrix} \Rightarrow [B'\mid B] = \left[ \begin{array}{ccc\|ccc} -1 & 0 & 1 & 1 & 0 & -1 \\ 1 & 1 & 0 & 0 & 1 & 0 \\ 2 & -2 & 2 & 2 & 0 & 2 \end{array} \right] \\ \xrightarrow{-R_1\to R_1} \left[ \begin{array}{ccc\|ccc} 1 & 0 & -1 & -1 & 0 & 1 \\ 1 & 1 & 0 & 0 & 1 & 0 \\ 2 & -2 & 2 & 2 & 0 & 2 \end{array} \right] \xrightarrow{R_2-R_1\to R_2, R_3-2R_1\to R_3} \left[ \begin{array}{ccc\|ccc} 1 & 0 & -1 & -1 & 0 & 1 \\ 0 & 1 & 1 & 1 & 1 & -1 \\ 0 & -2 & 4 & 4 & 0 & 0 \end{array} \right] \\ \xrightarrow{R_3+2R_2 \to R_3} \left[ \begin{array}{ccc\|ccc} 1 & 0 & -1 & -1 & 0 & 1 \\ 0 & 1 & 1 & 1 & 1 & -1 \\ 0 & 0 & 6 & 6 & 2 & -2 \end{array} \right] \xrightarrow{R_3/6 \to R_3} \left[ \begin{array}{ccc\|ccc} 1 & 0 & -1 & -1 & 0 & 1 \\ 0 & 1 & 1 & 1 & 1 & -1 \\ 0 & 0 & 1 & 1 & \frac{1}{3} & -\frac{1}{3} \end{array} \right] \\ \xrightarrow{R_1+R_3\to R_1, R_2-R_3\to R_2} \left[ \begin{array}{ccc\|ccc} 1 & 0 & 0 & 0 & \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & 0 & 0 & \frac{2}{3} & -\frac{2}{3} \\ 0 & 0 & 1 & 1 & \frac{1}{3} & -\frac{1}{3} \end{array} \right] \Rightarrow \bbox[red, 2pt]{P_{B,B'} = \begin{bmatrix} 0 & \frac{1}{3} & \frac{2}{3} \\ 0 & \frac{2}{3} & -\frac{2}{3} \\ 1 & \frac{1}{3} & -\frac{1}{3} \end{bmatrix}} \\ \textbf{(b) }[B\mid B'] = \left[ \begin{array}{ccc\|ccc} 1 & 0 & -1 & -1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 1 & 0 \\ 2 & 0 & 2 & 2 & -2 & 2 \end{array} \right] \xrightarrow{R_3-2R_1\to R_3} \left[ \begin{array}{ccc\|ccc} 1 & 0 & -1 & -1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 1 & 0 \\ 0 & 0 & 4 & 4 & -2 & 0 \end{array} \right] \\ \xrightarrow{R_3/4\to R_3} \left[ \begin{array}{ccc\|ccc} 1 & 0 & -1 & -1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & -\frac{1}{2} & 0 \end{array} \right] \xrightarrow{R_1+R_3\to R_1} \left[ \begin{array}{ccc\|ccc} 1 & 0 & 0 & 0 & -\frac{1}{2} & 1 \\ 0 & 1 & 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & -\frac{1}{2} & 0 \end{array} \right] \\ \Rightarrow \bbox[red, 2pt]{ P_{B',B} = \begin{bmatrix} 0 & -\frac{1}{2} & 1 \\ 1 & 1 & 0 \\ 1 & -\frac{1}{2} & 0 \end{bmatrix} } \\ \textbf{(c) }  \mathbf w= \begin{bmatrix}0\\ 0\\ 2 \end{bmatrix} \Rightarrow [B'\mid \mathbf w] =  \left[ \begin{array}{ccc\|c} -1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 \\ 2 & -2 & 2 & 2 \end{array} \right] \xrightarrow{-R_1\to R_1} \left[ \begin{array}{ccc\|c} 1 & 0 & -1 & 0 \\ 1 & 1 & 0 & 0 \\ 2 & -2 & 2 & 2 \end{array} \right] \xrightarrow{R_2-R_1\to R_2, R_3-2R_1\to R_3} \\ \left[ \begin{array}{ccc\|c} 1 & 0 & -1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & -2 & 4 & 2 \end{array} \right]  \xrightarrow{2R_2+R_3\to R_3}  \left[ \begin{array}{ccc\|c} 1 & 0 & -1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 6 & 2 \end{array} \right] \xrightarrow{R_3/6\to R_3} \left[ \begin{array}{ccc\|c} 1 & 0 & -1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1/3 \end{array} \right] \xrightarrow{R_1+R_3\to R_1, R_2-R_3\to R_2} \\\left[ \begin{array}{ccc\|c} 1 & 0 & 0 & 1/3 \\ 0 & 1 & 0 & -1/3 \\ 0 & 0 & 1 & 1/3 \end{array} \right] \Rightarrow  [\mathbf{w}]_{B'} =\bbox[red, 2pt]{\begin{bmatrix} \frac{1}{3} \\ -\frac{1}{3} \\ \frac{1}{3} \end{bmatrix} }$$

解答:$$\textbf{(a) }  M= \begin{bmatrix}0& 2& 3\\ 0& 2& 0\\ 1& 0& 1 \end{bmatrix} \Rightarrow \det(M-\lambda I) =-(\lambda-2)(\lambda^2-\lambda-3)=0 \Rightarrow \lambda =2,{1\pm \sqrt{13}\over 2}\\ \lambda_1=2 \Rightarrow (M-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}-2 & 2 & 3 \\0 & 0 & 0 \\1 & 0 & -1 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1=x_3\\ 2x_2+x_3=0} \\ \qquad \Rightarrow v= x_3 \begin{bmatrix}1\\ -1/2\\ 1 \end{bmatrix} \Rightarrow v_1= \begin{bmatrix}1\\ -1/2\\ 1 \end{bmatrix} \\ \lambda_2= {1-\sqrt{13}\over 2} \Rightarrow (M-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix} \frac{\sqrt{13}-1}{2} & 2 & 3 \\0 & \frac{\sqrt{13}+3}{2} & 0 \\1 & 0 & \frac{\sqrt{13}+1}{2} \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}=0 \\\qquad \Rightarrow \cases{x_2=0 \\ 2x_1+(1+\sqrt{13})x_3=0} \Rightarrow v= x_3 \begin{bmatrix}(-1-\sqrt{13})/2\\ 0\\1 \end{bmatrix} \Rightarrow v_2= \begin{bmatrix}(-1-\sqrt{13})/2\\ 0\\1 \end{bmatrix} \\ \lambda_3= {1+\sqrt{13}\over 2}  \Rightarrow (M-\lambda_3 I)v=0 \Rightarrow \begin{bmatrix} \frac{-\sqrt{13}-1}{2} & 2 & 3 \\0 & \frac{-\sqrt{13}+3}{2} & 0 \\ 1 & 0 & \frac{-\sqrt{13}+1}{2} \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}=0 \\ \qquad \Rightarrow \cases{2x_1=(\sqrt{13}-1)x_3\\ x_2=0} \Rightarrow v= x_3 \begin{bmatrix}(\sqrt{13}-1)/2\\ 0\\ 1 \end{bmatrix} \Rightarrow v_3= \begin{bmatrix}(\sqrt{13}-1)/2\\ 0\\ 1 \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{M = \begin{bmatrix} 1& (-1-\sqrt{13})/2 & (\sqrt{13}-1)/2\\ -1/2& 0& 0\\ 1& 1& 1\end{bmatrix} \begin{bmatrix} 2& 0& 0\\ 0& (1-\sqrt{13})/2& 0\\ 0& 0& (1+\sqrt{13})/2 \end{bmatrix}}\\ \qquad \bbox[red, 2pt]{\begin{bmatrix} 1& (-1-\sqrt{13})/2 & (\sqrt{13}-1)/2\\ -1/2& 0& 0\\ 1& 1& 1\end{bmatrix}^{-1}} \\ \Rightarrow M^7 =PD^7P^{-1} =P \begin{bmatrix} 128 & 0 & 0 \\0 & \frac{-97 \sqrt{13}+337}{2} & 0 \\0 & 0 & \frac{97 \sqrt{13}+337}{2}\end{bmatrix} P^{-1} \Rightarrow \bbox[red, 2pt]{M^7 = \begin{bmatrix}120 & 566 & 291 \\0 & 128 & 0 \\97 & 372 & 217 \end{bmatrix}} \\\textbf{(b) }  \text{Using Gram-Schmidt process, } \begin{bmatrix}0\\0\\1 \end{bmatrix} \Rightarrow \begin{bmatrix}0\\0\\1 \end{bmatrix}, \begin{bmatrix}2\\2\\0 \end{bmatrix} \Rightarrow \begin{bmatrix} \sqrt 2/2\\ \sqrt 2/2\\ 0 \end{bmatrix}, \begin{bmatrix}3\\0\\1 \end{bmatrix} \Rightarrow \begin{bmatrix}\sqrt 2/2\\ -\sqrt 2/2\\ 0 \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{Q= \begin{bmatrix} 0& \sqrt 2/2& \sqrt 2/2\\ 0& \sqrt 2/2& -\sqrt 2/2\\ 1& 0& 0\end{bmatrix} }\Rightarrow Q^T= \begin{bmatrix}0& 0& 1\\ \sqrt 2/2& \sqrt 2/2& 0\\ \sqrt 2/2& -\sqrt 2/2& 0 \end{bmatrix} \Rightarrow R=Q^T M \\ \Rightarrow \bbox[red, 2pt]{R= \begin{bmatrix} 1& 0& 1\\ 0& 2\sqrt 2& 3\sqrt 2/2\\ 0& 0& 3\sqrt 2/2 \end{bmatrix}}$$

解答:$$\textbf{(a) }  N\sim \text{Poisson}(\lambda t) \Rightarrow \text{The mean of $N$ is }\bbox[red, 2pt]{\lambda t} \\ \textbf{(b) } P(N = 0) = \frac{e^{-\lambda t}(\lambda t)^0}{0!} = \frac{e^{-\lambda t} \cdot 1}{1} = \bbox[red, 2pt]{e^{-\lambda t} } \\ \textbf{(c) } P(Z > t) = P(N = 0) = e^{-\lambda t} \Rightarrow  P(Z \le t) = 1 - P(Z > t) = \bbox[red, 2pt]{1 - e^{-\lambda t} } \\ \textbf{(d) } \text{From part (c), we established that the interarrival time $Z$ is Exponentially distributed }\\ \text{with rate parameter $\lambda$. The mean interarrival time is } \bbox[red, 2pt]{1\over \lambda}$$

解答:$$\textbf{(a) }  Y=e^X \Rightarrow F_Y(y) =P(Y\le y) =P(e^X\le y) =P(X\le \ln y)= F_X(\ln y) \\ \qquad \Rightarrow \bbox[red, 2pt] {F_Y(y) = \begin{cases} F_X(\ln y) & \text{if } y > 0 \\ 0 & \text{if } y \le 0 \end{cases} } \\ \textbf{(b) } f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy} [F_X(\ln y)]   = f_X(\ln y) \cdot \frac{d}{dy}(\ln y)  = f_X(\ln y) \cdot \frac{1}{y} \\ \qquad \Rightarrow \bbox[red, 2pt] {f_Y(y) = \begin{cases} \frac{1}{y} f_X(\ln y) & \text{if } y > 0 \\ 0 & \text{if } y \le 0 \end{cases} } \\ \textbf{(c) } f_X(x) = \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(x - m)^2}{2\sigma^2}\right) \Rightarrow  f_Y(y) = \frac{1}{y} f_X(\ln y)  \\\qquad = \frac{1}{y} \left( \frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{(\ln y - m)^2}{2\sigma^2}\right) \right) \Rightarrow  \bbox[red, 2pt]{f_Y(y) = \begin{cases} \frac{1}{y\sigma\sqrt{2\pi}} \exp\left(-\frac{(\ln y - m)^2}{2\sigma^2}\right) & \text{if } y > 0 \\ 0 & \text{if } y \le 0 \end{cases} } \\ \textbf{(d) }X\sim Exp(\lambda) \Rightarrow f_X(x) = \lambda e^{-\lambda x} \Rightarrow F_X(x)=1-e^{-\lambda x}, \text{ for }x\ge 0 \\ Y=e^{-\lambda X} \Rightarrow F_Y(y) =P(Y\le y) =P(e^{-\lambda X}\le y) =P(-\lambda X\le \ln y) =P\left(X \ge -\frac{\ln y}{\lambda}\right) \\  = 1 - P\left(X < -\frac{\ln y}{\lambda}\right) = 1 - F_X\left(-\frac{\ln y}{\lambda}\right)  = 1 - \left( 1 - \exp\left(-\lambda \left( -\frac{\ln y}{\lambda} \right)\right) \right) =y \\ \Rightarrow  f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy}(y) = 1 \quad \text{for } 0 < y \le 1 \\ \Rightarrow   {f_Y(y) = \begin{cases} 1 & \text{for } 0 < y \le 1 \\ 0 & \text{otherwise} \end{cases} } \Rightarrow \bbox[red, 2pt]{\text{$Y$ is a standard Uniform random variable}}$$

解答:$$\textbf{(a) }   f_X(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}  \Rightarrow \text{ the characteristic function of }X, \text{ denoted as }\varphi_X(t) \\ \Rightarrow  \varphi_X(t) = \mathbb{E}[e^{itX}] = \int_{-\infty}^{\infty} e^{itx} f_X(x) dx = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\frac{x^2}{2} + itx} dx = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\frac{1}{2}(x-it)^2 - \frac{t^2}{2}} dx \\=  e^{-\frac{t^2}{2}} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}(x-it)^2} dx =  \bbox[red, 2pt]{e^{-\frac{t^2}{2}} } \\ \textbf{(b) }  Y = \sigma X + m  \Rightarrow  \varphi_Y(t) = \mathbb{E}[e^{itY}]  = \mathbb{E}[e^{it(\sigma X + m)}]   = e^{itm} \mathbb{E}[e^{i(t\sigma)X}]  = e^{itm} \varphi_X(t\sigma) \\ =  e^{itm} \cdot e^{-\frac{1}{2}\sigma^2 t^2} =\bbox[red, 2pt]{ e^{itm - \frac{1}{2}\sigma^2 t^2} }$$

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解題僅供參考，碩士班歷年試題及詳解