104年專門職業及技術人員高等考試建築師、技師、第二次食品技師考試暨普通考試不動產經紀人、記帳士考試
等別:高等考試
類科:電機工程技師
科目:工程數學(包括線性代數、微分方程、複變函數與機率)
類科:電機工程技師
科目:工程數學(包括線性代數、微分方程、複變函數與機率)
解:
(一)$$det\left( A-\lambda I \right) =0\Rightarrow \left| \begin{matrix} 1-\lambda & 0 & 0 \\ 0 & 1-\lambda & 1 \\ 0 & 1 & 1-\lambda \end{matrix} \right| =-{ \left( \lambda -1 \right) }^{ 3 }+\left( \lambda -1 \right) =0\Rightarrow \lambda \left( \lambda -1 \right) \left( \lambda -2 \right) =0\\ \Rightarrow 特徵值\bbox[red,2pt]{\lambda=0,1,2}$$(二)$$det\left( A-\lambda I \right) =0\Rightarrow \left| \begin{matrix} 1-\lambda & 0 & 0 \\ 0 & 1-\lambda & 1 \\ 0 & 1 & 1-\lambda \end{matrix} \right| =-{ \left( \lambda -1 \right) }^{ 3 }+\left( \lambda -1 \right) =0\Rightarrow \lambda \left( \lambda -1 \right) \left( \lambda -2 \right) =0\\ \lambda =0\Rightarrow \left[ \begin{matrix} 1-\lambda & 0 & 0 \\ 0 & 1-\lambda & 1 \\ 0 & 1 & 1-\lambda \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \end{matrix} \right] =0\\ \; \; \; \; \; \Rightarrow \begin{cases} x_{ 1 }=0 \\ x_{ 1 }+x_{ 3 }=0 \end{cases}\Rightarrow 取u_{ 1 }=\left[ \begin{matrix} 0 \\ 1 \\ -1 \end{matrix} \right] \\ \lambda =1\Rightarrow \left[ \begin{matrix} 1-\lambda & 0 & 0 \\ 0 & 1-\lambda & 1 \\ 0 & 1 & 1-\lambda \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \end{matrix} \right] =0\\ \; \; \; \; \;\Rightarrow \begin{cases} x_{ 3 }=0 \\ x_{ 2 }=0 \end{cases}\Rightarrow 取u_{ 2 }=\left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right] \\ \lambda =2\Rightarrow \left[ \begin{matrix} 1-\lambda & 0 & 0 \\ 0 & 1-\lambda & 1 \\ 0 & 1 & 1-\lambda \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \end{matrix} \right] =\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 1 & -1 \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \end{matrix} \right] =0\\ \; \; \; \; \;\Rightarrow \begin{cases} x_{ 1 }=0 \\ x_{ 2 }=x_{ 3 } \end{cases}\Rightarrow 取u_{ 3 }=\left[ \begin{matrix} 0 \\ 1 \\ 1 \end{matrix} \right] \\ \Rightarrow 特徵向量為\bbox[red,2pt]{\left[ \begin{matrix} 0 \\ 1 \\ -1 \end{matrix} \right] ,\left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right] 及\left[ \begin{matrix} 0 \\ 1 \\ 1 \end{matrix} \right] } $$(三)$$取P=\left[ \begin{matrix} u_{ 1 } & u_{ 2 } & u_{ 3 } \end{matrix} \right] =\left[ \begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ -1 & 0 & 1 \end{matrix} \right] \Rightarrow { P }^{ -1 }=\left[ \begin{matrix} 0 & 1/2 & -1/2 \\ 1 & 0 & 0 \\ 0 & 1/2 & /21 \end{matrix} \right] \\ \Rightarrow { P }^{ -1 }AP=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{matrix} \right] =\Lambda \Rightarrow A=P\Lambda { P }^{ -1 }\\ 答:\bbox[red,2pt]{P=\left[ \begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ -1 & 0 & 1 \end{matrix} \right] ,\Lambda =\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{matrix} \right] }$$
解:$$使用留數定理來解本題\\ \int _{ -\infty }^{ \infty }{ \frac { 3x+2 }{ x\left( x-4 \right) \left( x^{ 2 }+9 \right) } } dx\Rightarrow \int _{ -\infty }^{ \infty }{ \frac { 3z+2 }{ z\left( z-4 \right) \left( z^{ 2 }+9 \right) } } dx\\ \Rightarrow z=0,4,\pm 3i為單極點(\text{single pole})\, 且0,4在實數線上,3i在上半圓\\ 因此原式=\pi i\left( \frac { 0+2 }{ \left( 0-4 \right) \left( 0+9 \right) } +\frac { 3\times 4+2 }{ 4\left( 4^{ 2 }+9 \right) } \right) +2\pi i\left( \frac { 3\times 3i+2 }{ 3i\left( 3i-4 \right) \left( 3i+3i \right) } \right) \\ =\pi i\left( \frac { -1 }{ 18 } +\frac { 7 }{ 50 } \right) +2\pi i\left( \frac { 9i+2 }{ 72-54i } \right) =\frac { 19\pi i }{ 225 } +\frac { -18\pi +4\pi i }{ 72-54i } =\bbox[red,2pt]{\frac { -378\pi }{ 2025 }} $$
解:$$f_{ X,Y }\left( x,y \right) =\begin{cases} x\left( y+1.5 \right) , & 0<x,y<1 \\ 0, & elsewhere \end{cases}\\ \Rightarrow \begin{cases} f_{ X }\left( x \right) =\int _{ 0 }^{ 1 }{ x\left( y+1.5 \right) dy } =\left. \left[ 0.5xy^{ 2 }+1.5xy \right] \right| _{ 0 }^{ 1 }=2x \\ f_{ Y }\left( y \right) =\int _{ 0 }^{ 1 }{ x\left( y+1.5 \right) dx } =\left. \left[ 0.5x^{ 2 }\left( y+1.5 \right) \right] \right| _{ 0 }^{ 1 }=0.5y+0.75 \end{cases}$$(一)$$E\left[ X \right] =\int { xf_{ X }\left( x \right) dx } =\int _{ 0 }^{ 1 }{ 2x^{ 2 } } dx=\left. \left[ \frac { 2 }{ 3 } x^{ 3 } \right] \right| _{ 0 }^{ 1 }=\bbox[red,2pt]{\frac { 2 }{ 3 }} $$(二)$$E\left[ Y \right] =\int { yf_{ Y }\left( y \right) dy } =\int _{ 0 }^{ 1 }{ 0.5y^{ 2 }+0.75y } dy=\left. \left[ \frac { 1 }{ 6 } y^{ 3 }+\frac { 3 }{ 8 } y^{ 2 } \right] \right| _{ 0 }^{ 1 }=\bbox[red,2pt]{\frac { 13 }{ 24 }} $$(三)$$E\left[ X^{ 2 } \right] =\int { x^{ 2 }f_{ X }\left( x \right) dx } =\int _{ 0 }^{ 1 }{ 2x^{ 3 } } dx=\left. \left[ \frac { 2 }{ 4 } x^{ 4 } \right] \right| _{ 0 }^{ 1 }=\bbox[red,2pt]{\frac { 1 }{ 2 }} $$(四)$$E\left[ XY \right] =\iint { xyf_{ X,Y }\left( x,y \right) } dxdy=\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ x^{ 2 }y\left( y+1.5 \right) } dx } dy=\int _{ 0 }^{ 1 }{ \left. \left[ \frac { 1 }{ 3 } x^{ 3 }y\left( y+1.5 \right) \right] \right| _{ 0 }^{ 1 } } dy\\ =\int _{ 0 }^{ 1 }{ \left( \frac { 1 }{ 3 } y\left( y+1.5 \right) \right) } dy=\left. \left[ \frac { 1 }{ 9 } y^{ 3 }+\frac { 1 }{ 4 } y^{ 2 } \right] \right| _{ 0 }^{ 1 }=\bbox[red,2pt]{\frac { 13 }{ 36 }} $$
解:$$C:由\left( 1,1,1 \right) 至\left( -2,1,3 \right) 的線段,即C:x=-3t+1,y=1,z=2t+1,0\le t\le 1\\ \Rightarrow dx=-3dt,dy=0,dz=2dt\Rightarrow \int _{ C }{ xyzdx-\cos { \left( yz \right) } dy+xzdz } \\ =\int _{ 0 }^{ 1 }{ \left( -3t+1 \right) \left( 2t+1 \right) \left( -3dt \right) +\left( -3t+1 \right) \left( 2t+1 \right) \left( 2dt \right) } =\int _{ 0 }^{ 1 }{ \left( 6{ t }^{ 2 }+t-1 \right) dt } \\ =\left. \left[ 2t^3+\frac{1}{2}t^2-t \right] \right| _0^1=2+\frac{1}{2}-1=\bbox[red,2pt]{\frac{3}{2}}$$
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