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:(a)μ=E(X)=2⋅13+3⋅12+11⋅16=16(4+9+11)=4E(X2)=4⋅13+9⋅12+121⋅16=16(8+27+121)=26σ2=var(X)=E(X2)−(E(X))2=26−16=10(b)μ=E(X)=1⋅0.4+3⋅0.1+4⋅0.2+5⋅0.3=3E(X2)=1⋅0.4+9⋅0.1+16⋅0.2+25⋅0.3=12⇒σ2=var(X)=E(X2)−(E(X))2=12−32=3
解答:E(X)=∫xf(x)dx=∫20x⋅x2dx=43E(X2)=∫x2f(x)dx=∫20x2⋅x2dx=2⇒var(X)=E(X2)−(E(X))2=2−169=29
解答:fXY(x,y)⇒{fX(x)=∫∞−∞fXY(x,y)dyfY(y)=∫∞−∞fXY(x,y)dx⇒FZ(z)=P(Z≤z)=P(XY≤z)=P(XY≤z,X≥0)+P(XY≤z,X≤0)=P(Y≤z/X,X≥0)+P(Y≥z,X≤0)=∫∞0fX(x)∫z/x−∞fY(y)dydx+∫0−∞fX(x)∫z/x−∞fY(y)dydx⇒pdf of Z=fZ(z)=ddzFZ(z)=∫∞0fX(x)fY(z/x)1|x|dx+∫0−∞fX(x)fY(z/x)1|x|dx=∫∞−∞fX(x)fY(z/x)1|x|dx, where {fX(x)=∫∞−∞fXY(x,y)dyfY(y)=∫∞−∞fXY(x,y)dx
解答:(a){fX(X=1)=0.1+0.2+0.2=0.5fX(X=3)=0.3+0.1+0.1=0.5⇒{fX(X=1)=0.5fX(X=3)=0.5{fY(Y=−3)=0.1+0.3=0.4fY(Y=2)=0.2+0.1=0.3fY(Y=4)=0.2+0.1=0.3⇒{fY(Y=−3)=0.4fY(Y=2)=0.3fY(Y=4)=0.3(b){μx=E(X)=1⋅0.5+3⋅0.5=2μy=E(Y)=(−3)⋅0.4+2⋅0.3+4⋅0.3=0.6,又E(XY)=1⋅(−3)⋅0.1+1⋅2⋅0.2+1⋅4⋅0.2+3⋅(−3)⋅0.3+3⋅2⋅0.1+3⋅4⋅0.1=0⇒Cov(X,Y=E(XY)−μxμy=0−2⋅0.6=−1.2(c){E(X2)=1⋅0.5+9⋅0.5=5E(Y2)=9⋅0.4+4⋅0.3+16⋅0.3=9.6⇒{Var(X)=E(X2)−μ2x=5−22=1Var(Y)=E(Y2)−μ2y=9.6−0.62=9.24⇒ρ=Cov(X,Y)σx⋅σy=−1.2√1⋅√9.24=−0.395(d)Cov(X,Y)≠0⇒NOT independent

解答:假設moment generating function of X:M(t)⇒M[k](0)=E(Xk)=0.8⇒M(t)=0.8et
解答:64+320+72−128−120−96=456−344=112解答:(a)A=[2012168]⇒det(A)=−32⇒A−1=1−32[8−12−1620]=[−1/43/81/2−5/8]
解答:(a)A=[6639]⇒det(A−λI)=(λ−3)(λ−12)=0⇒λ=3,12λ1=3⇒(A−λ1I)v=0⇒[3636][x1x2]=0⇒x1+2x2=0⇒v=x2[−21],取v1=[−21]λ2=12⇒λ=3,12λ2=12⇒(A−λ2I)v=0⇒[−663−3][x1x2]=0⇒x1=x2⇒v=x2[11],取v2=[11]⇒eigenvalues: 3,12 and eigenvectors: [−21],[11](b)A=[00−4242206]⇒det(A−λI)=−(λ−2)(λ−4)2=0⇒λ=2,4λ1=2⇒(A−λ1I)v=0⇒[−20−4222204][x1x2x3]=0⇒{x1+2x3=0x2=x3⇒v=x3[−211],取v1=[−211]λ2=4⇒(A−λ2I)v=0⇒[−40−4202202][x1x2x3]=0⇒x1=−x3⇒v=x2[010]+x3[−101],取v2=[010],v3=[−101]⇒eigenvalues: 2,4 and eigenvectors: [−211],[010],[−101]
解答:(a)cosθ=u⋅v‖
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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解題僅供參考, 其他歷年試題及詳解
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