國立成功大學114學年度碩士班招生考試試題
系 所:電腦與通信工程研究所
科目:機率與線性代數
解答:$$\textbf{a. }f_X(x) = \int_0^x 2\,dy =2x \Rightarrow E(X) = \int_0^1 xf_X(x)\,dx = \int_0^1 2x^2\,dx ={2\over 3} \\\quad \Rightarrow E(X^2) =\int_0^1 x^2f_X(x)\,dx = \int_0^1 2x^3\,dx ={1\over 2} \Rightarrow Var(X) =E(X^2)-(E(X))^2 ={1\over 2}-{4\over 9} ={1\over 18} \\\quad \Rightarrow \bbox[red, 2pt]{E(X)={2\over 3}, Var(X)={1\over 18}}\\ \textbf{b. }f_Y(y) = \int_y^1 2\,dx =2-2y \Rightarrow E(Y) = \int_0^1 yf_Y(y)\,dy = \int_0^1 (2y-2y^2)\,dy ={1\over 3} \\\quad \Rightarrow E(Y^2) =\int_0^1 y^2f_Y(y)\,dx = \int_0^1 (2y^2-2y^3)\,dy ={1\over 6} \Rightarrow Var(Y) =E(Y^2)-(E(Y))^2\\\quad ={1\over 6}-{1\over 9} ={1\over 18} \Rightarrow \bbox[red, 2pt]{E(Y)={1\over 3}, Var(Y)={1\over 18}} \\\textbf{c. }E(XY) =\int_0^1 \int_0^x xy\cdot 2\,dydx = \int_0^1 x^3\,dx ={1\over 4} \Rightarrow Cov(X,Y)=E(XY)- E(X)E(Y) \\\quad ={1\over 4} -{2\over 3}\cdot {1\over 3} = \bbox[red, 2pt]{1\over 36} \\\textbf{d. } Var(X+Y)= Var(X)+Var(Y)+2 Cov(X,Y) ={1\over 18}+{1\over 18}+ 2\cdot {1\over 36} = \bbox[red, 2pt]{1\over 6}$$
解答:$$\text{Harmonic number }H_6= 6 \left( {1\over 1}+ {1\over 2}+ {1\over 3}+ {1\over 4}+ {1\over 5}+ {1\over 6} \right) = \bbox[red, 2pt]{14.7}$$
解答:$$\textbf{a. } A= \begin{bmatrix} \frac{1}{3} & \frac{1}{4} \\ \frac{2}{3} & \frac{3}{4}\end{bmatrix} \Rightarrow \det(A- \lambda I) = \lambda^2-{13\over 12}\lambda+{ 1\over 12} =(\lambda-{1\over 12})(\lambda -1)=0 \Rightarrow \lambda={1\over 12},1\\ \lambda_1={1\over 12} \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} \frac{1}{4} & \frac{1}{4} \\ \frac{2}{3} & \frac{2}{3}\end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix} =0 \Rightarrow x_1=-x_2 \\\qquad \Rightarrow v= x_2 \begin{bmatrix} -1\\1 \end{bmatrix}, \text{ choose v_1} = \begin{bmatrix} -1\\1 \end{bmatrix} \\ \lambda_2=1 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix} \frac{-2}{3} & \frac{1}{4} \\\frac{2}{3} & \frac{-1}{4} \end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix} =0 \Rightarrow 8x_1= 3x_2 \\\qquad \Rightarrow v= x_2 \begin{bmatrix} 3/8\\1 \end{bmatrix}, \text{ choose v_2} =\begin{bmatrix} 3/8\\1 \end{bmatrix} \\ \Rightarrow P=[v_1 \; v_2] = \bbox[red, 2pt]{\begin{bmatrix} -1 & \frac{3}{8} \\1 & 1\end{bmatrix}} \\ \textbf{b. }A= PDP^{-1} = \begin{bmatrix} -1 & \frac{3}{8} \\1 & 1\end{bmatrix} \begin{bmatrix}\frac{1}{12} & 0 \\0 & 1 \end{bmatrix} \begin{bmatrix} \frac{-8}{11} & \frac{3}{11} \\ \frac{8}{11} & \frac{8}{11}\end{bmatrix} \Rightarrow A^n=PD^nP^{-1} \\= \begin{bmatrix} -1 & \frac{3}{8} \\1 & 1\end{bmatrix} \begin{bmatrix}\frac{1}{12^n} & 0 \\0 & 1 \end{bmatrix} \begin{bmatrix} \frac{-8}{11} & \frac{3}{11} \\ \frac{8}{11} & \frac{8}{11} \end{bmatrix} = \bbox[red, 2pt]{ {1\over 2^{2n}\cdot 3^n \cdot 11} \begin{bmatrix} 2^{2n}\cdot 3^{n+1}+8 & 2^{2n} \cdot 3^{n+1}-3\\ 2^{2n+3}\cdot 3^n -8& 2^{2n+3}\cdot 3^n+3\end{bmatrix}} \\ \textbf{c. }\lim_{n\to \infty} A^n = \begin{bmatrix} -1 & \frac{3}{8} \\1 & 1\end{bmatrix} \begin{bmatrix} 0 & 0 \\0 & 1 \end{bmatrix} \begin{bmatrix} \frac{-8}{11} & \frac{3}{11} \\ \frac{8}{11} & \frac{8}{11} \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix} \frac{3}{11} & \frac{3}{11} \\\frac{8}{11} & \frac{8}{11}\end{bmatrix} }$$
解答:$$\textbf{a. }A= \begin{bmatrix} 3&2& 2\\ 2& 3&-2 \end{bmatrix} \Rightarrow Y=A^TA = \begin{bmatrix}13& 12& 2 \\ 12& 13& -2\\ 2&-2 &8 \end{bmatrix} \\\Rightarrow \cases{\text{eigenvalue }\lambda_1=25 \Rightarrow \text{eigenvector }v_1= \begin{bmatrix}1\\1\\ 0 \end{bmatrix} \\ \text{eigenvalue }\lambda_2= 9 \Rightarrow \text{eigenvector }v_2= \begin{bmatrix}1/4\\ -1/4\\ 1 \end{bmatrix} \\ \text{eigenvalue }\lambda_3= 0 \Rightarrow \text{eigenvector }v_3= \begin{bmatrix}-2\\2 \\ 1 \end{bmatrix} } \\\Rightarrow V= \text{normalized }[ v_1\; v_2 \; v_3] = \begin{bmatrix}\sqrt 2/2 & \sqrt 2/6& -2/3\\ \sqrt 2/2&- \sqrt 2/6& 2/3\\ 0& 2\sqrt 2/3& 1/3 \end{bmatrix}\\ \text{nonzero eigenvalues }\cases{\sigma_1= \sqrt{\lambda_1} =5\\ \sigma_2= \sqrt{\lambda_2} =3} \Rightarrow \Sigma = \begin{bmatrix}\sigma_1 & 0 &0\\ 0& \sigma_2& 0 \end{bmatrix} = \begin{bmatrix}5& 0& 0\\ 0& 3& 0 \end{bmatrix} \\ u_1= {1\over \sigma_1} A v_1 ={1\over 5} \begin{bmatrix} 3&2& 2\\ 2& 3&-2 \end{bmatrix} \begin{bmatrix}\sqrt 2/2\\ \sqrt 2/2\\ 0 \end{bmatrix} = \begin{bmatrix}\sqrt 2/2\\ \sqrt 2/2 \end{bmatrix} \\\Rightarrow u_2= {1\over \sigma_2} A v_2 ={1\over 3} \begin{bmatrix} 3&2& 2\\ 2& 3&-2 \end{bmatrix} \begin{bmatrix}\sqrt 2/6\\ -\sqrt 2/6\\ 2\sqrt 2/6 \end{bmatrix} = \begin{bmatrix}\sqrt 2/2\\ -\sqrt 2/2 \end{bmatrix} \\ \Rightarrow U =[u_1 \; u_2] = \begin{bmatrix}{\sqrt 2\over 2} & {\sqrt 2\over 2}\\ {\sqrt 2\over 2}&-{\sqrt 2\over 2} \end{bmatrix} \Rightarrow A= U\Sigma V^T \\ \Rightarrow \bbox[red, 2pt]{A= \begin{bmatrix}{\sqrt 2\over 2} & {\sqrt 2\over 2}\\ {\sqrt 2\over 2}&-{\sqrt 2\over 2} \end{bmatrix} \begin{bmatrix}5 & 0& 0\\ 0& 3& 0 \end{bmatrix} \begin{bmatrix}{\sqrt 2\over 2}& {\sqrt 2\over 6}& -{2\over 2} \\ {\sqrt 2\over 2} & -{\sqrt 2\over 6}& {2\over 3} \\ 0& {2\sqrt 2\over 3}& {1\over 3}\end{bmatrix}} \\\textbf{b. }A= \begin{bmatrix} 3&2& 2\\ 2& 3&-2 \end{bmatrix} \Rightarrow AA^T = \begin{bmatrix}17& 8\\8& 17 \end{bmatrix} \Rightarrow (AA^T)^{-1} = \begin{bmatrix}{17\over 225}& -{8\over 225} \\ -{8\over 225} &{17\over 225}\end{bmatrix} \\\quad \Rightarrow A^+= A^T (AA^T)^{-1} = \begin{bmatrix}3& 2\\2& 3\\ 2& -2 \end{bmatrix} \begin{bmatrix}{17\over 225}& -{8\over 225} \\ -{8\over 225} &{17\over 225}\end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}{7\over 45} & {2\over 45} \\ {2\over 45} &{7\over 45} \\ {2\over 9} &-{2\over 9}\end{bmatrix}} \\\textbf{c. }\cases{A= \begin{bmatrix} 3&2& 2\\ 2& 3&-2 \end{bmatrix} \\b= \begin{bmatrix}1\\ 0 \end{bmatrix}} \Rightarrow \cases{ A^TA = \left[ \begin{matrix}13 & 12 & 2\\12 & 13 & -2\\2 & -2 & 8\end{matrix} \right] \\ A^Tb= \begin{bmatrix}3\\ 2\\2 \end{bmatrix}} \Rightarrow B= [A^TA \mid A^Tb] = \left[ \begin{array}{rrr|r} 13&12&2 &3\\ 12&13& -2& 2\\ 2&-2& 8& 2 \end{array} \right] \\ \Rightarrow rref(B)= \left[ \begin{array}{rrr|r} 1 & 0 & 2 & \frac{3}{5}\\0 & 1 & -2 & - \frac{2}{5}\\0 & 0 & 0 & 0 \end{array} \right] \Rightarrow \text{a least-squares solution }\hat x = \bbox[red, 2pt]{\begin{bmatrix}{3\over 5}\\ -{2\over 5} \\ 0 \end{bmatrix}} \\\textbf{d. } A\hat x =\begin{bmatrix} 3&2& 2\\ 2& 3&-2 \end{bmatrix} \begin{bmatrix}{3 \over 5}\\ -{2\over 5} \\ 0 \end{bmatrix} =\begin{bmatrix}1\\ 0 \end{bmatrix} \Rightarrow ||b-A\hat x|| = \bbox[red, 2pt]0$$
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解題僅供參考，其他碩士班題及詳解
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