115年中央大學環工碩士班-工程數學詳解

國立中央大學115學年度碩士班考試入學試題

系所:環境工程 科目:工程數學

解答:$$L\{y''\} +L\{y'\} -6L\{y\} =0 \Rightarrow (s^2Y(s) - s - 1) + (sY(s) - 1) - 6Y(s) = 0  \\ \Rightarrow Y(s)={s+2\over s^2+s-6} ={s+2 \over (s+3)(s-2)} ={1/5\over s+3}+{4/5\over s-2} \\ \Rightarrow y(t)= {1\over 5}L^{-1} \left\{ {1\over s+3}\right\} + {4\over 5}L^{-1} \left\{{1\over s-2} \right\} \Rightarrow \bbox[red, 2pt]{y(t)={1\over 5}e^{-3t}+{4\over 5}e^{2t}}$$

解答:$$ A = \begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}  \Rightarrow \det(A-\lambda I) =0 \Rightarrow \lambda^2-1=0 \Rightarrow \lambda=\pm 1 \\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}1&-1\\3& -3 \end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix}=0 \Rightarrow x_1=x_2 \Rightarrow v_1= \begin{bmatrix}1\\1 \end{bmatrix} \\ \lambda_2=-1 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}3&-1\\3& -1 \end{bmatrix} \begin{bmatrix}x_1\\x_2 \end{bmatrix}=0 \Rightarrow x_2=3x_1 \Rightarrow v_2= \begin{bmatrix}1\\3 \end{bmatrix} \\ y=c_1e^{\lambda_1 t}v_1+ c_2 e^{\lambda_2 t}v_2 = c_1 e^{t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 1 \\ 3 \end{pmatrix} \Rightarrow \cases{y_1(t)=c_1e^t+c_2e^{-t} \\y_2(t)=c_1e^t+3c_2e^{-t}} \\ \textbf{(a) }\bbox[red, 2pt]{\pm 1} \; \textbf{(b) }\bbox[red, 2pt]{\begin{bmatrix}1\\1 \end{bmatrix}, \begin{bmatrix}1\\3 \end{bmatrix}} \; \textbf{(c) } \bbox[red, 2pt]{\cases{y_1(t)=c_1e^t+c_2e^{-t} \\y_2(t)=c_1e^t+3c_2e^{-t}}}$$

解答:$$B為上三角矩陣\Rightarrow \lambda_1+ \lambda_2+ \lambda_3= 3+3+4=\bbox[red, 2pt]{10}$$

解答:$$ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 3 & 3 \\ 2 & 1 & 4 \end{bmatrix}  \Rightarrow \det(A-\lambda I) =  \lambda^3 - 8\lambda^2 + 13\lambda - 6 = 0  \Rightarrow  (\lambda - 1)(\lambda - 1)(\lambda - 6) = 0 \\ \Rightarrow \lambda=1(代數重數為 2),6(代數重數為 1)\\ \lambda=1 \Rightarrow A-\lambda I=   \begin{bmatrix} 0 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 1 & 3 \end{bmatrix} \Rightarrow RREF(A-\lambda I)= \begin{bmatrix}1&0& 1 \\0& 1& 1\\0&0& 0\end{bmatrix} \Rightarrow rank(A-\lambda I)=2\\ \Rightarrow  \text{幾何重數} = \text{零化度(Nullity)} = n - \text{Rank}(A - I) = 3 - 2 = 1  \ne 代數重數 \\ \Rightarrow \bbox[red, 2pt]{\text{$A$不可對角化}}$$

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解題僅供參考，碩士班歷年試題及詳解