臺灣綜合大學系統109學年度學士班轉學生聯合招生考試
科目名稱:工程數學
類組代碼:D37


解答:A=\left(\begin{matrix}1 & 0 & 1 \\0 & 1 & 0 \\ 1 & 0 & 1\end{matrix}\right) \Rightarrow \det(A-\lambda I)= -\lambda (\lambda-1) (\lambda-2)=0 \Rightarrow \lambda=0,1,2\\ \lambda_1=0 \Rightarrow (A-\lambda_1 I)\mathbf x=0 \Rightarrow \left(\begin{matrix}1 & 0 & 1 \\0 & 1 & 0 \\ 1 & 0 & 1\end{matrix}\right) \left(\begin{matrix}x_1\\x_2 \\ x_3 \end{matrix}\right) =0 \Rightarrow \cases{x_1=-x_3\\ x_2=0},取v_1=\left(\begin{matrix}-1\\0 \\ 1 \end{matrix}\right) \\ \lambda_2=1 \Rightarrow (A-\lambda_2 I)\mathbf x=0 \Rightarrow \left(\begin{matrix}0 & 0 & 1 \\0 & 0 & 0 \\ 1 & 0 & 0\end{matrix}\right) \left(\begin{matrix}x_1\\x_2 \\ x_3 \end{matrix}\right) =0 \Rightarrow \cases{x_1=0\\ x_3=0},取v_2=\left(\begin{matrix}0\\ 1 \\ 0 \end{matrix}\right) \\ \lambda_3=2 \Rightarrow (A-\lambda_3 I)\mathbf x=0 \Rightarrow \left(\begin{matrix}-1 & 0 & 1 \\0 & -1 & 0 \\ 1 & 0 & -1\end{matrix}\right) \left(\begin{matrix}x_1 \\x_2 \\ x_3 \end{matrix}\right) =0 \Rightarrow \cases{x_1=x_3\\ x_2=0},取v_3=\left(\begin{matrix} 1\\0 \\ 1 \end{matrix} \right) \\ \Rightarrow \left(\begin{matrix}x'(t) \\y'(t) \\ z'(t) \end{matrix}\right) =A \left( \begin{matrix}x(t) \\y(t) \\ z(t) \end{matrix}\right) \Rightarrow \left( \begin{matrix}x(t) \\y(t) \\ z(t) \end{matrix}\right) =(C_1v_1e^{\lambda_1 t}+C_2v_2e^{\lambda_2 t}+ C_3v_3e^{\lambda_3 t}) \\ =C_1\left(\begin{matrix}-1\\0 \\ 1 \end{matrix}\right)e^{0t} +C_2\left(\begin{matrix}0\\ 1 \\ 0 \end{matrix}\right) e^t+ C_3\left(\begin{matrix} 1\\0 \\ 1 \end{matrix} \right) e^{2t} \Rightarrow \bbox[red,2pt]{\cases{x(t)= -C_1+ C_3e^{2t} \\ y(t)= C_2e^t\\ z(t)=C_1 +C_3e^{2t}}}
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