【ID】1179 【题型】解答题 【类型】考研真题 【来源】2000年全国硕士研究生招生考试试题

(1) 求 $\left(\begin{array}{l}x_{n+1} \\ y_{n+1}\end{array}\right)$ 与 $\left(\begin{array}{l}x_{n} \\ y_{n}\end{array}\right)$ 的关系式并写成矩阵形式: $\left(\begin{array}{l}x_{n+1} \\ y_{n+1}\end{array}\right)=\boldsymbol{A}\left(\begin{array}{l}x_{n} \\ y_{n}\end{array}\right)$;
(2) 验证 $\boldsymbol{\eta}_{1}=\left(\begin{array}{l}4 \\ 1\end{array}\right), \boldsymbol{\eta}_{2}=\left(\begin{array}{c}-1 \\ 1\end{array}\right)$ 是 $\boldsymbol{A}$ 的两个线性无关的特征向量, 并求出相应的特征值;
(3) 当 $\left(\begin{array}{l}x_{1} \\ y_{1}\end{array}\right)=\left(\begin{array}{l}\frac{1}{2} \\ \frac{1}{2}\end{array}\right)$ 时,求 $\left(\begin{array}{l}x_{n+1} \\ y_{n+1}\end{array}\right)$.

【详解】(1)由题意, $\frac{1}{6} x_{n}+y_{n}$ 是非熟练工人数, $\frac{2}{5}\left(\frac{1}{6} x_{n}+y_{n}\right)$ 是年终由非熟练工人 变成的熟练工人数, $\frac{5}{6} x_{n}$ 是年初支援其他部门后的熟练工人数, 根据年终熟练工的人数列 出等式(1), 根据年终非熟练工人人数列出等式(2)得
$$\left\{\begin{array}{l} x_{n+1}=\frac{5}{6} x_{n}+\frac{2}{5}\left(\frac{1}{6} x_{n}\right. \\ y_{n+1}=\frac{3}{5}\left(\frac{1}{6} x_{n}+y_{n}\right) \end{array}\right.$$
(1) $\Rightarrow\left\{\begin{array}{l}x_{n+1}=\frac{5}{6} x_{n}+\frac{1}{15} x_{n}+\frac{2}{5} y_{n} \\ y_{n+1}=\frac{1}{10} x_{n}+\frac{3}{5} y_{n}\end{array}\right.$
$$\Rightarrow\left\{\begin{array}{l} x_{n+1}=\frac{9}{10} x_{n}+\frac{2}{5} y_{n} \\ y_{n+1}=\frac{1}{10} x_{n}+\frac{3}{5} y_{n} \end{array}, \text { 即 }\left(\begin{array}{l} x_{n+1} \\ y_{n+1} \end{array}\right)=\left(\begin{array}{cc} \frac{9}{10} & \frac{2}{5} \\ \frac{1}{10} & \frac{3}{5} \end{array}\right)\left(\begin{array}{l} x_{n} \\ y_{n} \end{array}\right)\right.$$

$$A=\left(\begin{array}{cc} \frac{9}{10} & \frac{2}{5} \\ \frac{1}{10} & \frac{3}{5} \end{array}\right)$$
(2) 把 $\eta_{1}, \eta_{2}$ 作为列向量写成矩阵的形式 $\left(\eta_{1}, \eta_{2}\right)$, 因为其行列式
$$\left|\left(\eta_{1}, \eta_{2}\right)\right|=\left|\begin{array}{cc} 4 & -1 \\ 1 & 1 \end{array}\right|=5 \neq 0$$

$$A \eta_{1}=\left(\begin{array}{cc} \frac{9}{10} & \frac{2}{5} \\ \frac{1}{10} & \frac{3}{5} \end{array}\right)\left(\begin{array}{l} 4 \\ 1 \end{array}\right)=\left(\begin{array}{l} 4 \\ 1 \end{array}\right)=\eta_{1}, A \eta_{2}=\left(\begin{array}{c} -\frac{1}{2} \\ \frac{1}{2} \end{array}\right)=\frac{1}{2} \eta_{2},$$

(3)因为
$$\left(\begin{array}{l} x_{n+1} \\ y_{n+1} \end{array}\right)=A\left(\begin{array}{l} x_{n} \\ y_{n} \end{array}\right)=A^{2}\left(\begin{array}{l} x_{n-1} \\ y_{n-1} \end{array}\right) \cdots=A^{n}\left(\begin{array}{l} x_{1} \\ y_{1} \end{array}\right)=A^{n}\left(\begin{array}{l} \frac{1}{2} \\ \frac{1}{2} \end{array}\right)$$

$$P=\left(\eta_{1}, \eta_{2}\right)=\left(\begin{array}{cc} 4 & -1 \\ 1 & 1 \end{array}\right),$$

\begin{aligned} &\left(\begin{array}{cccc} 4 & -1 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{array}\right) \rightarrow\left(\begin{array}{cccc} 1 & 1 & 0 & 1 \\ 4 & -1 & 1 & 0 \end{array}\right) \\ &\rightarrow\left(\begin{array}{cccc} 1 & 1 & 0 & 1 \\ 0 & -\frac{1}{4} & 1 & \frac{1}{4} \end{array}\right) \rightarrow\left(\begin{array}{cccc} 1 & 1 & 0 & 1 \\ 0 & 1 & -\frac{4}{5} & \frac{1}{5} \end{array}\right) \rightarrow\left(\begin{array}{cccc} 1 & 0 & \frac{4}{5} & \frac{4}{5} \\ 0 & 1 & -\frac{4}{5} & \frac{1}{5} \end{array}\right) \end{aligned}

$$\left(\begin{array}{l} x_{n+1} \\ y_{n+1} \end{array}\right)=A^{n}\left(\begin{array}{l} \frac{1}{2} \\ \frac{1}{2} \end{array}\right)=\frac{1}{10}\left[\begin{array}{l} 8-3\left(\frac{1}{2}\right)^{n} \\ 2+3\left(\frac{1}{2}\right)^{n} \end{array}\right]$$