【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
某试验性生产线每年一月份进行熟练工与非熟练工的人数统计, 然后将 $\frac{1}{6}$ 熟练工支援其他生产部 门, 其缺额由招收新的非熟练工补齐. 新、老非熟练工经过培训及实践至年终考核有 $\frac{2}{5}$ 成为熟练
工. 设第 $n$ 年一月份统计的熟练工和非熟练工所占百分比分别为 $x_{n}$ 和 $y_{n}$, 记成向量 $\left(\begin{array}{l}x_{n} \\ y_{n}\end{array}\right)$.
(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 $\left(\begin{array}{l}x_{n+1} \\ y_{n+1}\end{array}\right)$.
答案:
【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) $\Rightarrow\left\{\begin{array}{l}x_{n+1}=\frac{5}{6} x_{n}+\frac{1}{15} x_{n}+\frac{2}{5} y_{n} \\ y_{n+1}=\frac{1}{10} x_{n}+\frac{3}{5} y_{n}\end{array}\right.$
$$
\Rightarrow\left\{\begin{array}{l}
x_{n+1}=\frac{9}{10} x_{n}+\frac{2}{5} y_{n} \\
y_{n+1}=\frac{1}{10} x_{n}+\frac{3}{5} y_{n}
\end{array}, \text { 即 }\left(\begin{array}{l}
x_{n+1} \\
y_{n+1}
\end{array}\right)=\left(\begin{array}{cc}
\frac{9}{10} & \frac{2}{5} \\
\frac{1}{10} & \frac{3}{5}
\end{array}\right)\left(\begin{array}{l}
x_{n} \\
y_{n}
\end{array}\right)\right.
$$
可见
$$
A=\left(\begin{array}{cc}
\frac{9}{10} & \frac{2}{5} \\
\frac{1}{10} & \frac{3}{5}
\end{array}\right)
$$
(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
$$
\left|\left(\eta_{1}, \eta_{2}\right)\right|=\left|\begin{array}{cc}
4 & -1 \\
1 & 1
\end{array}\right|=5 \neq 0
$$
矩阵为满秩, 由矩阵的秩和向量的关系可见 $\eta_{1}, \eta_{2}$ 线性无关.

$$
A \eta_{1}=\left(\begin{array}{cc}
\frac{9}{10} & \frac{2}{5} \\
\frac{1}{10} & \frac{3}{5}
\end{array}\right)\left(\begin{array}{l}
4 \\
1
\end{array}\right)=\left(\begin{array}{l}
4 \\
1
\end{array}\right)=\eta_{1}, A \eta_{2}=\left(\begin{array}{c}
-\frac{1}{2} \\
\frac{1}{2}
\end{array}\right)=\frac{1}{2} \eta_{2},
$$
由特征值、特征向量的定义, 得 $\eta_{1}$ 为 $A$ 的属于特征值 $\lambda_{1}=1$ 的特征向量, $\eta_{2}$ 为 $A$ 的属于特征
值 $\lambda_{2}=\frac{1}{2}$ 特征向量.
(3)因为
$$
\left(\begin{array}{l}
x_{n+1} \\
y_{n+1}
\end{array}\right)=A\left(\begin{array}{l}
x_{n} \\
y_{n}
\end{array}\right)=A^{2}\left(\begin{array}{l}
x_{n-1} \\
y_{n-1}
\end{array}\right) \cdots=A^{n}\left(\begin{array}{l}
x_{1} \\
y_{1}
\end{array}\right)=A^{n}\left(\begin{array}{l}
\frac{1}{2} \\
\frac{1}{2}
\end{array}\right)
$$
因此只要计算 $A^{n}$ 即可. 令
$$
P=\left(\eta_{1}, \eta_{2}\right)=\left(\begin{array}{cc}
4 & -1 \\
1 & 1
\end{array}\right),
$$
则由 $P^{-1} A P=\left(\begin{array}{ll}\lambda_{1} & \\ & \lambda_{2}\end{array}\right)$, 有 $A=P\left(\begin{array}{ll}\lambda_{1} & \\ & \lambda_{2}\end{array}\right) P^{-1}$,

于是
其中求逆矩阵的过程为:
$$
\begin{aligned}
&\left(\begin{array}{cccc}
4 & -1 & 1 & 0 \\
1 & 1 & 0 & 1
\end{array}\right) \rightarrow\left(\begin{array}{cccc}
1 & 1 & 0 & 1 \\
4 & -1 & 1 & 0
\end{array}\right) \\
&\rightarrow\left(\begin{array}{cccc}
1 & 1 & 0 & 1 \\
0 & -\frac{1}{4} & 1 & \frac{1}{4}
\end{array}\right) \rightarrow\left(\begin{array}{cccc}
1 & 1 & 0 & 1 \\
0 & 1 & -\frac{4}{5} & \frac{1}{5}
\end{array}\right) \rightarrow\left(\begin{array}{cccc}
1 & 0 & \frac{4}{5} & \frac{4}{5} \\
0 & 1 & -\frac{4}{5} & \frac{1}{5}
\end{array}\right)
\end{aligned}
$$
所以 $\left(\begin{array}{cc}4 & -1 \\ 1 & 1\end{array}\right)^{-1}=\left[\begin{array}{cc}\frac{4}{5} & \frac{4}{5} \\ -\frac{4}{5} & \frac{1}{5}\end{array}\right]=\frac{1}{5}\left[\begin{array}{cc}4 & 4 \\ -4 & 1\end{array}\right]$ 因此
$$
\left(\begin{array}{l}
x_{n+1} \\
y_{n+1}
\end{array}\right)=A^{n}\left(\begin{array}{l}
\frac{1}{2} \\
\frac{1}{2}
\end{array}\right)=\frac{1}{10}\left[\begin{array}{l}
8-3\left(\frac{1}{2}\right)^{n} \\
2+3\left(\frac{1}{2}\right)^{n}
\end{array}\right]
$$

解析:

视频讲解

提示1:如果发现题目有错或排版有误或您有更好的解题方法,请点击“编辑”功能进行更新。
提示2: Kmath一直以来坚持内容免费,这导致我们亏损严重。 如果看到这条信息的每位读者能慷慨打赏 10 元, 我们一个月内就能脱离亏损, 并保证在接下来的一整年里继续免费提供优质内容。捐赠
关闭