中国科学院真题 第20109题
📝 题目
1．（15 分）$\left\{\begin{array}{l}x_{n+1}=x_{n}+4 y_{n} \\ y_{n+1}=2 x_{n}+y_{n}\end{array}\right.$ 已知 $x_{0}=1, y_{0}=0$ ，求 $x_{100}, y_{100}$ ．
💡 答案解析
1．解 由 $\left(\begin{array}{ll}x_{n+1} & y_{n+1}\end{array}\right)=\left(\begin{array}{ll}x_{n} & y_{n}\end{array}\right)\left(\begin{array}{ll}1 & 2 \\ 4 & 1\end{array}\right)$ ，得 $\left(\begin{array}{ll}x_{n} & y_{n}\end{array}\right)=\left(\begin{array}{ll}1 & 0\end{array}\right)\left(\begin{array}{ll}1 & 2 \\ 4 & 1\end{array}\right)^{n}$ ，又
$$
\left|\lambda I-\left(\begin{array}{ll}
1 & 2 \\
4 & 1
\end{array}\right)\right|=(\lambda-1+2 \sqrt{2})(\lambda-1-2 \sqrt{2})=0
$$
得 $\lambda_{1}=1-2 \sqrt{2}, \lambda_{2}=1+2 \sqrt{2}$ ，易解得对应的特征向量分别为 $\boldsymbol{\eta}_{1}=\binom{-1}{\sqrt{2}}, \boldsymbol{\eta}_{2}=\binom{1}{\sqrt{2}}$ ，故
$$
\left(\begin{array}{ll}
1 & 2 \\
4 & 1
\end{array}\right)=\left(\begin{array}{cc}
-1 & 1 \\
\sqrt{2} & \sqrt{2}
\end{array}\right)\left(\begin{array}{cc}
1-2 \sqrt{2} & 0 \\
0 & 1+2 \sqrt{2}
\end{array}\right)\left(\begin{array}{cc}
-1 & 1 \\
\sqrt{2} & \sqrt{2}
\end{array}\right)^{-1}
$$
所以，
$$
\left(\begin{array}{ll}
x_{100} & y_{100}
\end{array}\right)=\left(\begin{array}{ll}
1 & 0
\end{array}\right)\left(\begin{array}{cc}
-1 & 1 \\
\sqrt{2} & \sqrt{2}
\end{array}\right)\left(\begin{array}{cc}
(1-2 \sqrt{2})^{100} & 0 \\
0 & (1+2 \sqrt{2})^{100}
\end{array}\right)\left(\begin{array}{cc}
-1 & 1 \\
\sqrt{2} & \sqrt{2}
\end{array}\right)^{-1}
$$
即
$$
\left(\begin{array}{ll}
x_{100} & y_{100}
\end{array}\right)=\left(\frac{1}{2}\left[(1-2 \sqrt{2})^{100}+(1+2 \sqrt{2})^{100}\right] \quad \frac{\sqrt{2}}{4}\left[(1+2 \sqrt{2})^{100}-(1-2 \sqrt{2})^{100}\right]\right)
$$
📋 详细解题步骤
步骤 1/1
目标：1. 将递推写成向量形式 (x_{n+1}, y_{n+1}) = (x_n, y_n) M；2. 求M的特征值和特征向量；3. 将M对角化：M=PDP^{-1}；4. 计算M^n，代回得x_{100}, y_{100}。
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