题目
(2024,1)已知数列x_{n)},y_{n)},z_{n)}满足x_(0)=-1,y_(0)=0,z_(0)=2,且}x_(n)=-2x_(n-1)+2z_(n-1),y_(n)=-2y_(n-1)-2z_(n-1),z_(n)=-6x_(n-1)-3y_(n-1)+3z_(n-1).记alpha_(n)=}x_(n)y_(n)z_(n)(n=1,2,···).
(2024,1)已知数列$\{x_{n}\}$,$\{y_{n}\}$,$\{z_{n}\}$满足$x_{0}=-1$,$y_{0}=0$,$z_{0}=2$,且
$\begin{cases}x_{n}=-2x_{n-1}+2z_{n-1},\\y_{n}=-2y_{n-1}-2z_{n-1},\\z_{n}=-6x_{n-1}-3y_{n-1}+3z_{n-1}.\end{cases}$
记$\alpha_{n}=\begin{pmatrix}x_{n}\\y_{n}\\z_{n}\end{pmatrix}$,写出满足$\alpha_{n}=A\alpha_{n-1}$的矩阵A,并求$A^{n}$及$x_{n},y_{n},z_{n}(n=1,2,···)$.
题目解答
答案
1. **确定矩阵 $A$**:
根据递推关系,得
\[
A = \begin{pmatrix} -2 & 0 & 2 \\ 0 & -2 & -2 \\ -6 & -3 & 3 \end{pmatrix}.
\]
2. **求特征值和特征向量**:
特征值为 $\lambda_1 = -2$, $\lambda_2 = 0$, $\lambda_3 = 1$。
3. **对角化**:
令 $P = \begin{pmatrix} 1 & 1 & 2 \\ -2 & -1 & -2 \\ 0 & 1 & 3 \end{pmatrix}$,计算 $P^{-1}$ 后得
\[
A^n = PD^nP^{-1}.
\]
4. **答案**:
\[
\boxed{
\begin{cases}
x_n = \frac{1}{3} \left[ 3^n - (-2)^n \right], \\
y_n = \frac{2}{3} \left[ (-2)^n - 3^n \right], \\
z_n = 3^n - \frac{2}{3} (-2)^n \end{cases}
\]
5. **总结**:
�{2024,1)
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\begin{cases}
x_n = \frac{1}{3} \left[ 3^n - (-2)^n \right], \\
y_n = \frac{2}{3} \left[ (-2)^n - 3^n \right], \\
z_n = 3^n - \frac{2}{3} (-2)^n \end{cases}
\]
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