题目
编号：A2016223
已知矩阵 $A = \begin{bmatrix}
0 & -1 & 1\\
2 & -3 & 0\\
0 & 0 & 0
\end{bmatrix}$.
$(Ⅰ)$ 求 $A^{99}$;
$(Ⅱ)$ 设 $3$ 阶矩阵 $B=(\alpha_{1}, \alpha_{2}, \alpha_{3})$ 满足 $B^{2} = BA$. 记 $B^{100} = (\beta_{1}, \beta_{2}, \beta_{3})$, 将 $\beta_{1}$, $\beta_{2}$, $\beta_{3}$ 分别表示为 $\alpha_{1}$, $\alpha_{2}$, $\alpha_{3}$ 的线性组合.
解析
第 $(Ⅰ)$ 问
第 $(Ⅰ)$ 问要求解 $A^{99}$, 很显然，我们不能真的做 $99$ 次矩阵 $A$ 与矩阵 $A$ 的乘法运算，而且，矩阵 $A$ 也不可能在进行少数几次与自己的乘法运算之后就变成零矩阵或单位矩阵等特殊矩阵，因此，我们只能借助于矩阵的 $A$ 的相似对角矩阵来求解 $A^{99}$.
由题可知：
$$
|\lambda E – A| = 0 \Rightarrow
$$
$$
{\color{Red}
\begin{bmatrix}
\lambda & 1 & -1\\
-2 & \lambda+3 & 0\\
0 & 0 & \lambda
\end{bmatrix} = 0}. \quad {\color{White}①}
\Rightarrow
$$
$$
\lambda^{2} (\lambda + 3) + 2 \lambda = 0 \Rightarrow
$$
$$
\lambda[\lambda (\lambda + 3) + 2] = 0 \Rightarrow
$$
$$
\lambda(\lambda^{2} + 3 \lambda + 2) = 0 \Rightarrow
$$
$$
\left\{\begin{matrix}
\lambda_{1} = 0;\\
\lambda_{2} = -1;\\
\lambda_{3} = -2.
\end{matrix}\right.
$$
又由 $(\lambda_{i} E – A)x = O$ 可知：
当 $\lambda_{1} = 0$ 时，由上面的 $①$ 式可得：
$$
\lambda_{1} E – A \Rightarrow
$$
$$
\begin{bmatrix}
\lambda_{1} & 1 & -1\\
-2 & \lambda_{1}+3 & 0\\
0 & 0 & \lambda_{1}
\end{bmatrix}
\Rightarrow
$$
$$
\begin{bmatrix}
0 & 1 & -1\\
-2 & 3 & 0\\
0 & 0 & 0
\end{bmatrix}
\Rightarrow
\begin{bmatrix}
0 & 1 & -1\\
-2 & 0 & 3\\
0 & 0 & 0
\end{bmatrix} \Rightarrow
$$
$$
X_{1} =
\begin{bmatrix}
\frac{3}{2}\\
1\\
1
\end{bmatrix}
\Rightarrow
\begin{bmatrix}
3\\
2\\
2
\end{bmatrix}.
$$
当 $\lambda_{2} = -1$ 时，由上面的 $①$ 式可得：
$$
\lambda_{2} E – A \Rightarrow
$$
$$
\begin{bmatrix}
\lambda_{2} & 1 & -1\\
-2 & \lambda_{2}+3 & 0\\
0 & 0 & \lambda_{2}
\end{bmatrix}
\Rightarrow
$$
$$
\begin{bmatrix}
-1 & 1 & -1\\
-2 & 2 & 0\\
0 & 0 & -1
\end{bmatrix}
\Rightarrow
\begin{bmatrix}
1 & -1 & 1\\
0 & 0 & -2\\
0 & 0 & 0
\end{bmatrix} \Rightarrow
$$
$$
X_{2} =
\begin{bmatrix}
1\\
1\\
0
\end{bmatrix}.
$$
当 $\lambda_{3} = -2$ 时，由上面的 $①$ 式可得：
$$
\lambda_{3} E – A \Rightarrow
$$
$$
\begin{bmatrix}
\lambda_{3} & 1 & -1\\
-2 & \lambda_{3}+3 & 0\\
0 & 0 & \lambda_{3}
\end{bmatrix}
\Rightarrow
$$
$$
\begin{bmatrix}
-2 & 1 & -1\\
-2 & 1 & 0\\
0 & 0 & -2
\end{bmatrix}
\Rightarrow
\begin{bmatrix}
-2 & 1 & -1\\
0 & 0 & 1\\
0 & 0 & 0
\end{bmatrix} \Rightarrow
$$
$$
X_{3} =
\begin{bmatrix}
\frac{1}{2}\\
1\\
0
\end{bmatrix}
\Rightarrow
\begin{bmatrix}
1\\
2\\
0
\end{bmatrix}.
$$
综上，若令 $P = (X_{1}, X_{2}, X_{3}) =$ $\begin{bmatrix}
3 & 1 & 1\\
2 & 1 & 2\\
2 & 0 & 0
\end{bmatrix}$, $\Lambda = \begin{bmatrix}
\lambda_{1} & & \\
& \lambda_{2} & \\
& & \lambda_{3}
\end{bmatrix} =$ $\begin{bmatrix}
0 & & \\
& -1 & \\
& & -2
\end{bmatrix}$, 则：
$$
P^{-1} A P = \Lambda \Rightarrow
$$
$$
A = P \Lambda P^{-1}.
$$
又：
$$
(P \vdots E) \Rightarrow
$$
$$
\begin{bmatrix}
3 & 1 & 1 & \vdots & 1 & 0 & 0\\
2 & 1 & 2 & \vdots & 0 & 1 & 0\\
2 & 0 & 0 & \vdots & 0 & 0 & 1
\end{bmatrix}
\Rightarrow
$$
$$
(E \vdots P^{-1}) =
\begin{bmatrix}
1 & 0 & 0 & \vdots & 0 & 0 & \frac{1}{2}\\
0 & 1 & 0 & \vdots & 2 & -1 & -2\\
0 & 0 & 1 & \vdots & -1 & 1 & \frac{1}{2}
\end{bmatrix}.
$$
接着：
$$
A = P \Lambda P^{-1} \Rightarrow
$$
$$
A^{99} = (P \Lambda P^{-1})^{99} \Rightarrow
$$
$$
A^{99} = (P \Lambda P^{-1})(P \Lambda P^{-1}) \cdots (P \Lambda P^{-1}) \cdots _{99个P \Lambda P^{-1}} \Rightarrow
$$
$$
A^{99} = P \Lambda (P^{-1} P) \Lambda (P^{-1} \cdots P) \Lambda P^{-1} \cdots _{99个P \Lambda P^{-1}} \Rightarrow
$$
$$
A^{99} = P \Lambda ^{99} P^{-1}.
$$
于是：
$$
A^{99} =
$$
$$
\begin{bmatrix}
3 & 1 & 1\\
2 & 1 & 2\\
2 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
0 & & \\
& (-1)^{99} & \\
& & (-2)^{99}
\end{bmatrix}
\begin{bmatrix}
0 & 0 & \frac{1}{2}\\
2 & -1 & -2\\
-1 & 1 & \frac{1}{2}.
\end{bmatrix}
\Rightarrow
$$
$$
\begin{bmatrix}
3 & 1 & 1\\
2 & 1 & 2\\
2 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
0 & & \\
& -1 & \\
& & -2^{99}
\end{bmatrix}
\begin{bmatrix}
0 & 0 & \frac{1}{2}\\
2 & -1 & -2\\
-1 & 1 & \frac{1}{2}.
\end{bmatrix}
\Rightarrow
$$
$$
\begin{bmatrix}
0 & -1 & -2^{99}\\
0 & -1 & -2^{100}\\
0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
0 & 0 & \frac{1}{2}\\
2 & -1 & -2\\
-1 & 1 & \frac{1}{2}.
\end{bmatrix}
\Rightarrow
$$
$$
\begin{bmatrix}
-2+2^{99} & 1-2^{99} & 2-2^{98}\\
-2+2^{100} & 1-2^{100} & 2-2^{99}\\
0 & 0 & 0
\end{bmatrix}.
$$
第 $(Ⅱ)$ 问
由题可知：
$$
B^{2} = BA \Rightarrow
$$
$$
B^{100} = B^{98}B^{2} \Rightarrow
$$
$$
B^{100} = B^{98}BA \Rightarrow
$$
$$
B^{100} = B^{99}A \Rightarrow
$$
$$
B^{100} = B^{97}B^{2}A \Rightarrow
$$
$$
B^{100} = B^{97}BAA \Rightarrow
$$
$$
B^{100} = B^{98}A^{2} \Rightarrow
$$
$$
\cdots \cdots
$$
$$
B^{100} = BA^{99} \Rightarrow
$$
$$
B^{100} = (\alpha_{1}, \alpha_{2}, \alpha_{3}) A^{99} \Rightarrow
$$
$$
(\beta_{1}, \beta_{2}, \beta_{3}) = (\alpha_{1}, \alpha_{2}, \alpha_{3}) A^{99} \Rightarrow
$$
$$
(\beta_{1}, \beta_{2}, \beta_{3}) = (\alpha_{1}, \alpha_{2}, \alpha_{3}) \begin{bmatrix}
-2+2^{99} & 1-2^{99} & 2-2^{98}\\
-2+2^{100} & 1-2^{100} & 2-2^{99}\\
0 & 0 & 0
\end{bmatrix} \Rightarrow
$$
$$
\left\{\begin{matrix}
\beta_{1} = (-2+2^{99}) \alpha_{1} + (-2+2^{100}) \alpha_{2} + 0 \alpha_{3};\\
\beta_{2} = (1-2^{99}) \alpha_{1} + ( 1-2^{100}) \alpha_{2} + 0 \alpha_{3};\\
\beta_{3} = (2-2^{98}) \alpha_{1} + (2-2^{99}) \alpha_{2} + 0 \alpha_{3}.
\end{matrix}\right.
$$