题号:
6283
题型:
解答题
来源:
2023《线性代数》方阵n次方计算方法总结与典型例题求解
设 $A=\left(\begin{array}{llll}0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 3 \\ 3 & 2 & 1 & 0\end{array}\right)$ ，求 $A^n(n \geq 1)$.
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答案：
答案：
先计算几项找规律
$$
\begin{aligned}
& A^2=A A=\left(\begin{array}{llll}
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 2 \\
0 & 0 & 0 & 3 \\
3 & 2 & 1 & 0
\end{array}\right)\left(\begin{array}{llll}
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 2 \\
0 & 0 & 0 & 3 \\
3 & 2 & 1 & 0
\end{array}\right)=\left(\begin{array}{cccc}
3 & 2 & 1 & 0 \\
6 & 4 & 2 & 0 \\
9 & 6 & 3 & 0 \\
0 & 0 & 0 & 10
\end{array}\right) \\
& A^3=A^2 A=\left(\begin{array}{cccc}
3 & 2 & 1 & 0 \\
6 & 4 & 2 & 0 \\
9 & 6 & 3 & 0 \\
0 & 0 & 0 & 10
\end{array}\right)\left(\begin{array}{llll}
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 2 \\
0 & 0 & 0 & 3 \\
3 & 2 & 1 & 0
\end{array}\right)=\left(\begin{array}{cccc}
0 & 0 & 0 & 10 \\
0 & 0 & 0 & 20 \\
0 & 0 & 0 & 30 \\
30 & 20 & 10 & 0
\end{array}\right) \\
& A^4=A^3 A=\left(\begin{array}{cccc}
0 & 0 & 0 & 10 \\
0 & 0 & 0 & 20 \\
0 & 0 & 0 & 30 \\
30 & 20 & 10 & 0
\end{array}\right)\left(\begin{array}{cccc}
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 2 \\
0 & 0 & 0 & 3 \\
3 & 2 & 1 & 0
\end{array}\right)=\left(\begin{array}{cccc}
30 & 20 & 10 & 0 \\
60 & 40 & 20 & 0 \\
90 & 60 & 30 & 0 \\
0 & 0 & 0 & 100
\end{array}\right) \\
&
\end{aligned}
$$
$$
\begin{aligned}
A^5=A^4 A & =\left(\begin{array}{cccc}
30 & 20 & 10 & 0 \\
60 & 40 & 20 & 0 \\
90 & 60 & 30 & 0 \\
0 & 0 & 0 & 100
\end{array}\right)\left(\begin{array}{llll}
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 2 \\
0 & 0 & 0 & 3 \\
3 & 2 & 1 & 0
\end{array}\right) \\
& =\left(\begin{array}{cccc}
0 & 0 & 0 & 100 \\
0 & 0 & 0 & 200 \\
0 & 0 & 0 & 300 \\
300 & 200 & 100 & 0
\end{array}\right) \\
A^6=A^5 A & =\left(\begin{array}{cccc}
0 & 0 & 0 & 100 \\
0 & 0 & 0 & 200 \\
0 & 0 & 0 & 300 \\
300 & 200 & 100 & 0
\end{array}\right)\left(\begin{array}{llll}
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 2 \\
0 & 0 & 0 & 3 \\
3 & 2 & 1 & 0
\end{array}\right) \\
& =\left(\begin{array}{cccc}
300 & 200 & 100 & 0 \\
600 & 400 & 200 & 0 \\
900 & 600 & 300 & 0 \\
0 & 0 & 0 & 1000
\end{array}\right)
\end{aligned}
$$
进行奇偶分类并找到规律:
$$
\begin{aligned}
A^1 & =\left(\begin{array}{llll}
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 2 \\
0 & 0 & 0 & 3 \\
3 & 2 & 1 & 0
\end{array}\right), A^3=\left(\begin{array}{cccc}
0 & 0 & 0 & 1 \times 10^1 \\
0 & 0 & 0 & 2 \times 10^1 \\
0 & 0 & 0 & 3 \times 10^1 \\
3 \times 10^1 & 2 \times 10^1 & 1 \times 10^1 & 0
\end{array}\right) \\
A^5 & =\left(\begin{array}{cccc}
0 & 0 & 0 & 1 \times 10^2 \\
0 & 0 & 0 & 2 \times 10^2 \\
0 & 0 & 0 & 3 \times 10^2 \\
3 \times 10^2 & 2 \times 10^2 & 1 \times 10^2 & 0
\end{array}\right)
\end{aligned}
$$
$$
, \cdots, A^{2 k+1}=\left|\begin{array}{cccc}
0 & 0 & 0 & 1 \times 10^k \\
0 & 0 & 0 & 2 \times 10^k \\
0 & 0 & 0 & 3 \times 10^k \\
3 \times 10^k & 2 \times 10^k & 1 \times 10^k & 0
\end{array}\right|
$$
$$
\begin{aligned}
& A^2=\left(\begin{array}{cccc}
3 & 2 & 1 & 0 \\
6 & 4 & 2 & 0 \\
9 & 6 & 3 & 0 \\
0 & 0 & 0 & 10^1
\end{array}\right), A^4=\left(\begin{array}{cccc}
3 \times 10^1 & 2 \times 10^1 & 1 \times 10^1 & 0 \\
6 \times 10^1 & 4 \times 10^1 & 2 \times 10^1 & 0 \\
9 \times 10^1 & 6 \times 10^1 & 3 \times 10^1 & 0 \\
0 & 0 & 0 & 10^2
\end{array}\right) \\
& A^6=\left(\begin{array}{cccc}
3 \times 10^2 & 2 \times 10^2 & 1 \times 10^2 & 0 \\
6 \times 10^2 & 4 \times 10^2 & 2 \times 10^2 & 0 \\
9 \times 10^2 & 6 \times 10^2 & 3 \times 10^2 & 0 \\
0 & 0 & 0 & 10^3
\end{array}\right) \\
&, \cdots, A^{2 k}=\left(\begin{array}{cccc}
3 \times 10^{k-1} & 2 \times 10^{k-1} & 1 \times 10^{k-1} & 0 \\
6 \times 10^{k-1} & 4 \times 10^{k-1} & 2 \times 10^{k-1} & 0 \\
9 \times 10^{k-1} & 6 \times 10^{k-1} & 3 \times 10^{k-1} & 0 \\
0 & 0 & 0 & 10^k
\end{array}\right)
\end{aligned}
$$
由此可以猜想出，当 $n$ 为奇数 $(n \geq 1)$ 时，有
$$
A^n=\left(\begin{array}{cccc}
0 & 0 & 0 & 10^{\frac{n-1}{2}} \\
0 & 0 & 0 & 2 \cdot 10^{\frac{n-1}{2}} \\
0 & 0 & 0 & 3 \cdot 10^{\frac{n-1}{2}} \\
3 \cdot 10^{\frac{n-1}{2}} & 2 \cdot 10^{\frac{n-1}{2}} & 10^{\frac{n-1}{2}} & 0
\end{array}\right) .
$$
当 $n$ 为偶数 $(n \geq 2)$ 时，有
$$
A^n=\left(\begin{array}{cccc}
3 \cdot 10^{\frac{n}{2}-1} & 2 \cdot 10^{\frac{n}{2}-1} & 10^{\frac{n}{2}-1} & 0 \\
6 \cdot 10^{\frac{n}{2}-1} & 4 \cdot 10^{\frac{n}{2}-1} & 2 \cdot 10^{\frac{n}{2}-1} & 0 \\
9 \cdot 10^{\frac{n}{2}-1} & 6 \cdot 10^{\frac{n}{2}-1} & 3 \cdot 10^{\frac{n}{2}-1} & 0 \\
0 & 0 & 0 & 10^{\frac{n}{2}}
\end{array}\right) .
$$
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