2023《线性代数》方阵n次方计算方法总结与典型例题求解-Kmath科数- 中考、高考、考验数学题库

设 $A=\left(\begin{array}{cc}3 & 4 \\ 4 & -3\end{array}\right)$ ，求 $A^n$.

答案：

先计算 $A^2, A^3, A^4, A^5$ ，找规律.
$$
\begin{aligned}
& A^2=A A=\left(\begin{array}{cc}
3 & 4 \\
4 & -3
\end{array}\right)\left(\begin{array}{cc}
3 & 4 \\
4 & -3
\end{array}\right)=\left(\begin{array}{cc}
25 & 0 \\
0 & 25
\end{array}\right)=\left(\begin{array}{cc}
5^2 & 0 \\
0 & 5^2
\end{array}\right) \\
& A^3=A^2 A=\left(\begin{array}{cc}
5^2 & 0 \\
0 & 5^2
\end{array}\right)\left(\begin{array}{cc}
3 & 4 \\
4 & -3
\end{array}\right)=\left(\begin{array}{cc}
5^2 \times 3 & 5^2 \times 4 \\
5^2 \times 4 & -5^2 \times 3
\end{array}\right) \\
& A^4=A^3 A=\left(\begin{array}{cc}
5^2 \times 3 \\
5^2 \times 4 \\
5^2 \times 4 \\
4 & -3
\end{array}\right)=\left(\begin{array}{cc}
5^4 & 0 \\
0 & 5^4 \\
0
\end{array}\right) \\
& A^5=A^4 A=\left(\begin{array}{cc}
5^4 & 0 \\
0 & 5^4
\end{array}\right)\left(\begin{array}{cc}
3 & 4 \\
4 & -3
\end{array}\right)=\left(\begin{array}{cc}
5^4 \times 3 & 5^4 \times 4 \\
5^4 \times 4 & -5^4 \times 3
\end{array}\right)
\end{aligned}
$$

如果将它们放一起观察，没有明显规律，但如果将它们的奇偶分开放一起找规律，即
$$
\begin{aligned}
& A^2=\left(\begin{array}{cc}
5^2 & 0 \\
0 & 5^2
\end{array}\right), A^4=\left(\begin{array}{cc}
5^4 & 0 \\
0 & 5^4
\end{array}\right), \cdots, A^{2 k}=\left(\begin{array}{cc}
5^{2 k} & 0 \\
0 & 5^{2 k}
\end{array}\right) \\
& A^3=\left(\begin{array}{cc}
5^2 \times 3 & 5^2 \times 4 \\
5^2 \times 4 & -5^2 \times 3
\end{array}\right), A^5=\left(\begin{array}{cc}
5^4 \times 3 & 5^4 \times 4 \\
5^4 \times 4 & -5^4 \times 3
\end{array}\right) \\
& \cdots, A^{2 k+1}=\left(\begin{array}{cc}
5^{2 k} \times 3 & 5^{2 k} \times 4 \\
5^{2 k} \times 4 & -5^{2 k} \times 3
\end{array}\right)
\end{aligned}
$$

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