$$A^n=\left(\sum_{i=1}^n a_i b_i\right)^{n-1} A$$

【参考证明】：按题意可设 $A=\left(\begin{array}{cccc}a_1 b_1 & a_1 b_2 & \cdots & a_1 b_n \\ a_2 b_1 & a_2 b_2 & \cdots & a_2 b_n \\ \vdots & \vdots & \ddots & \vdots \\ a_n b_1 & a_n b_2 & \cdots & a_n b_n\end{array}\right)$ ， 因\begin{aligned}
& A=\left(\begin{array}{cccc}
a_1 b_1 & a_1 b_2 & \cdots & a_1 b_n \\
a_2 b_1 & a_2 b_2 & \cdots & a_2 b_n \\
\vdots & \vdots & \ddots & \vdots \\
a_n b_1 & a_n b_2 & \cdots & a_n b_n
\end{array}\right)=\left(\begin{array}{c}
a_1 \\
a_2 \\
\vdots \\
a_n
\end{array}\right)\left(b_1, b_2, \cdots, b_n\right) \\
& =\boldsymbol{\alpha} \boldsymbol{\beta}^T \\
& \text { 所以 } \boldsymbol{A}^n=\left(\boldsymbol{\alpha} \boldsymbol{\beta}^T\right)^n=\left(\boldsymbol{\alpha} \boldsymbol{\beta}^T\right)\left(\boldsymbol{\alpha} \boldsymbol{\beta}^T\right) \cdots\left(\boldsymbol{\alpha} \boldsymbol{\beta}^T\right) \\
& =\boldsymbol{\alpha} \boldsymbol{\beta}^T \boldsymbol{\alpha} \boldsymbol{\beta}^T \cdots \boldsymbol{\alpha} \boldsymbol{\beta}^T=\alpha \underbrace{\left(\boldsymbol{\beta}^T \boldsymbol{\alpha}\right)\left(\boldsymbol{\beta}^T \boldsymbol{\alpha}\right) \cdots\left(\boldsymbol{\beta}^T \boldsymbol{\alpha}\right)}_{n-1 \uparrow} \boldsymbol{\beta}^T \\
& =\underbrace{\left(\boldsymbol{\beta}^T \alpha\right)\left(\boldsymbol{\beta}^T \alpha\right) \cdots\left(\boldsymbol{\beta}^T \alpha\right)}_{n-1 \uparrow} \alpha \boldsymbol{\beta}^T=\left(\left(b_1, b_2, \cdots, b_n\right)\left(\begin{array}{c}
a_1 \\
a_2 \\
\vdots \\
a_n
\end{array}\right)\right)^{n-1} \boldsymbol{A} \\
& =\left(\sum_{i=1}^n a_i b_i\right)^{n-1} A \\
&
\end{aligned}

【参考解答】：按方法 2，当然可以先尝试着计算 $A^2$ ，可得:
\begin{aligned} & A^2=\left(\begin{array}{cccc} 2 & -1 & -2 & 3 \\ -4 & 2 & 4 & -6 \\ 0 & 0 & 0 & 0 \\ 6 & -3 & -6 & 9 \end{array}\right)\left(\begin{array}{cccc} 2 & -1 & -2 & 3 \\ -4 & 2 & 4 & -6 \\ 0 & 0 & 0 & 0 \\ 6 & -3 & -6 & 9 \end{array}\right) \\ & =\left(\begin{array}{cccc} 26 & -9 & -26 & 39 \\ -52 & 26 & 52 & -78 \\ 0 & 0 & 0 & 0 \\ 78 & -39 & -78 & 117 \end{array}\right)=13\left(\begin{array}{cccc} 2 & -1 & -2 & 3 \\ -4 & 2 & 4 & -6 \\ 0 & 0 & 0 & 0 \\ 6 & -3 & -6 & 9 \end{array}\right) \\ & =13 A \end{aligned}

$$A^n=13^{n-1} A=13^{n-1}\left(\begin{array}{cccc} 2 & -1 & -2 & 3 \\ -4 & 2 & 4 & -6 \\ 0 & 0 & 0 & 0 \\ 6 & -3 & -6 & 9 \end{array}\right) .$$

$$A=\left(\begin{array}{cccc} 2 & -1 & -2 & 3 \\ -4 & 2 & 4 & -6 \\ 0 & 0 & 0 & 0 \\ 6 & -3 & -6 & 9 \end{array}\right)=\left(\begin{array}{c} 1 \\ -2 \\ 0 \\ 3 \end{array}\right)(2,-1,-2,3)=\alpha \beta^T$$

$$A^n=\left((2,-1,-2,3)\left(\begin{array}{c} 1 \\ -2 \\ 0 \\ 3 \end{array}\right)\right)^{n-1} A=13^{n-1}\left(\begin{array}{cccc} 2 & -1 & -2 & 3 \\ -4 & 2 & 4 & -6 \\ 0 & 0 & 0 & 0 \\ 6 & -3 & -6 & 9 \end{array}\right)$$