Finding Person 1's values whilst only knowing Person 7's using iteration The general formula is $a_{n+1}=a_n-\sqrt{3}b_n$ and $b_{n+1}=\sqrt{3}a_n+b_n$.
Person 7's numbers are $a_7=8$ and $b_7=32$.
Firstly, I tried to solve it without iteration and just kept changing my values until I reached Person 1's numbers, but then I realised that I had probably got it wrong, and it was taking way too long. So afterwards, I attempted to create a general equation that had all values for $a$ and $b$ from 1 to 7. I'm not sure if that was a useless step, as I am unable to go anywhere with this information.
$$32-8\sqrt{3}=\sqrt{3}(a_1+a_2+a_3+a_4+a_5+a_6)+3(b_2+b_3+b_4+ b_5+ b_6)+4b_1$$
The above was my final step before I was at a loss and decided to come here for help.
 A: How about solving for $a_1$ and $b_1$ from the following equation system: \begin{equation} \begin{bmatrix} a_{n+1} \\ b_{n+1} \end{bmatrix} = \begin{bmatrix} 1 & -\sqrt{3} \\ \sqrt{3} & 1 \end{bmatrix}\begin{bmatrix} a_{n} \\ b_{n} \end{bmatrix} \text{ for } n = 1, 2, 3, \cdots? \end{equation}
Then you can find out $\begin{bmatrix} 1 & -\sqrt{3} \\ \sqrt{3} & 1 \end{bmatrix}^{-6}$ and plug $a_{6+1=7} = 8, b_{6+1=7} = 32$ to find \begin{equation} \begin{bmatrix} a_{1} \\ b_{1} \end{bmatrix} = \begin{bmatrix} 1 & -\sqrt{3} \\ \sqrt{3} & 1 \end{bmatrix}^{-6} \begin{bmatrix} a_{7} \\ b_{7} \end{bmatrix} = \begin{bmatrix} 1/64 & 0 \\ 0 & 1/64 \end{bmatrix} \begin{bmatrix} 8 \\ 32 \end{bmatrix} = \cdots  \end{equation}
Of course you need to know how to find the (true) inverse of a square matrix to understand how the $1/64$ appears.
A: (This process works because of how unique the numbers are)
Consider the complex number $ c_n = a_n + i b_n$.
Find a relationship between $ c_n$ and $ c_{n+1}$.

 $ c_{n+1} = c_n \times 2 e^{ \frac{i \pi } { 3 } }$.

Hence, find a relationship between $c_1$ and $c_7$.

 $c_7 = 2^6 c_1$.

Hence, find $ a_1, b_1$.
