Tomi's Challenges - #1

Let $A$ be $\begin{bmatrix} 2 & -2\sqrt{3}\\ 2\sqrt{3} & 2 \end{bmatrix}^{2021}$ . $\newline$ The most natural way to solve this problem is to think about what $A$ does to the vector's length. $\newline$ Notice that $A$ looks suspiciously like the rotation matrix $\newline$

$\begin{bmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta) \end{bmatrix}$ $\newline$

Let's rewrite $A$ . $\newline$ $\begin{bmatrix} 2 & -2\sqrt{3}\\ 2\sqrt{3} & 2 \end{bmatrix}^{2021}=4^{2021}\begin{bmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix}^{2021}$ $\newline$ Notice that $\newline$ $\begin{bmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix}^{2021}=\begin{bmatrix} \cos(60°) & -\sin(60°)\\ \sin(60°) & \cos(60°) \end{bmatrix}^{2021}$ $\newline$

So we know that this matrix rotates a vector by $60°$ 2021 times. Since rotating a vector has no effect on its length, the length of the vector which is being multiplied by the matrix is 1, the length of $y$ is $\newline$

$\left |4^{2021}\begin{bmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix}^{2021}\begin{bmatrix} \frac{2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}} \end{bmatrix} \right | = \left | 4^{2021}\begin{bmatrix} \frac{2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}} \end{bmatrix} \right | =4^{2021}\times{1}=4^{2021}$ . $\newline$ Powers of 4 with positive exponents always end with 4 or 6. For odd exponents, they end with $4$ , and for even exponents they end with $6$ . So the length of the vector ends with $4$ .