Looking ahead

As David observed, it suffices to find a matrix $A$ with $A^3=I_2$ , since $A^{2019}=(A^3)^{673}$ . Since $A^3-I_2=(A-I_2)(A^2+A+I_2)$ , it suffices to find a matrix with $A^2+A+I_2=0$ . By Hamilton-Cayley, it suffices to find a matrix with $tr(A)=-1$ and $det(A)=1$ (and $a=2019$ , of course). Such a matrix can be constructed by inspection, for example, $A=\begin{bmatrix} 2019& 2019\times 2020 +1\\-1 & -2020 \end{bmatrix}$ .

(This proof can be presented more elegantly in terms of complex eigenvalues, but I attempted to keep things as basic as possible.)

Cubing the rotation matrix $R = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ for $\theta = \frac{2 \pi}{3}$ will give the identity matrix $I_2$ . In other words, $B^3 = I_2$ for $B = \begin{bmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{bmatrix}$ .

This also means that $(CBC^{-1})^3 = I_2$ for any invertible matrix $C = \begin{bmatrix} a' & b' \\ c' & d' \end{bmatrix}$ .

Now, if $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = CBC^{-1}$ , then ${A}^3 = I_2$ , which in turn means ${A}^{2019} = ({A}^{3})^{673} = (I_2)^{673} = I_2$ . All we need now is to find values for $a'$ , $b'$ , $c'$ , and $d'$ such that $a = 2019$ , and the statement will be proved true.

Using $C^{-1} = \begin{bmatrix} \frac{d'}{a'd' - b'c'} & \frac{-b'}{a'd' - b'c'} \\ \frac{-c'}{a'd' - b'c'} & \frac{a'}{a'd' - b'c'} \end{bmatrix}$ and a little bit of matrix multiplication, we find that $a = \frac{1}{a'd' - b'c'}(-\frac{1}{2}a'd' + \frac{\sqrt{3}}{2}b'd' + \frac{\sqrt{3}}{2}a'c' + \frac{1}{2}b'c')$ , $b = \frac{1}{a'd' - b'c'}(\frac{1}{2}a'b' - \frac{\sqrt{3}}{2}b'^2 - \frac{\sqrt{3}}{2}a'^2 - \frac{1}{2}a'b')$ , $c = \frac{1}{a'd' - b'c'}(-\frac{1}{2}c'd' + \frac{\sqrt{3}}{2}b'^2 + \frac{\sqrt{3}}{2}c'^2 + \frac{1}{2}c'd')$ , and $d = \frac{1}{a'd' - b'c'}(\frac{1}{2}b'c' - \frac{\sqrt{3}}{2}b'd' - \frac{\sqrt{3}}{2}a'c' - \frac{1}{2}a'd')$ .

There are several real solutions to obtain $a = 2019$ , one being $a' = 1$ , $b' = \frac{4039\sqrt{3}}{3}$ , $c' = 0$ , and $d' = 1$ . These values give us $b = \frac{8156762 \sqrt{3}}{3}$ , $c = \frac{\sqrt{3}}{2}$ , and $d = -2020$ .

Therefore, $\begin{bmatrix} 2019 & \frac{8156762 \sqrt{3}}{3} \\ \frac{\sqrt{3}}{2} & -2020 \end{bmatrix}^{2019} = I_2$ , which is sufficient to show that the statement is true .