Q:

Confused about finding non-brute force way to solve for matrix to the 2019th power

I am attempting to solve this problem, it has four parts. I solved part a (a trivial matrix problem), but the next three parts appear to be a bit confusing to me. I just would like some help getting started so I can see and observe this matrix and come up with a solution. 
The Questions Note:
The idea here is NOT to use brute force computation to get $A^{2019}$ matrix, instead use some observations which can significantly reduce computational work and will also give you an insight into such problems.
Obviously this is some huge numbered matrix, but I do not understand what observation will reduce this? My first thought was just to use a calculator and calculate $A^{2019}$ and then just multiply that by each vector. But that appears to be not the point of the question. 
$$
Let A = \left[\begin{array}{rrr}
  -4 & -6 & -12 \\
   -2 & -1 & -4 \\
  2 & 3 & 6 
 \end{array}\right]
$$
And Let $u$ = [6 5 -3], $v$ = [-2 0 1], and $w$ = [-2 -1 1].
b). Compute $A^{2019}\mathbb v$

A:

I assume that you're asking about $A^{2019}v$.
To that end, observe that $Av = 2v$ (I'll leave it to you to verify that this is true).  
It follows that 
$$
A^2v = AAv = A(2v) = 2Av = 4v.
$$
Similarly, $A^3 v = 8v$, and in general we have $A^k v = 2^k v$ for any positive integer $k$.

