Matrix to power $2012$

How to calculate $A^{2012}$?

$A = \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$

How can one calculate this? It must be tricky or something, cause there was only 1 point for solving this.

Solution 1:

Observe that $A^2$ is $A$.

So $A^{\large2012}$ is $A$ too.

Solution 2:

$$A = \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$$ $$A^2=\left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]\cdot \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$$

$$A^2 = \left[\begin{array}{ccc}{9-2-4}&{-3+2}&{-6+2+2}\\{6-4}&{-2+2}&{-4+2}\\{6-2}&{-2+1}&{-4+2+1}\end{array}\right]$$ $$A^2 = \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$$ $$A^2=A$$

$$A=A^2=A^3=\cdots =A^n,n\in N$$

so $$A^{2012}=A$$