Given this matrix $A$, find $A^{144}$ I know this is a very common question to ask when one is making their way into Linear Algebra (i.e. given a matrix, find the result of that matrix to the nth-power).
I'm given this matrix: $$
   A=
  \left[ {\begin{array}{cc}
   0 & -1 \\
   1 & 0 \\
  \end{array} } \right]$$
I'm asked to compute, by hand, the matrix $A^{144}$
What I tried:
I calculated $A^2$, $A^3$, $A^4$ and $A^5$ and tried to find a pattern, so that I could first find the more general $A^n$ expression, and then make $n=144$. This is what I got:
$$
   A^2=
  \left[ {\begin{array}{cc}
   -1 & 0 \\
   0 & -1 \\
  \end{array} } \right]$$
$$
   A^3=
  \left[ {\begin{array}{cc}
   0 & 1 \\
   -1 & 0 \\
  \end{array} } \right]$$
$$
   A^4=
  \left[ {\begin{array}{cc}
   1 & 0 \\
   0 & 1 \\
  \end{array} } \right]$$
$$
   A^5=
  \left[ {\begin{array}{cc}
   0 & -1 \\
   1 & 0 \\
  \end{array} } \right]$$
So there seems to be kind of a "circular pattern", where $A^6=A^2$ and therefore, since $144/6=24$, I'm suspecting that: $$¿\,\boxed{A^{144}=A^2} \,?$$
However, I was hoping to find the more general $A^n$ matrix first and confirm the above result.
But, I can't seem to find a function such that:
$$f(n) =
\left\{
 \begin{array}{ll}
  0  & \mbox{if n is odd}  \\
  1 & \mbox{if n is even} 
 \end{array}
\right.
$$
This function would allow me to sort out the (1,1) and (2,2) elements of $A^n$ being 0 when n is odd (and the (1,2) and (2,1) being 0 when n is even).
I'm a bit confused at this point, your help is greatly appreciated.
 A: You have a weird problem, you say you can't find a function such that it equals $0$ on odd integers and $1$ on even ones, yet saying this is defining such a function.
Otherwise, you saw that $A^4 = I$ (the identity matrix), so just make a division :
$144 = 4\times36$ so that $A^{144} = (A^4)^{36} = I^{36} = I$
A nice interpretation in $\mathbb{R}^2$ is that your matrix represents a rotation by $\frac{\pi}{2}$ around the origin in the canonical basis. Then, rotating $144$ times results in not rotating at all...
A: As you wrote in you post: $A^4 = I$. Therefore: $A^{4k} = I$ for all $k \in \mathbb{N}$. 
Because $144 = 4 \times 36$, you have : 

$$A^{144} = I.$$

From the results in your post, you can deduce that:

$$ \forall k \in \mathbb{N}, \; A^{4k} = I, \; A^{4k+1} = A, \; A^{4k+2} = - I \; \text{and} \; A^{4k+3} = -A. $$

A: Your analysis shows that $A^n = A^{n \text{ mod } 4}$. Since you know $A^0,A^1, A^2, A^3$ you then have a general formula.
 Note that $144 \text{ mod } 4 = 0$.
A: Note that $A^4 = I$, and $A^{144} = (A^4)^{36}$, and you're done. If you really want to go with that $A^6 = A^2$ instead, then we have that 
$$A^{144} = (A^6)^{24} \\
= (A^2)^{24} = A^{48} \\
= (A^6)^8 = (A^2)^8 \\
= A^{16}= A^4\times A^{12}\\
= A^4 \times(A^6)^2 = A^4\times(A^2)^2\\
= A^8 = A^2\times A^6\\
= A^2\times A^2 = A^4$$
This seems a much more roundabout way to get the same result.
A: Even when there isn't a function, you can just define and use it as you just did. But regardless, your answer can be found in modular arithmetic. 
$f(n) \equiv n+1 \pmod{2}$
Here, this represents the remainder when divided by 2. If you need further help, let me know. 
