How to calculate matrix raised to a high power.

Evaluate the eigenvalues and also the matrix P and D so $ A=PDP^{-1} $

then $ A^{n} =PD^{n}P^{-1} $

since the matrix D is diagonal then $ D^{n} $ has the element

$$ a12=a21=0 $$ and $ a11= \lambda _{1}^{n} $

$ a22= \lambda _{2}^{n} $

here lambda are the 2 eigenvalues

Another approach you may use is to break $13$ down into powers of two. We see that $13=8+4+1$. You can compute $A^8$, $A^4$ and then find your desired product by $A^8 \cdot A^4 \cdot A = A^{13}$.

Powers of two are easier to handle than other powers, and so this method is used often when computing large powers.

This is because $A^2=A\cdot A$, $A^4 = A^2 \cdot A^2$, and $A^8=A^4 \cdot A^4$. Therefore three multiplications are needed to find $A^8$ and $A^4$. Then two more multiplications are needed to compute $A^{13} = A^8 \cdot A^4 \cdot A$. Thus only five multiplications are needed. This is in contrast to the necessary 12 products that need to be computed in a direct method.

Hint: Have you tried multiplying your matrix by itself $M^2=?$