TITLE: Find the $2013$th power of a given $3\times 3$ matrix
QUESTION [6 upvotes]: Question from my linear algebra homework I'm struggling with:
Let $D = \begin{bmatrix} -2 & 5 & 4 \\-1 & 0 & 0 \\0 & 4 & 3 \end{bmatrix}$
We are asked:

Find $D^5+3D^2-D+I$
Find $D^{2013}$
Write $D^{-1}$ as a polynomial of $D$

I solved questions 1) and 3) but can't solve 2)...

REPLY [5 votes]: The characteristic polynomial of $D$ is given by
$$\chi_D(\lambda) = \det(\lambda I_3 - D) = \lambda^3 - \lambda^2 - \lambda + 1 = (\lambda - 1)^2 (\lambda+1)$$
By Cayley Hamilton theorem, we have
$$\chi_D(D) = D^3 - D^2 - D + I_3 = (D - I_3)^2(D + I_3) = (D^2 - I_3)(D - I_3) = 0$$
Since $(\lambda + 1)\chi_D(\lambda) = (\lambda^2 - 1)^2$, we have
$(D^2 - I_3)^2 = 0$ and hence
$$\begin{align}
D^{2013} 
= & (I_3 + (D^2 - I_3))^{1006} D\\
= & (I_3 + 1006 (D^2 - I_3)) D\quad\color{blue}{\longleftarrow \text{throwing away terms}\propto (D^2 - I_3)^2}\\
= & D + 1006 (D^2 - I_3)((D - I_3) + I_3)\\
= & D + 1006 (D^2 - I_3)\quad\color{blue}{\longleftarrow \text{throwing away terms}\propto (D^2 - I_3)(D-I_3)}\\
= & \left[\begin{array}{rrr}
-2014  & 6041 & 4028 \\
2011 & -6036  &  -4024 \\
-4024  & 12076  & 8051
\end{array}\right]
\end{align}
$$

REPLY [2 votes]: Bull headed approach. 2013 written in binary form is 11111011101.
Therefore, repeatedly square D 10 times which gives you $D$, $D^2$, $D^4$, $D^8$, $D^{16}$, $D^{32}$, $D^{64}$, $D^{128}$, $D^{256}$, $D^{512}$ and $D^{1024}$.
Then multiply $D^{1024} \times D^{512} \times D^{256} \times D^{128} \times D^{64} \times D^{16} \times D^{8} \times D^4$.
So, you will have to perform 17 multiplications, which is doable.