How to compute matrices to high powers?

Question

How can I find the following matrix raised to the power of $279$ without using eigenvalues?

$$\begin{pmatrix}0 & 0 & -1\\\ 1 & 0 & -1\\\ 0 & 1 & -1\end{pmatrix}$$

Answer 1

Note that this matrix is a companion matrix, so we can read off its characteristic polynomial to be $$ x^3+x^2+x+1 $$ By Cayley-Hamilton, $$ A^3+A^2+A+I=0\implies (A-I)(A^3+A^2+A+I)=0\\ \implies A^4=I $$ So, $$ A^{279}=A^{276}A^3=A^3 $$

Answer 2

The observation that $A^4=I$ is indeed useful.

$$279 = 4(69)+3$$

Hence $$A^{279}=A^{4(69)+3}=(A^4)^{69}A^3=I^{69}A^3=A^3.$$

Remark: Your matrix is a companion matrix.

$$I+A+A^2+A^3=0$$

$$(I-A)(I+A+A^2+A^3)=0$$

$$I-A^4=0$$

$$A^4=I$$