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Problem 929

AIME late

Algebra Difficulty 5.8 Prove it

2. From $z^{2}+z+1=0$ we get $z^{3}=1$

$A=I_{3}+B$ , where $B=\left(\begin{array}{ccc}0 & 0 & 0 \\ z^{2} & 0 & 0 \\ z & z^{2} & 0\end{array}\right)$

$I_{3} B=B I_{3} \Rightarrow A^{n}=\left(I_{3}+B\right)^{n}=I_{3}+C_{n}^{1} B+C_{n}^{2} B^{2}+\cdots+B^{n}$

$B^{2}=\left(\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ z & 0 & 0\end{array}\right)$ and $B^{k}=O_{3}$ , for any $k \geq 3$ .

$A^{n}=I_{3}+n B+\frac{n(n-1)}{2} B^{2}=\left(\begin{array}{ccc}1 & 0 & 0 \\ n z^{2} & 1 & 0 \\ \frac{n(n+1)}{2} z & n z^{2} & 1\end{array}\right)$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

2. (7p) Let $z \in \mathbb{C}$ with the property $z^{2}+z=-1$ and the matrix $A=\left(\begin{array}{ccc}1 & 0 & 0 \\ z^{2} & 1 & 0 \\ z & z^{2} & 1\end{array}\right) \in M_{3}(\mathbb{C})$ . Calculate $A^{n}$ , where $n$ is a non-zero natural number.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.