Let A=pmatrix2&-1&-11&0&-11&-1&0pmatrix and B=A-I.

Question

Let $A=\begin{pmatrix}2&-1&-1\1&0&-1\1&-1&0\end{pmatrix}$ and $B=A-I$. If $\omega=\dfrac{\sqrt3i-1}2$, then the number of elements in the set $\{n\in\{1,\dots,100\}: A^n+(\omega B)^n=A+B\}$ is equal to

Answer: 17

Accepted Answer

Direct computation shows $A^2=A$ (idempotent), so $A^n=A$ for all $n\ge1$.

$B=A-I=\begin{pmatrix}1&-1&-1\1&-1&-1\1&-1&-1\end{pmatrix}$, and $B^2=-B$, so $B^3=B$, i.e. $B^n=B$ whenever $n$ is odd, and $B^n=-B$ when $n$ is even ($n\ge2$).

Also $\omega=\dfrac{-1+\sqrt3i}2$ is a primitive cube root of unity, so $\omega^n=1$ requires $3\mid n$.

For $A^n+(\omega B)^n=A+B$: since $A^n=A$ always, we need $(\omega B)^n=B$, i.e. $\omega^nB^n=B$. This requires $\omega^n=1$ (so $3\mid n$) and $B^n=B$ (so $n$ odd).

So $n$ must be an odd multiple of $3$: $n=3,9,15,\dots,99$ --- an arithmetic sequence with common difference $6$, from $3$ to $99$: $\dfrac{99-3}6+1=17$ terms.

Solutions · 1