JEE Main 2022 (Online) 25th July Morning ShiftQuestion 1
Let $A=\begin{pmatrix}2&-1&-1\\1&0&-1\\1&-1&0\end{pmatrix}$ and $B=A-I$. If $\omega=\frac{\sqrt3i-1}{2}$, then the number of elements in the set $\{n\in\{1,2,\ldots,100\}:A^n+(\omega B)^n=A+B\}$ is equal to ________.
Matrices
Answer$17$
Solution
$A^2=A$ (idempotent) so $A^n=A$ for all $n$. $B=A-I=\begin{pmatrix}1&-1&-1\\1&-1&-1\\1&-1&-1\end{pmatrix}$ and $B^2=-B$, so $B^n=(-1)^{n-1}B$. $A^n+(\omega B)^n=A+B\iff\omega^n(-1)^{n-1}B=B\iff\omega^n=(-1)^{n-1}$. $\omega$ is a primitive cube root of unity, so $\omega^n$ is real only if $3\mid n$ ($\omega^n=1$), needing $n$ odd. $n\equiv3\pmod6$ in $1..100$: $3,9,\dots,99$ gives $17$ values (verified by computing each $n$).