JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Matrices

A direct multiplication shows that $A$ is idempotent, $A^2=A$, so $A^n=A$ for every $n\ge 1$. For $B=A-I$ all three rows equal $(1,-1,-1)$, and squaring gives $B^2=-B$; consequently $B^n=B$ when $n$ is odd and $B^n=-B$ when $n$ is even, the sign simply alternating rather than repeating with period three. The number $\omega =\frac{-1+\sqrt{3}i}{2}$ is a primitive cube root of unity, so ${\omega }^3=1$ and ${\omega }^n$ runs through $1,\omega ,{\omega }^2$ as $n$ takes residues $0,1,2$ modulo $3$. Since $A^n+(\omega B{)}^n=A+{\omega }^n{B}^n$, the required equality $A^n+(\omega B{)}^n=A+B$ reduces to ${\omega }^n{B}^n=B$. As $B\ne O$, this forces the scalar condition ${\omega }^n(\pm 1)=1$, where the $\pm 1$ is the sign carried by $B^n$. A unit complex number ${\omega }^n$ multiplied by a real $\pm 1$ can equal $1$ only when ${\omega }^n$ is itself real, that is ${\omega }^n=1$, which needs $n\equiv 0(\mathrm{mod}3)$; the sign must then be $+1$, which needs $n$ odd. Hence $n$ must be an odd multiple of $3$, namely $n=3,9,15,\dots ,99$, an arithmetic progression with common difference $6$ containing $\frac{99-3}{6}+1=17$ terms.