Verified Solution • 2025–26 Session moderate

QJEEM22Maad68: JEE Main 2022 Mathematics - Class 12 JEEM22Maad68 Solution: Let and . If , then the number of elements in the is equal to…

Let A=(2amp;−1amp;−11amp;0amp;−11amp;−1amp;0)A=\left(\begin{array}{rrr}2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0\end{array}\right) and B=A−IB=A-I . If ω=3i−12\omega=\frac{\sqrt{3} i-1}{2} , then the number of elements in the set⁡{n∈{1,2,…,100}:An+(ωB)n=A+B}\operatorname{set}\left\{n \in\{1,2, \ldots, 100\}: A^{n}+(\omega B)^{n}=A+B\right\} is equal to ____________.

Step-by-Step Verified Solution

Here A=(2amp;−1amp;−11amp;0amp;−11amp;−1amp;0)A=\left(\begin{array}{ccc}2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0\end{array}\right)

We get A2=AA^{2}=A and similarly for

B=A−I=[1amp;−1amp;−11amp;−1amp;−11amp;−1amp;−1]B=A-I=\left[\begin{array}{lll}1 & -1 & -1 \\1 & -1 & -1 \\1 & -1 & -1\end{array}\right]

We get B2=−B⇒B3=BB^{2}=-B \Rightarrow B^{3}=B

∴An+(ωB)n=A+(ωB)n for n∈N\therefore A^{n}+(\omega B)^{n}=A+(\omega B)^{n} \quad \text { for } n \in \mathrm{N}

For ωn\omega^{n} to be unity nn shall be multiple of 3 and for BnB^{n} to be B.nB . n shell be 3,5,7,…993,5,7, \ldots 99

∴n={3,9,15,…..99}\therefore n=\{3,9,15, \ldots . .99\}

Number of elements =17=17

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