JEE MAIN - Mathematics (2022 - 25th July Morning Shift - No. 16)
Let $$A=\left(\begin{array}{rrr}2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0\end{array}\right)$$ and $$B=A-I$$. If $$\omega=\frac{\sqrt{3} i-1}{2}$$, then the number of elements in the $$\operatorname{set}\left\{n \in\{1,2, \ldots, 100\}: A^{n}+(\omega B)^{n}=A+B\right\}$$ is equal to ____________.
Answer
17
Explanation
Here $A=\left(\begin{array}{ccc}2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0\end{array}\right)$
We get $A^{2}=A$ and similarly for
$$ B=A-I=\left[\begin{array}{lll} 1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1 \end{array}\right] $$
We get $B^{2}=-B \Rightarrow B^{3}=B$
$$ \therefore A^{n}+(\omega B)^{n}=A+(\omega B)^{n} \quad \text { for } n \in \mathrm{N} $$
For $\omega^{n}$ to be unity $n$ shall be multiple of 3 and for $B^{n}$ to be $B . n$ shell be $3,5,7, \ldots 99$
$\therefore n=\{3,9,15, \ldots . .99\}$
Number of elements $=17$
We get $A^{2}=A$ and similarly for
$$ B=A-I=\left[\begin{array}{lll} 1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1 \end{array}\right] $$
We get $B^{2}=-B \Rightarrow B^{3}=B$
$$ \therefore A^{n}+(\omega B)^{n}=A+(\omega B)^{n} \quad \text { for } n \in \mathrm{N} $$
For $\omega^{n}$ to be unity $n$ shall be multiple of 3 and for $B^{n}$ to be $B . n$ shell be $3,5,7, \ldots 99$
$\therefore n=\{3,9,15, \ldots . .99\}$
Number of elements $=17$
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