JEE Main 2022 — Mathematics Algebra

Given, $A=[\begin{matrix}2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0\end{matrix}]$A=[211​−10−1​−1−10​]

$$\Rightarrow {A}^{2}=[\begin{matrix}2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0\end{matrix}][\begin{matrix}2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0\end{matrix}]=[\begin{matrix}2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0\end{matrix}]=A$$⇒A2=[211​−10−1​−1−10​][211​−10−1​−1−10​]=[211​−10−1​−1−10​]=A

$$\Rightarrow {A}^{n}=A$$⇒An=A

Now $\forall n\in {1,2,\ldots ,100}$∀n∈1,2,…,100

Now, $B=A-I=[\begin{matrix}1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1\end{matrix}]$B=A−I=[111​−1−1−1​−1−1−1​]

$${B}^{2}=[\begin{matrix}1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1\end{matrix}][\begin{matrix}1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1\end{matrix}]=-[\begin{matrix}1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1\end{matrix}]=-B$$B2=[111​−1−1−1​−1−1−1​][111​−1−1−1​−1−1−1​]=−[111​−1−1−1​−1−1−1​]=−B

$$\Rightarrow {B}^{3}=-{B}^{2}=B$$⇒B3=−B2=B

$$\Rightarrow {B}^{5}=B$$⇒B5=B

$$\Rightarrow {B}^{99}=B$$⇒B99=B

Also, ${\omega }^{3k}=1$ω3k=1

So, $n=$n= common of ${1,3,5,\ldots ,99}$1,3,5,…,99 and ${3,6,9,\ldots ,99}=17$3,6,9,…,99=17