Let \(\mathrm{S}=\left\{\mathrm{m} \in \mathbf{Z}: \mathrm{A}^{\mathrm{m}^2}+\mathrm{A}^{\mathrm{m}}=3…
Solution
2 & -1 \\ 1 & 0
\end{array}\right]$
Now finding characteristic equation
$\begin{aligned}
& \left|\begin{array}{cc}
2-\lambda & -1 \\ 1 & -\lambda
\end{array}\right|=0 \\ & \Rightarrow(2-\lambda)(-\lambda)-(-1)(1)=-2 \lambda+\lambda^2+1=0 \\ & \Rightarrow \lambda^2-2 \lambda+1=0 \\ & \Rightarrow(\lambda-1)^2=0 \\ & \Rightarrow \lambda=1
\end{aligned}$
Since $A$ satisfies $(A-I)^2=0$
$\begin{aligned}
& \therefore \quad A=I+N \text { where } \\ & N=A-I \\ & N=\left[\begin{array}{ll}
1 & -1 \\ 1 & -1
\end{array}\right] \\ & N^2=0 \\ & A^m=(I+N)^m=I+m N \\ & A^m \cdot A^m=(I+m N)(I+m N)=I+2 m N+m^2 N^2
\end{aligned}$
$\text { Since } N^2=0$
$\Rightarrow \quad A^{m^2}=I+2 m N$
Now putting in given condition
$\begin{aligned}
& I+m^2 N+I+m N=3 I-A^{-6} \\ & A^{-1}=\left[\begin{array}{cc}
0 & 1 \\ -1 & 2
\end{array}\right] \\ & A^{-6}=\left(A^{-1}\right)^6=I+(-6) N
\end{aligned}$
$\therefore \quad$ Putting in (i)
$\begin{aligned}
& \left(m^2+m\right) N=I-(I-6 N) \\ & \left(m^2+m\right) N=6 N
\end{aligned}$
Since $N \neq 0$
$\begin{aligned}
& \Rightarrow m^2+m=6 \\ & \Rightarrow m^2+m-6=0 \\ & \Rightarrow(m-2)(m+3)=0 \\ & \Rightarrow m=2,-3
\end{aligned}$
$\therefore \quad$ Number of elements in $S$ is 2
Asked in: JEE Main 2025 (29 Jan Shift 1)