JEE Main 2025 — Mathematics Algebra

$A=\left[\begin{array}{cc}
2 & -1 \
1 & 0
\end{array}\right]$Now finding characteristic equation$Nowfindingcharacteristicequation\begin{aligned}
& \left|\begin{array}{cc}
2-\lambda & -1 \
1 & -\lambda
\end{array}\right|=0 \
& \Rightarrow(2-\lambda)(-\lambda)-(-1)(1)=-2 \lambda+\lambda^2+1=0 \
& \Rightarrow \lambda^2-2 \lambda+1=0 \
& \Rightarrow(\lambda-1)^2=0 \
& \Rightarrow \lambda=1
\end{aligned}$Since$SinceA$satisfies$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$$\Rightarrow \quad A^{m^2}=I+2 m N$Now putting in given condition$Nowputtingingivencondition\begin{aligned}
& I+m^2 N+I+m N=3 I-A^{-6} \
& A^{-1}=\left[\begin{array}{cc}
0 & 1 \
-1 & 2
\end{array}\right] \
& A^{-6}=\left(A^{-1}\right)^6=I+(-6) N
\end{aligned}$$\therefore \quad$Putting in (i)$Puttingin(i)\begin{aligned}
& \left(m^2+m\right) N=I-(I-6 N) \
& \left(m^2+m\right) N=6 N
\end{aligned}$Since$SinceN \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$Number of elements in$NumberofelementsinS$ is 2