Mathematics - Matrices Question with Solution | TestHub

$A=\left[\begin{array}{cc}2& -1\\ 1& 0\end{array}\right]$ Now finding characteristic equation $\begin{array}{rl}& \left|\begin{array}{cc}2-\lambda & -1\\ 1& -\lambda \end{array}\right|=0\\ & \Rightarrow (2-\lambda )(-\lambda )-(-1)(1)=-2\lambda +{\lambda }^2+1=0\\ & \Rightarrow {\lambda }^2-2\lambda +1=0\\ & \Rightarrow (\lambda -1{)}^2=0\\ & \Rightarrow \lambda =1\end{array}$ Since $A$ satisfies $(A-I{)}^2=0$ $\begin{array}{rl}& \therefore A=I+N\text{ where }\\ & N=A-I\\ & N=\left[\begin{array}{ll}1& -1\\ 1& -1\end{array}\right]\\ & N^2=0\\ & A^m=(I+N{)}^m=I+mN\\ & A^m\cdot A^m=(I+mN)(I+mN)=I+2mN+m^2{N}^2\end{array}$ $\text{ Since }{N}^2=0$ $\Rightarrow A^{m^2}=I+2mN$ Now putting in given condition $\begin{array}{rl}& I+m^{2}N+I+mN=3I-A^{-6}\\ & A^{-1}=\left[\begin{array}{cc}0& 1\\ -1& 2\end{array}\right]\\ & A^{-6}={\left(A^{-1}\right)}^6=I+(-6)N\end{array}$ $\therefore$ Putting in (i) $\begin{array}{rl}& \left(m^2+m\right)N=I-(I-6N)\\ & \left(m^2+m\right)N=6N\end{array}$ Since $N\ne 0$ $\begin{array}{rl}& \Rightarrow m^2+m=6\\ & \Rightarrow m^2+m-6=0\\ & \Rightarrow (m-2)(m+3)=0\\ & \Rightarrow m=2,-3\end{array}$ $\therefore$ Number of elements in $S$ is 2