Let $S=\{m\in\mathbb{Z}\,:\,A^{m^{2}}+A^{m}=3I-A^{-6}\}$, where $A=\begin{pmatrix}2&-1\\1&0\end{pmatrix}$. Then $n(S)$ is equal to:
Step-by-Step Solution
Key Concept: Characteristic polynomial $(\lambda-1)^{2}=0$ gives $(A-I)^{2}=0$. With $N=A-I$, $N^{2}=0$, so $A^{k}=(I+N)^{k}=I+kN$ (the series truncates immediately).
Characteristic: $\det(A-\lambda I)=\lambda^{2}-2\lambda+1=(\lambda-1)^{2}.$ Let $N=A-I=\begin{pmatrix}1&-1\\1&-1\end{pmatrix}.$ Then $N^{2}=0.$
Hence $A^{k}=(I+N)^{k}=I+kN$ for every $k\in\mathbb{Z}$ (including $k=-1$: $A^{-1}=I-N$, so $A^{-6}=I-6N$).
LHS $=A^{m^{2}}+A^{m}=2I+(m^{2}+m)N.$
RHS $=3I-(I-6N)=2I+6N.$
Equating coefficients of $N$ (noting $N\ne 0$): $m^{2}+m-6=0\Rightarrow (m-2)(m+3)=0\Rightarrow m\in\{2,-3\}.$
So $n(S)=2.$
Correct Answer: 2