JEE Main 2025 Jan 29 Shift 1, Mathematics Q22: Algebra Of Matrices

Compute the first few powers of $A$:

$A^2=\left[\begin{array}{c}3-2\\ 2-1\end{array}\right],A^3=\left[\begin{array}{c}4-3\\ 3-2\end{array}\right],A^4=\left[\begin{array}{c}5-4\\ 4-3\end{array}\right].$

A pattern emerges: $A^m=\left[\begin{array}{c}m+1-m\\ m-(m-1)\end{array}\right]$ for $m\ge 1$ (and the formula extends to all integers $m$, since $\det A=1$).

Replace $m$ by $m^2$ for the second matrix and add:

$A^{m^2}+A^m=\left[\begin{array}{c}{m}^2+m+2-m^2-m\\ m^2+m-(m^2+m)\end{array}\right].$

The RHS: $A^{-6}=\left[\begin{array}{c}-56\\ -67\end{array}\right]$ (since $A^{-1}=\left[\begin{array}{c}01\\ -12\end{array}\right]$ and iterate), so $3I-A^{-6}=\left[\begin{array}{c}8-6\\ 6-4\end{array}\right].$

Matching, $m^2+m+2=8\Rightarrow m^2+m-6=0\Rightarrow (m+3)(m-2)=0.$ Both $m=-3$ and $m=2$ satisfy the remaining entries, so $n(S)=2.$