JEE Main 2026 Jan 21 Shift 2, Mathematics Q14: Algebra Of Matrices

Notice $A-I=\left[\begin{array}{c}2-4\\ 1-2\end{array}\right]$ is nilpotent of index 2 (it has rank 1 and trace 0; $(A-I{)}^2=0$). Writing $A=I+N$ with $N^2=0$, the binomial theorem gives $A^n=I+nN$, so $A^n=\left[\begin{array}{c}2n+1-4n\\ n-2n+1\end{array}\right]$.

For $n=15$: $A^{15}=\left[\begin{array}{c}31-60\\ 15-29\end{array}\right]$, so $A^{15}+B=\left[\begin{array}{c}2-11\\ 2-11\end{array}\right]$.

The homogeneous system has both rows giving $2x-11y=0$, i.e. $x=\frac{11y}{2}$. With $y=2$ we get $x=11$, matching option (3).