JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Matrices

For a $2\times 2$ matrix, $M^2-(\mathrm{tr}M)M+(\det M)I=O$. Matching $A^2-4A+I=O$ gives $\mathrm{tr}A=4$, so $1+\alpha =4$ and $\alpha =3$; matching $B^2-5B-6I=O$ gives $\det B=-6$, so $6-3\beta =-6$ and $\beta =4$. Thus $A=\left[\begin{array}{cc}1& 2\\ 1& 3\end{array}\right]$ and $B=\left[\begin{array}{cc}3& 3\\ 4& 2\end{array}\right]$. Then $B-A=\left[\begin{array}{cc}2& 1\\ 3& -1\end{array}\right]$, $B+A=\left[\begin{array}{cc}4& 5\\ 5& 5\end{array}\right]$ and $(B-A)(B+A)=\left[\begin{array}{cc}13& 15\\ 7& 10\end{array}\right]$, whose transpose is $\left[\begin{array}{cc}13& 7\\ 15& 10\end{array}\right]$, not $\left[\begin{array}{cc}13& 15\\ 7& 10\end{array}\right]$; so (S1) is false. For (S2), $\det (A+B)=\det \left[\begin{array}{cc}4& 5\\ 5& 5\end{array}\right]=-5$, and since $\det (\mathrm{adj}M)=(\det M{)}^{n-1}=(-5{)}^1=-5$ for a $2\times 2$ matrix, (S2) is true. Hence only (S2) is correct.