JEE Main 2026 — Matrices Question with Solution

For a $2\times 2$ matrix $M$, the characteristic equation is given by $M^2-\text{Tr}(M)M+\det (M)I=O$.

For matrix $A=\left[\begin{array}{cc}1& 2\\ 1& \alpha \end{array}\right]$, we are given $A^2-4A+I=O$.
Comparing the trace, we get $\text{Tr}(A)=1+\alpha =4⟹\alpha =3$.
Thus, $A=\left[\begin{array}{cc}1& 2\\ 1& 3\end{array}\right]$.

For matrix $B=\left[\begin{array}{cc}3& 3\\ \beta & 2\end{array}\right]$, we are given $B^2-5B-6I=O$.
Comparing the determinant, we get $\det (B)=6-3\beta =-6⟹3\beta =12⟹\beta =4$.
Thus, $B=\left[\begin{array}{cc}3& 3\\ 4& 2\end{array}\right]$.

Evaluating Statement (S1):
$B-A=\left[\begin{array}{cc}3& 3\\ 4& 2\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 1& 3\end{array}\right]=\left[\begin{array}{cc}2& 1\\ 3& -1\end{array}\right]$

$B+A=\left[\begin{array}{cc}3& 3\\ 4& 2\end{array}\right]+\left[\begin{array}{cc}1& 2\\ 1& 3\end{array}\right]=\left[\begin{array}{cc}4& 5\\ 5& 5\end{array}\right]$

$(B-A)(B+A)=\left[\begin{array}{cc}2& 1\\ 3& -1\end{array}\right]\left[\begin{array}{cc}4& 5\\ 5& 5\end{array}\right]=\left[\begin{array}{cc}2(4)+1(5)& 2(5)+1(5)\\ 3(4)-1(5)& 3(5)-1(5)\end{array}\right]=\left[\begin{array}{cc}13& 15\\ 7& 10\end{array}\right]$

$[(B-A)(B+A){]}^T={\left[\begin{array}{cc}13& 15\\ 7& 10\end{array}\right]}^T=\left[\begin{array}{cc}13& 7\\ 15& 10\end{array}\right]$
Since $\left[\begin{array}{cc}13& 7\\ 15& 10\end{array}\right]\ne \left[\begin{array}{cc}13& 15\\ 7& 10\end{array}\right]$, Statement (S1) is incorrect.

Evaluating Statement (S2):
For any $2\times 2$ matrix $M$, $\det (\text{adj}(M))=\det (M{)}^{2-1}=\det (M)$.

$\det (\text{adj}(A+B))=\det (A+B)$

$\det (A+B)=\det \left[\begin{array}{cc}4& 5\\ 5& 5\end{array}\right]=4(5)-5(5)=20-25=-5$

Thus, $\det (\text{adj}(A+B))=-5$. Statement (S2) is correct.

Therefore, only (S2) is correct.

Answer: only (S2) is correct