JEE Main 2026 — Mathematics Algebra

For a $2 \times 2$2×2 matrix $M$M, the characteristic equation is given by $M^2 - \text{Tr}(M)M + \det(M)I = O$M2−Tr(M)M+det(M)I=O.

For matrix $A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix}$A=[11​2α​], we are given $A^2 - 4A + I = O$A2−4A+I=O.

Comparing the trace, we get $\text{Tr}(A) = 1 + \alpha = 4 \implies \alpha = 3$Tr(A)=1+α=4⟹α=3.

Thus, $A = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix}$A=[11​23​].

For matrix $B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}$B=[3β​32​], we are given $B^2 - 5B - 6I = O$B2−5B−6I=O.

Comparing the determinant, we get $\det(B) = 6 - 3\beta = -6 \implies 3\beta = 12 \implies \beta = 4$det(B)=6−3β=−6⟹3β=12⟹β=4.

Thus, $B = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix}$B=[34​32​].

Evaluating Statement (S1):

$$B - A = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}$$B−A=[34​32​]−[11​23​]=[23​1−1​]

$$B + A = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}$$B+A=[34​32​]+[11​23​]=[45​55​]

$$(B-A)(B+A) = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix} = \begin{bmatrix} 2(4)+1(5) & 2(5)+1(5) \\ 3(4)-1(5) & 3(5)-1(5) \end{bmatrix} = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$$(B−A)(B+A)=[23​1−1​][45​55​]=[2(4)+1(5)3(4)−1(5)​2(5)+1(5)3(5)−1(5)​]=[137​1510​]

$$[(B-A)(B+A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}^T = \begin{bmatrix} 13 & 7 \\ 15 & 10 \end{bmatrix}$$[(B−A)(B+A)]T=[137​1510​]T=[1315​710​]

Since $\begin{bmatrix} 13 & 7 \\ 15 & 10 \end{bmatrix} \neq \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$[1315​710​]=[137​1510​], Statement (S1) is incorrect.

Evaluating Statement (S2):

For any $2 \times 2$2×2 matrix $M$M, $\det(\text{adj}(M)) = \det(M)^{2-1} = \det(M)$det(adj(M))=det(M)2−1=det(M).

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

$$\det(A+B) = \det \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix} = 4(5) - 5(5) = 20 - 25 = -5$$det(A+B)=det[45​55​]=4(5)−5(5)=20−25=−5

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

Therefore, only (S2) is correct.

Answer: only (S2) is correct