Let A = bmatrix 1 & 2 1 & α bmatrix and B = bmatrix 3 & 3 β & 2...

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 $A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix}$, we are given $A^2 - 4A + I = O$.
Comparing the trace, we get $\text{Tr}(A) = 1 + \alpha = 4 \implies \alpha = 3$.
Thus, $A = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix}$.

For matrix $B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}$, we are given $B^2 - 5B - 6I = O$.
Comparing the determinant, we get $\det(B) = 6 - 3\beta = -6 \implies 3\beta = 12 \implies \beta = 4$.
Thus, $B = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix}$.

Evaluating Statement (S1):
$B - A = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}$

$B + A = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}$

$(B-A)(B+A) = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix} = \begin{bmatrix} 2(4)+1(5) & 2(5)+1(5) \\ 3(4)-1(5) & 3(5)-1(5) \end{bmatrix} = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$

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

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

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

$\det(A+B) = \det \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix} = 4(5) - 5(5) = 20 - 25 = -5$

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

Therefore, only (S2) is correct.

Answer: only (S2) is correct