(S1): $[(B-A)(B+A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$
and
(S2): $\det(\text{adj}(A+B)) = -5$,
Matrix A Determination
Given $A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix}$ and the equation $A^2 - 4A + I = O$. The Cayley-Hamilton theorem states that $A^2 - (\text{tr}(A))A + (\det(A))I = O$. Comparing the given equation with the theorem:
- Trace: $\text{tr}(A) = 1 + \alpha = 4$, which implies $\alpha = 3$.
- Determinant: $\det(A) = (1)(\alpha) - (2)(1) = \alpha - 2 = 1$. Substituting $\alpha = 3$, we get $3 - 2 = 1$, confirming consistency.
Therefore, matrix $A = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix}$.
Matrix B Determination
Given $B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}$ and the equation $B^2 - 5B - 6I = O$. Applying the Cayley-Hamilton theorem ($B^2 - (\text{tr}(B))B + (\det(B))I = O$):
- Trace: $\text{tr}(B) = 3 + 2 = 5$. This matches the coefficient of $B$.
- Determinant: $\det(B) = (3)(2) - (3)(\beta) = 6 - 3\beta$. From the equation, $\det(B)$ must be $-6$.
- Equating determinants: $6 - 3\beta = -6 \implies 3\beta = 12 \implies \beta = 4$.
Therefore, matrix $B = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix}$.
Statement (S1) Analysis
Statement (S1) claims that $[(B-A)(B+A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$.
- Calculate $B-A$: $B-A = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}$.
- Calculate $B+A$: $B+A = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}$.
- Compute the product $(B-A)(B+A)$: $(B-A)(B+A) = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix} = \begin{bmatrix} (8+5) & (10+5) \\ (12-5) & (15-5) \end{bmatrix} = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$.
- Compute the transpose: $[(B-A)(B+A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}^T = \begin{bmatrix} 13 & 7 \\ 15 & 10 \end{bmatrix}$.
The calculated transpose $\begin{bmatrix} 13 & 7 \\ 15 & 10 \end{bmatrix}$ does not match the matrix $\begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$ in statement (S1). Hence, (S1) is incorrect.
Statement (S2) Analysis
Statement (S2) claims that $\det(\text{adj}(A+B)) = -5$.
- Calculate $A+B$: $A+B = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} + \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}$.
- Calculate the determinant of $A+B$: $\det(A+B) = (4)(5) - (5)(5) = 20 - 25 = -5$.
- Apply the property $\det(\text{adj}(M)) = (\det(M))^{n-1}$ for an $n \times n$ matrix $M$. For the 2x2 matrix $A+B$: $\det(\text{adj}(A+B)) = (\det(A+B))^{2-1} = \det(A+B)$.
- Result: $\det(\text{adj}(A+B)) = -5$.
This matches the value in statement (S2). Hence, (S2) is correct.
Final Conclusion
Based on the analysis, statement (S1) is incorrect, while statement (S2) is correct. Therefore, only (S2) is correct.
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