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(B) Given $A^2 - 4A + I = O$. By Cayley-Hamilton theorem,the characteristic equation of $A$ is $\lambda^2 - 	ext{tr}(A)\lambda + \det(A) = 0$. Here $

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only (S2) is correct

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(B) Given $A^2 - 4A + I = O$. By Cayley-Hamilton theorem,the characteristic equation of $A$ is $\lambda^2 - 	ext{tr}(A)\lambda + \det(A) = 0$. Here $	ext{tr}(A) = 1 + \alpha$ and $\det(A) = \alpha - 2$. So,$\lambda^2 - (1 + \alpha)\lambda + (\alpha - 2) = 0$. Comparing with $A^2 - 4A + I = O$,we get $1 + \alpha = 4 \Rightarrow \alpha = 3$ and $\alpha - 2 = 1 \Rightarrow \alpha = 3$. Thus,$A = \begin{bmatrix} 1 & 2 \ 1 & 3 \end{bmatrix}$. For $B^2 - 5B - 6I = O$,the characteristic equation is $\lambda^2 - 	ext{tr}(B)\lambda + \det(B) = 0$. Here $	ext{tr}(B) = 3 + 2 = 5$ and $\det(B) = 6 - 3\beta$. Comparing with $B^2 - 5B - 6I = O$,we get $	ext{tr}(B) = 5$ (which is $5=5$) and $\det(B) = -6$. So,$6 - 3\beta = -6 \Rightarrow 3\beta = 12 \Rightarrow \beta = 4$. Thus,$B = \begin{bmatrix} 3 & 3 \ 4 & 2 \end{bmatrix}$. Now,$A + B = \begin{bmatrix} 1+3 & 2+3 \ 1+4 & 3+2 \end{bmatrix} = \begin{bmatrix} 4 & 5 \ 5 & 5 \end{bmatrix}$. $\det(A + B) = (4)(5) - (5)(5) = 20 - 25 = -5$. For (S2): $\det(	ext{adj}(A + B)) = (\det(A + B))^{n-1} = (-5)^{2-1} = -5$. Thus,(S2) is correct. For (S1): $B - A = \begin{bmatrix} 2 & 1 \ 3 & -1 \end{bmatrix}$,$B + A = \begin{bmatrix} 4 & 5 \ 5 & 5 \end{bmatrix}$. $(B - A)(B + A) = \begin{bmatrix} 2 & 1 \ 3 & -1 \end{bmatrix} \begin{bmatrix} 4 & 5 \ 5 & 5 \end{bmatrix} = \begin{bmatrix} 8+5 & 10+5 \ 12-5 & 15-5 \end{bmatrix} = \begin{bmatrix} 13 & 15 \ 7 & 10 \end{bmatrix}$. Transpose $[(B - A)(B + A)]^T = \begin{bmatrix} 13 & 7 \ 15 & 10 \end{bmatrix}$. This does not match the given matrix in (S1). Thus,(S1) is wrong.

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