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

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$225$

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(B) Given $A^{2} - 4A + 2I = O$. By Cayley-Hamilton theorem, the characteristic equation is $\lambda^{2} - 	ext{Tr}(A)\lambda + 	ext{det}(A) = 0$. Comparing, $	ext{Tr}(A) = 4 \Rightarrow \alpha + 2 = 4 \Rightarrow \alpha = 2$.Similarly, for $B^{2} - 3B + I = O$, $	ext{Tr}(B) = 3 \Rightarrow 1 + \beta = 3 \Rightarrow \beta = 2$.Now, $A^{2} = 4A - 2I$. Then $A^{3} = 4A^{2} - 2A = 4(4A - 2I) - 2A = 14A - 8I$.$A^{3} = 14 \begin{bmatrix} 2 & 2 \ 1 & 2 \end{bmatrix} - \begin{bmatrix} 8 & 0 \ 0 & 8 \end{bmatrix} = \begin{bmatrix} 28 & 28 \ 14 & 28 \end{bmatrix} - \begin{bmatrix} 8 & 0 \ 0 & 8 \end{bmatrix} = \begin{bmatrix} 20 & 28 \ 14 & 20 \end{bmatrix}$.For $B$, $B^{2} = 3B - I$. Then $B^{3} = 3B^{2} - B = 3(3B - I) - B = 8B - 3I$.$B^{3} = 8 \begin{bmatrix} 1 & 1 \ 1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 0 \ 0 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 8 \ 8 & 16 \end{bmatrix} - \begin{bmatrix} 3 & 0 \ 0 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 8 \ 8 & 13 \end{bmatrix}$.$A^{3} - B^{3} = \begin{bmatrix} 20 & 28 \ 14 & 20 \end{bmatrix} - \begin{bmatrix} 5 & 8 \ 8 & 13 \end{bmatrix} = \begin{bmatrix} 15 & 20 \ 6 & 7 \end{bmatrix}$.$	ext{det}(A^{3} - B^{3}) = (15 	imes 7) - (20 	imes 6) = 105 - 120 = -15$.Since $	ext{det}(	ext{adj}(M)) = (	ext{det}(M))^{n-1}$ for a $2 	imes 2$ matrix, $	ext{det}(	ext{adj}(A^{3} - B^{3})) = 	ext{det}(A^{3} - B^{3}) = -15$.Therefore, $(	ext{det}(	ext{adj}(A^{3} - B^{3})))^{2} = (-15)^{2} = 225$.

For some $\alpha, \beta \in R$, let $A = \begin{bmatrix} \alpha & 2 \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 1 \\ 1 & \beta \end{bmatrix}$ be such that $A^{2} - 4A + 2I = B^{2} - 3B + I = O$. Then $(\text{det}(\text{adj}(A^{3} - B^{3})))^{2}$ is equal to ....

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