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

### Related Formula For a 2 times 2$2 \times 2$ matrix M$M$, Cayley-Hamilton equation states: M^2 - operatornameTr(M)M + operatornamedet(M)I = O$$M^2 - \operatorname{Tr}(M)M + \operatorname{det}(M)I = O$$ Also, |operatornameadj(M)| = |M|^n-1$|\operatorname{adj}(M)| = |M|^{n-1}$, and for a 2 times 2$2 \times 2$ matrix, |operatornameadj(M)| = |M|$|\operatorname{adj}(M)| = |M|$. ### Core Logic Using the characteristic equation A^2 - operatornameTr(A)A + det(A)I = O$A^2 - \operatorname{Tr}(A)A + \det(A)I = O$: Comparing with A^2 - 4A + 2I = O$A^2 - 4A + 2I = O$: operatornameTr(A) = 4 Rightarrow alpha + 2 = 4 Rightarrow alpha = 2$\operatorname{Tr}(A) = 4 \Rightarrow \alpha + 2 = 4 \Rightarrow \alpha = 2$ Comparing with B^2 - 3B + I = O$B^2 - 3B + I = O$: operatornameTr(B) = 3 Rightarrow 1 + beta = 3 Rightarrow beta = 2$\operatorname{Tr}(B) = 3 \Rightarrow 1 + \beta = 3 \Rightarrow \beta = 2$ ### Step 1: Compute A^3 and B^3 via reduction A = beginbmatrix 2 & 2 \\ 1 & 2 endbmatrix$$A = \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix}$$ A^2 = 4A - 2I$A^2 = 4A - 2I$ A^3 = A(4A - 2I) = 4A^2 - 2A = 4(4A - 2I) - 2A = 14A - 8I$$A^3 = A(4A - 2I) = 4A^2 - 2A = 4(4A - 2I) - 2A = 14A - 8I$$ A^3 = 14 beginbmatrix 2 & 2 \\ 1 & 2 endbmatrix - 8 beginbmatrix 1 & 0 \\ 0 & 1 endbmatrix = beginbmatrix 28 & 28 \\ 14 & 28 endbmatrix - beginbmatrix 8 & 0 \\ 0 & 8 endbmatrix = beginbmatrix 20 & 28 \\ 14 & 20 endbmatrix$$A^3 = 14 \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix} - 8 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 28 & 28 \\ 14 & 28 \end{bmatrix} - \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix}$$ Similarly, B = beginbmatrix 1 & 1 \\ 1 & 2 endbmatrix$B = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$ B^2 = 3B - I$B^2 = 3B - I$ B^3 = 3B^2 - B = 3(3B - I) - B = 8B - 3I$$B^3 = 3B^2 - B = 3(3B - I) - B = 8B - 3I$$ B^3 = 8 beginbmatrix 1 & 1 \\ 1 & 2 endbmatrix - beginbmatrix 3 & 0 \\ 0 & 3 endbmatrix = beginbmatrix 8 & 8 \\ 8 & 16 endbmatrix - beginbmatrix 3 & 0 \\ 0 & 3 endbmatrix = beginbmatrix 5 & 8 \\ 8 & 13 endbmatrix$$B^3 = 8 \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 8 \\ 8 & 16 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix}$$ ### Step 2: Difference and Determinant A^3 - B^3 = beginbmatrix 20 & 28 \\ 14 & 20 endbmatrix - beginbmatrix 5 & 8 \\ 8 & 13 endbmatrix = beginbmatrix 15 & 20 \\ 6 & 7 endbmatrix$$A^3 - B^3 = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix} - \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} = \begin{bmatrix} 15 & 20 \\ 6 & 7 \end{bmatrix}$$ det(A^3 - B^3) = (15 times 7) - (20 times 6) = 105 - 120 = -15$$\det(A^3 - B^3) = (15 \times 7) - (20 \times 6) = 105 - 120 = -15$$ ### Step 3: Final Answer For a 2 times 2$2 \times 2$ matrix, |operatornameadj(M)| = |M|$|\operatorname{adj}(M)| = |M|$. Thus, det(operatornameadj(A^3 - B^3)) = -15$\det(\operatorname{adj}(A^3 - B^3)) = -15$. (det(operatornameadj(A^3 - B^3)))^2 = (-15)^2 = 225$$(\det(\operatorname{adj}(A^3 - B^3)))^2 = (-15)^2 = 225$$ ### Pattern Recognition Do not manually multiply matrices to the 3rd power. Cayley-Hamilton strictly reduces M^3$M^3$ down to a linear combination cM + dI$cM + dI$. Expanding this requires only basic scalar arithmetic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices