Let B = $$\begin{bmatrix} 1 & 3 \\ 1 & 5 \end{bmatrix}$$ and A be a 2 x 2 matrix such that AB^{-1} = A^{-1}. If BCB^{-1} = A and C^4 + \alpha C^2 + \beta I = 0, then 2\beta - \alpha is equal to :
Step-by-Step Solution
Key Concept: Given AB^-1 = A^-1, we have A^2 = B. Given BCB^-1 = A, we have C = B^-1AB. Since A^2 = B, C^2 = (B^-1AB)(B^-1AB) = B^-1A^2B = B^-1BB = B. Then C^4 = B^2. Substituting into the equation C^4 + aC^2 + BI = 0 gives B^2 + aB + BI = 0. Using the characteristic equation of B, B^2 - tr(B)B + det(B)I = 0, we compare coefficients to find a and b.
Given $AB^{-1} = A^{-1} \implies A^2 = B$. Given $BCB^{-1} = A \implies C = B^{-1}AB$. Then $C^2 = (B^{-1}AB)(B^{-1}AB) = B^{-1}A^2B = B^{-1}BB = B$. Thus $C^4 = B^2$. The equation $C^4 + \alpha C^2 + \beta I = 0$ becomes $B^2 + \alpha B + \beta I = 0$. For $B = \begin{bmatrix} 1 & 3 \\ 1 & 5 \end{bmatrix}$, the characteristic equation is $B^2 - tr(B)B + det(B)I = 0$. $tr(B) = 1+5=6$, $det(B) = 5-3=2$. So $B^2 - 6B + 2I = 0$. Comparing $B^2 + \alpha B + \beta I = 0$ with $B^2 - 6B + 2I = 0$, we get $\alpha = -6$ and $\beta = 2$. Then $2\beta - \alpha = 2(2) - (-6) = 4 + 6 = 10$.
Correct Answer: 4