Let $$B = \begin{bmatrix} 1 & 3 \\ 1 & 5 \end{bmatrix}$$ and $$A$$ be a $$2 \times 2$$ matrix such that $$AB^{-1} = A^{-1}$$. If $$BCB^{-1} = A$$ and $$C^4 + \alpha C^2 + \beta I = O$$, then $$2\beta - \alpha$$ is equal to

From $$AB^{-1} = A^{-1}$$, multiply by $$A$$ on the right: $$A B^{-1} A = I \implies B = A^2$$.

 Given $$BCB^{-1} = A$$. Since $$B=A^2$$, we have $$A^2 C A^{-2} = A$$.

 This implies $$C$$ is similar to $$A$$ (specifically $$C = A^{-2} A A^2 = A$$). So $$C = A$$.

 Since $$A^2 = B$$, then $$C^2 = B$$.

 Use the Characteristic Equation of $$B$$: $$|B - \lambda I| = 0$$.

$$\begin{vmatrix} 1-\lambda & 3 \\ 1 & 5-\lambda \end{vmatrix} = (1-\lambda)(5-\lambda) - 3 = \lambda^2 - 6\lambda + 2 = 0$$

: By Cayley-Hamilton, $$B^2 - 6B + 2I = O$$. Substitute $$B = C^2$$:

$$(C^2)^2 - 6(C^2) + 2I = 0 \implies C^4 - 6C^2 + 2I = 0$$

\Compare with $$C^4 + \alpha C^2 + \beta I = 0$$: $$\alpha = -6, \beta = 2$$.

 $$2\beta - \alpha = 2(2) - (-6) = 4 + 6 = 10$$.

Correct Option: D (10)