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

<p>$$\begin{aligned} & B=\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right] \\ & A B^{-1}=A^{-1} \\ & \Rightarrow A^2=B \end{aligned}$$</p> <p>Also, $B C B^{-1}=A$</p> <p>$$\begin{aligned} \Rightarrow C & =B^{-1} A B \\ \Rightarrow C^4 & =\left(B^{-1} A B\right)\left(B^{-1} A B\right)\left(B^{-1} A B\right)\left(B^{-1} A B\right) \\ & =B^{-1} A^4 B \\ & =B^{-1} B^2 B \\ \Rightarrow \quad C^4 & =B^2 \end{aligned}$$</p> <p>Also, $C^2=\left(B^{-1} A B\right)\left(B^{-1} A B\right)$</p> <p>$$\begin{aligned} & =B^{-1} A^2 B \\ & =B^{-1} B B \end{aligned}$$</p> <p>$$\begin{aligned} & \Rightarrow C^2=B \\ & \Rightarrow C^4+\alpha C^2+\beta I=0 \\ & \Rightarrow B^2+\alpha B+\beta I=0 \\ & B^2=\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right]=\left[\begin{array}{ll} 4 & 18 \\ 6 & 28 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ll} 4 & 18 \\ 6 & 28 \end{array}\right]+\alpha\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right]+\left[\begin{array}{ll} \beta & 0 \\ 0 & \beta \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ & \Rightarrow 4+\alpha+\beta=0 \end{aligned}$$</p> <p>and $18+3 \alpha=0$</p> <p>$$\begin{aligned} & \Rightarrow \alpha=-6 \\ & \Rightarrow \beta=2 \\ & \Rightarrow 2 \beta-\alpha=10 \end{aligned}$$</p>