Let B = [ 1 3 1 5 ] and A be a 2 × 2 matrix such that A B - 1 = A - 1 . If B C B - 1 = A and C 4 + C 2 + I = O , then is equal to [2024]

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Solution

Q.
Let $B = [\begin{matrix} 1 & 3 \\ 1 & 5 \end{matrix}]$ and $A$ be a $2 \times 2$ matrix such that $A B^{- 1} = A^{- 1} .$ If $B C B^{- 1} = A$ and $C^{4} + α C^{2} + β I = O,$ then $2 β - α$ is equal to [2024]

1 2

2 8

3 10

4 16

Ans.
(3)

$B = [\begin{matrix} 1 & 3 \\ 1 & 5 \end{matrix}]$

We have, $B C B^{- 1} = A$

$\Rightarrow (B C B^{- 1}) (B C B^{- 1}) = A^{2} \Rightarrow B C B^{- 1} B C B^{- 1} = A^{2}$

$\Rightarrow B C^{2} B^{- 1} = A^{2} \Rightarrow B^{- 1} B C^{2} B^{- 1} = B^{- 1} A^{2}$

$\Rightarrow C^{2} B^{- 1} B = B^{- 1} A^{2} B$

$\Rightarrow C^{2} = B^{- 1} A^{2} B$

$\Rightarrow C^{2} = B^{- 1} A A B$ ...(i)

$\Rightarrow C^{2} \cdot C^{2} = B^{- 1} A^{2} B B^{- 1} A^{2} B$

$\Rightarrow C^{4} = B^{- 1} A^{4} B$ ...(ii)

Now, $A B^{- 1} = A^{- 1}$

$\Rightarrow A^{2} B^{- 1} = I$

$\Rightarrow A^{2} = B$ ...(iii)

Using (i), (ii) and (iii), we get

$C^{2} = B^{- 1} B B = B$ and $C^{4} = B^{- 1} B^{2} B = B^{2}$

So, $C^{4} + α C^{2} + β I = 0 \Rightarrow B^{2} + α B + β I = 0$

$= [\begin{matrix} 1 & 3 \\ 1 & 5 \end{matrix}] [\begin{matrix} 1 & 3 \\ 1 & 5 \end{matrix}] + [\begin{matrix} α & 3 α \\ α & 5 α \end{matrix}] + [\begin{matrix} β & 0 \\ 0 & β \end{matrix}] = 0$

$\Rightarrow [\begin{matrix} 4 & 18 \\ 6 & 28 \end{matrix}] + [\begin{matrix} α & 3 α \\ α & 5 α \end{matrix}] + [\begin{matrix} β & 0 \\ 0 & β \end{matrix}] = 0$

$\Rightarrow 4 + α + β = 0$ and $6 + α = 0$

$\Rightarrow α = - 6, β = 2 \Rightarrow 2 β - α = 10$

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