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JEE Main 2024
09-04-2024 S2
Question
Let $B=\left[\begin{array}{ll}1 & 3 \\ 1 & 5\end{array}\right]$ and $A$ be a $2 \times 2$ matrix such that $\mathrm{AB}^{-1}=\mathrm{A}^{-1}$. If $\mathrm{BCB}^{-1}=\mathrm{A}$ and $\mathrm{C}^4+\alpha \mathrm{C}^2+\beta \mathrm{I}=0$, then $2 \beta-\alpha$ is equal to :
Select the correct option:
A
16
B
2
C
8
D
10
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$ \begin{aligned} & \mathrm{BCB}^{-1}=\mathrm{A} \\ & \Rightarrow\left(\mathrm{BCB}^{-1}\right)\left(\mathrm{BCB}^{-1}\right)=\mathrm{A} \cdot \mathrm{~A} \\ & \Rightarrow \mathrm{BCICB}^{-1}=\mathrm{A}^2 \\ & \Rightarrow \mathrm{BC}^2 \mathrm{~B}^{-1}=\mathrm{A}^2 \\ & \Rightarrow \mathrm{~B}^{-1}\left(\mathrm{BC}^2 \mathrm{~B}^{-1}\right) \mathrm{B}=\mathrm{B}^{-1}(\mathrm{~A} \cdot \mathrm{~A}) \mathrm{B} \end{aligned} $
From equation (1)
$ \begin{aligned} & C^2=A^{-1} \cdot A \cdot B \\ & C^2=B \end{aligned} $
Also $\mathrm{AB}^{-1}=\mathrm{A}^{-1}$
$ \begin{aligned} & \Rightarrow \mathrm{AB}^{-1} \cdot \mathrm{~A}=\mathrm{A}^{-1} \mathrm{~A}=\mathrm{I} \\ & \Rightarrow \mathrm{~A}^{-1}\left(\mathrm{AB}^{-1} \mathrm{~A}\right)=\mathrm{A}^{-1} \mathrm{I} \\ & \mathrm{~B}^{-1} \mathrm{~A}=\mathrm{A}^{-1} \end{aligned} $
Now characteristics equation of $C^2$ is
$ \begin{aligned} & \left|\mathrm{C}_2-\lambda \mathrm{I}\right|=0 \\ & |B-\lambda \mathrm{I}|=0 \\ & \Rightarrow\left|\begin{array}{cc} 1-\lambda & 3 \\ 1 & 5-\lambda \end{array}\right|=0 \end{aligned} $
$\begin{aligned} & \Rightarrow(1-\lambda)(5-1)-3=0 \Rightarrow\left(\lambda^2-6 \lambda+5\right)-3=0 \\ & \Rightarrow \lambda^2-6 \lambda+2=0 \\ & \Rightarrow \beta^2-6 \mathrm{~B}+2 \mathrm{I}=0 \\ & \Rightarrow \mathrm{C}^4-6 \mathrm{C}^2+2 \mathrm{I}=0 \\ & \alpha=-6 \\ & \beta=2 \\ & \therefore 2 \beta-\alpha=4+6=10\end{aligned}$
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