Let $B=\left[\begin{array}{ll}1 & 3 \\ 1 & 5\end{array}\right]$ and $A$ be a $2 \times 2$ matrix such that…

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
  1. 16
  2. 2
  3. 8
  4. 10

Solution

$\begin{aligned} & \mathrm{BCB}^{-1}=\mathrm{A} \\ & \Rightarrow\left(\mathrm{BCB}^{-1}\right)\left(\mathrm{BCB}^{-1}\right)=\mathrm{A} \cdot \mathrm{A} \\ & \Rightarrow \mathrm{BCI} \mathrm{CB}^{-1}=\mathrm{A}^2 \\ & \Rightarrow \mathrm{BC}^2 \mathrm{~B}^{-1}=\mathrm{A}^2 \\ & \Rightarrow \mathrm{B}^{-1}\left(\mathrm{BC}^2 \mathrm{~B}^{-1}\right) \mathrm{B}=\mathrm{B}^{-1} \text { (A.A)B } \end{aligned}$
From equation (1) $\begin{aligned} & \mathrm{C}^2=\mathrm{A}^{-1} \cdot \mathrm{A} \cdot \mathrm{B} \\ & \mathrm{C}^2=\mathrm{B} \end{aligned}$ $\begin{aligned} & \text { Also } \mathrm{AB}^{-1}=\mathrm{A}^{-1} \\ & \Rightarrow \mathrm{AB}^{-1} \cdot \mathrm{A}=\mathrm{A}^{-1} \mathrm{~A}=\mathrm{I} \\ & \Rightarrow \mathrm{A}^{-1}\left(\mathrm{AB}^{-1} \mathrm{~A}\right)=\mathrm{A}^{-1} \mathrm{I} \\ & \mathrm{B}^{-1} \mathrm{~A}=\mathrm{A}^{-1} \end{aligned}$
Now characteristics equation of $\mathrm{C}^2$ is $\begin{aligned} & \left|C_2-\lambda I\right|=0 \\ & |B-\lambda I|=0 \end{aligned}$ $\begin{aligned} & \Rightarrow\left|\begin{array}{cc}1-\lambda & 3 \\ 1 & 5-\lambda\end{array}\right|=0 \\ & \Rightarrow(1-\lambda)(5-1)-3=0 \Rightarrow\left(\lambda^2-6 \lambda+5\right)-3=0 \\ & \Rightarrow \lambda^2-6 \lambda+2=0 \\ & \Rightarrow \beta^2-6 \mathrm{~B}+2 \mathrm{I}=0 \\ & \Rightarrow \mathrm{C}^4-6 \mathrm{C}^2+2 \mathrm{I}=0 \\ & \alpha=-6 \\ & \beta=2 \\ & \therefore 2 \beta-\alpha=4+6=10\end{aligned}$

Asked in: JEE Main 2024 (09 Apr Shift 2)

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