JEE MAIN - Mathematics (2025 - 2nd April Morning Shift - No. 1)
Let $\mathrm{A}=\left[\begin{array}{cc}\alpha & -1 \\ 6 & \beta\end{array}\right], \alpha>0$, such that $\operatorname{det}(\mathrm{A})=0$ and $\alpha+\beta=1$. If I denotes $2 \times 2$ identity matrix, then the matrix $(I+A)^8$ is :
$\left[\begin{array}{cc}257 & -64 \\ 514 & -127\end{array}\right]$
$\left[\begin{array}{cc}766 & -255 \\ 1530 & -509\end{array}\right]$
$\left[\begin{array}{cc}1025 & -511 \\ 2024 & -1024\end{array}\right]$
$\left[\begin{array}{ll}4 & -1 \\ 6 & -1\end{array}\right]$
Explanation
Let $|A|=0 \Rightarrow \alpha \beta-(-6)=0 \Rightarrow \alpha \beta=-6$ and $\alpha+\beta=1 \Rightarrow \alpha \beta$ are roots of the equation
$$\begin{aligned} & x^2-x-6=0 \Rightarrow x=3,-2 . \text { Since } \alpha>0 \\ & \Rightarrow \quad \alpha=3, \beta=-2 \\ & \Rightarrow \quad A=\left[\begin{array}{ll} 3 & -1 \\ 6 & -2 \end{array}\right] \Rightarrow I+A=\left[\begin{array}{ll} 4 & -1 \\ 6 & -1 \end{array}\right] \\ & (I+A)^2=\left[\begin{array}{ll} 10 & -3 \\ 18 & -5 \end{array}\right] \Rightarrow(I+A)^4=\left[\begin{array}{ll} 46 & -15 \\ 90 & -29 \end{array}\right] \\ & \Rightarrow \quad(I+A)^8=\left[\begin{array}{cc} 766 & -255 \\ 1530 & -509 \end{array}\right] \end{aligned}$$
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