JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
Let \(A=\left[\begin{array}{cc}\alpha & -1 \\ 6 & \beta\end{array}\right], \alpha \gt 0\), such that \(\operatorname{det}(A)=0\) and \(\alpha+\beta=1\). If I denotes \(2 \times 2\) identity matrix, then the matrix \((1+\mathrm{A})^8\) is:
- A \(\left[\begin{array}{ll}4 & -1 \\ 6 & -1\end{array}\right]\)
- B \(\left[\begin{array}{cc}257 & -64 \\ 514 & -127\end{array}\right]\)
- C \(\left[\begin{array}{cc}1025 & -511 \\ 2024 & -1024\end{array}\right]\)
- D \(\left[\begin{array}{cc}766 & -255 \\ 1530 & -509\end{array}\right]\)
Answer & Solution
Correct Answer
(D) \(\left[\begin{array}{cc}766 & -255 \\ 1530 & -509\end{array}\right]\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & |\mathrm{A}|=0 \\ & \alpha \beta+6=0 \\ & \alpha \beta=-6 \\ & \alpha+\beta=1 \\ & \Rightarrow \alpha=3, \beta=-2 \\ & \mathrm{~A}=\left[\begin{array}{ll}3 & -1 \\ 6 & -2\end{array}\right] \\ & \mathrm{A}^2=\left[\begin{array}{ll}3 & -1 \\ 6 &…
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