JEE Main 2025 (Online) 2nd April Morning ShiftQuestion 1
Let $A=\begin{bmatrix}\alpha&-1\\6&\beta\end{bmatrix}$, $\alpha>0$, such that $\det(A)=0$ and $\alpha+\beta=1$. If $I$ denotes $2\times2$ identity matrix, then the matrix $(I+A)^{8}$ is :
Matrices
Solution
$\det A=\alpha\beta+6=0$, $\alpha+\beta=1$: $\alpha,\beta$ roots of $t^2-t-6=0$, i.e. $3,-2$; $\alpha>0\Rightarrow\alpha=3,\beta=-2$. $A=\begin{bmatrix}3&-1\\6&-2\end{bmatrix}$, $\operatorname{tr}A=1$, $\det A=0$, so $A^2=A$. $(I+A)^8=I+(2^8-1)A=I+255A=\begin{bmatrix}766&-255\\1530&-509\end{bmatrix}$. Option (C).