JEE Main 2022 (Online) 27th July Morning ShiftQuestion 1
Let $A=\begin{pmatrix}1&2\\-2&-5\end{pmatrix}$. Let $\alpha,\beta\in\mathbb{R}$ be such that $\alpha A^2+\beta A=2I$. Then $\alpha+\beta$ is equal to
Matrices
Solution
$\operatorname{tr}A=-4$, $\det A=-5+4=-1$. Cayley-Hamilton: $A^2+4A-I=O\Rightarrow A^2=I-4A$. $\alpha A^2+\beta A=\alpha I+(\beta-4\alpha)A=2I\Rightarrow \alpha=2,\ \beta=4\alpha=8$ ($A$ is not scalar). $\alpha+\beta=10$. Option (D).