JEE Main 2024 (Online) 4th April Evening ShiftQuestion 1
Let $A$ be a $2\times2$ symmetric matrix such that $A\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\7\end{bmatrix}$ and the determinant of $A$ be $1$. If $A^{-1}=\alpha A+\beta I$, where $I$ is an identity matrix of order $2\times2$, then $\alpha+\beta$ equals $\_\_\_\_\_\_\_\_$ .
Matrices
Answer$5$
Solution
$A=\begin{bmatrix}p&q\\q&r\end{bmatrix}$ with $p+q=3$, $q+r=7$, $pr-q^2=1$ gives $A=\begin{bmatrix}1&2\\2&5\end{bmatrix}$. Cayley-Hamilton: $A^2-6A+I=O$, so $A^{-1}=-A+6I$ and $\alpha+\beta=5$.