JEE Main 2023 (Online) 8th April Evening ShiftQuestion 1
If $A=\begin{bmatrix}1&5\\\lambda&10\end{bmatrix}$, $A^{-1}=\alpha A+\beta I$ and $\alpha+\beta=-2$, then $4\alpha^{2}+\beta^{2}+\lambda^{2}$ is equal to :
Matrices
Solution
$A^{-1}=\alpha A+\beta I\Rightarrow I=\alpha A^2+\beta A$. Cayley-Hamilton: $A^2=11A-(10-5\lambda)I$. $I=\alpha\,(11A-(10-5\lambda)I)+\beta A$, so $11\alpha+\beta=0$ and $-\alpha(10-5\lambda)=1$. With $\alpha+\beta=-2$: $\beta=-11\alpha\Rightarrow-10\alpha=-2$, $\alpha=\tfrac15$, $\beta=-\tfrac{11}5$, then $10-5\lambda=-5\Rightarrow\lambda=3$. $4\alpha^2+\beta^2+\lambda^2=\tfrac4{25}+\tfrac{121}{25}+9=5+9=14$. Option (C).