JEE 2023 — Mathematics PYQ

$A=\left[\begin{array}{cc}1& amp;5\\ \lambda & amp;10\end{array}\right]$

$\Rightarrow |A-xI|=0$

$\Rightarrow \left|\begin{array}{cc}1-x& amp;5\\ \lambda & amp;10-x\end{array}\right|=0$

$\Rightarrow (1-x)(10-x)-5\lambda =0$

$\Rightarrow 10-11x+x^2-5\lambda =0$

Also, $\Rightarrow A^{-1}=\alpha A+\beta I$

$\Rightarrow \alpha A^2+\beta A-I=0$

and $A^2-11A+(10-5\lambda )I=0$

On solving, we get
$\alpha =\frac{1}{5},\beta =-\frac{11}{5}$

So, $5\lambda -10=5\Rightarrow \lambda =3$

$\therefore 4{\alpha }^2+{\beta }^2+{\lambda }^2$

$=4\left(\frac{1}{25}\right)+\left(\frac{121}{25}\right)+9$

$=\frac{125}{25}+9=14$