If A = 1 5 λ 10 ,   A - 1 = α A + β I and α + β = - 2 , then 4 α 2 +…

Given,

$A=\left[\begin{array}{cc}1& 5\\ \lambda & 10\end{array}\right], A^{-1}=\alpha A+\beta I$ and $\alpha +\beta =-2$,

Now solving by using characteristic equation, we get
$|A-k|=0\Rightarrow \left|\begin{array}{cc}1-k& 5\\ \lambda & 10-k\end{array}\right|=0$

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

$\Rightarrow A^2-11A+(10-5\lambda )I=0$ {by putting $k=A$}

$\Rightarrow A^{-1}=\frac{1}{10-5\lambda }(-A+11I)$

Now on comparing with $A^{-1}=\alpha A+\beta I$ we get,

$\Rightarrow \alpha =-\frac{1}{10-5\lambda }, \beta =\frac{11}{10-5\lambda }$

And given,

$\alpha +\beta =-2\Rightarrow \lambda =3, \alpha =\frac{1}{5}, \beta =\frac{-11}{5}$

Hence, the value of $4{\alpha }^2+{\beta }^2+{\lambda }^2=14$