If A = [ cases 1 & 5 λ & 10 cases ], A ^ - 1 = + and α + β = - 2 ,...

Given,

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

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

$⇒ {k}^{2} - 11k + 10 - 5\lambda = 0$

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

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

Now on comparing with ${A}^{- 1} = \alphaA + \betaI$ we get,

$⇒ \alpha = - \dfrac{1}{10 - 5\lambda}, \beta = \dfrac{11}{10 - 5\lambda}$

And given,

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

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