Let A = [ cases 1 & 2 - 1 & 4 cases ] . If A ^ - 1 = + , α, β ∈ R,...

Given,

A = \left[\begin{cases}1 & 2 \\ - 1 & 4\end{cases}\right] ⇒ \left|A\right| = 6

So, {A}^{- 1} = \dfrac{adj A}{\left|A\right|} = \dfrac{1}{6}\left[\begin{cases}4 & - 2 \\ 1 & 1\end{cases}\right] = \left[\begin{cases}\dfrac{2}{3} & - \dfrac{1}{3} \\ \dfrac{1}{6} & \dfrac{1}{6}\end{cases}\right]

Here, \left[$\begin{cases}\dfrac{2}{3} & - \dfrac{1}{3} \\ \dfrac{1}{6} & \dfrac{1}{6}\end{cases}$\right] = \left[$\begin{cases}\alpha & 0 \\ 0 & \alpha\end{cases}$\right] + \left[$\begin{cases}\beta & 2\beta \\ - \beta & 4\beta\end{cases}$\right]</p><p>\left$\begin{array}{ll}\alpha + \beta = \dfrac{2}{3} \\ \beta = - \dfrac{1}{6}\end{array}$\right\} ⇒ \alpha = \dfrac{2}{3} + \dfrac{1}{6} = \dfrac{5}{6}

Hence, $4\left(\alpha - \beta\right) = 4\left(1\right) = 4$