Let A = [ cases 1 & - 1 2 & α cases ] and B = [ cases β & 1 1 & 0...

Given A = \left[\begin{cases}1 & - 1 \\ 2 & \alpha\end{cases}\right] and B = \left[\begin{cases}\beta & 1 \\ 1 & 0\end{cases}\right], \alpha, \beta \in R,

So, A + B = \left[\begin{cases}\beta + 1 & 0 \\ 3 & \alpha\end{cases}\right]

Now {\left(A + B\right)}^{2} = \left[\begin{cases}\beta + 1 & 0 \\ 3 & \alpha\end{cases}\right]\left[\begin{cases}\beta + 1 & 0 \\ 3 & \alpha\end{cases}\right]

= \left[\begin{cases}{\left(\beta + 1\right)}^{2} & 0 \\ 3\left(\beta + 1\right) + 3\alpha & {\alpha}^{2}\end{cases}\right]

Also, {A}^{2} = \left[\begin{cases}1 & - 1 \\ 2 & \alpha\end{cases}\right]\left[\begin{cases}1 & - 1 \\ 2 & \alpha\end{cases}\right]= \left[\begin{cases}- 1 & - 1 - \alpha \\ 2 + 2\alpha & {\alpha}^{2} - 2\end{cases}\right]

Now solving {\left(A + B\right)}^{2} = {A}^{2} + \left[\begin{cases}2 & 2 \\ 2 & 2\end{cases}\right]

⇒ \left[\begin{cases}{\left(\beta + 1\right)}^{2} & 0 \\ 3\left(\alpha + \beta + 1\right) & {\alpha}^{2}\end{cases}\right] = \left[\begin{cases}1 & - \alpha + 1 \\ 2\alpha + 4 & {\alpha}^{2}\end{cases}\right]

Now on comparing both side we get, $\alpha = 1 = {\alpha}_{1}$

And {B}^{2} = \left[\begin{cases}\beta & 1 \\ 1 & 0\end{cases}\right]\left[\begin{cases}\beta & 1 \\ 1 & 0\end{cases}\right] = \left[\begin{cases}{\beta}^{2} + 1 & \beta \\ \beta & 1\end{cases}\right]

Now using ${\left(A + B\right)}^{2} = {B}^{2}$

⇒ \left[\begin{cases}{\beta}^{2} + 1 & \beta \\ \beta & 1\end{cases}\right] = \left[\begin{cases}{\left(\beta + 1\right)}^{2} & 0 \\ 3\left(\beta + 1\right) + 3\alpha & {\alpha}^{2}\end{cases}\right]

Again on comparing both side we get, $\beta = 0, \alpha = - 1 = {\alpha}_{2}$

So, \left|{\alpha}_{1} - {\alpha}_{2}\right| = \left|1 - \left(- 1\right)\right| = 2