$${(A + B)^2} = {A^2} + {B^2} + AB + BA$$

$$ = {A^2} + \left[ {\matrix{ 2 & 2 \cr 2 & 2 \cr } } \right]$$

$$\therefore$$ $${B^2} + AB + BA = \left[ {\matrix{ 2 & 2 \cr 2 & 2 \cr } } \right]$$ ..... (1)

$$AB = \left[ {\matrix{ 1 & { - 1} \cr 2 & \alpha \cr } } \right]\left[ {\matrix{ \beta & 1 \cr 1 & 0 \cr } } \right] = \left[ {\matrix{ {\beta - 1} & 1 \cr {\alpha + 2\beta } & 2 \cr } } \right]$$

$$BA = \left[ {\matrix{ \beta & 1 \cr 1 & 0 \cr } } \right]\left[ {\matrix{ 1 & { - 1} \cr 2 & \alpha \cr } } \right] = \left[ {\matrix{ {\beta + 2} & {\alpha - \beta } \cr 1 & { - 1} \cr } } \right]$$

$${B^2} = \left[ {\matrix{ \beta & 1 \cr 1 & 0 \cr } } \right]\left[ {\matrix{ \beta & 1 \cr 1 & 0 \cr } } \right] = \left[ {\matrix{ {{\beta ^2} + 1} & \beta \cr \beta & 1 \cr } } \right]$$

By (1) we get

$$\left[ {\matrix{ {{\beta ^2} + 2\beta } + 2 & {\alpha + 1} \cr {\alpha + 3\beta + 1} & 2 \cr } } \right] = \left[ {\matrix{ 2 & 2 \cr 2 & 2 \cr } } \right]$$

$$\therefore$$ $$\alpha = 1\,\,\beta = 0\,\, \Rightarrow {\alpha _1} = 1$$

Similarly if $${A^2} + AB + BA = 0$$ then

$$\left( {{A^2} = \left[ {\matrix{ 1 & { - 1} \cr 2 & \alpha \cr } } \right]\left[ {\matrix{ 1 & { - 1} \cr 2 & \alpha \cr } } \right] = \left[ {\matrix{ { - 1} & { - 1 - \alpha } \cr {2 + 2\alpha } & {{\alpha ^2} - 2} \cr } } \right]} \right)$$

$$\left[ {\matrix{ {2\beta } & {\alpha - \beta + 1 - 1 - \alpha } \cr {\alpha + 2\beta + 1 + 2 + 2\alpha } & {{\alpha ^2} - 2 + 1} \cr } } \right] = \left[ {\matrix{ 0 & 0 \cr 0 & 0 \cr } } \right]$$

$$ \Rightarrow \beta = 0$$ and $$\alpha = - 1\,\, \Rightarrow {\alpha _2} = - 1$$

$$\therefore$$ $$|{\alpha _1} - {\alpha _2}| = |2| = 2.$$