Let $$A=\left[\begin{array}{cc}1 & -1 \\ 2 & \alpha\end{array}\right]$$ and $$B=\left[\begin{array}{cc}\beta & 1 \\ 1 & 0\end{array}\right], \alpha, \beta \in \mathbf{R}$$. Let $\alpha_{1}$ be the value of $\alpha$ which satisfies $$(\mathrm{A}+\mathrm{B})^{2}=\mathrm{A}^{2}+\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]$$ and $\alpha_{2}$ be the value of $\alpha$ which satisfies $(\mathrm{A}+\mathrm{B})^{2}=\mathrm{B}^{2}$. Then $\left|\alpha_{1}-\alpha_{2}\right|$ is equal to ___________.

<p>${(A + B)^2} = {A^2} + {B^2} + AB + BA$</p> <p>$= {A^2} + \left[ {\matrix{ 2 & 2 \cr 2 & 2 \cr } } \right]$</p> <p>$\therefore$ ${B^2} + AB + BA = \left[ {\matrix{ 2 & 2 \cr 2 & 2 \cr } } \right]$ ..... (1)</p> <p>$$AB = \left[ {\matrix{ 1 & { - 1} \cr 2 & \alpha \cr } } \right]\left[ {\matrix{ \beta & 1 \cr 1 & 0 \cr } } \right] = \left[ {\matrix{ {\beta - 1} & 1 \cr {\alpha + 2\beta } & 2 \cr } } \right]$$</p> <p>$$BA = \left[ {\matrix{ \beta & 1 \cr 1 & 0 \cr } } \right]\left[ {\matrix{ 1 & { - 1} \cr 2 & \alpha \cr } } \right] = \left[ {\matrix{ {\beta + 2} & {\alpha - \beta } \cr 1 & { - 1} \cr } } \right]$$</p> <p>$${B^2} = \left[ {\matrix{ \beta & 1 \cr 1 & 0 \cr } } \right]\left[ {\matrix{ \beta & 1 \cr 1 & 0 \cr } } \right] = \left[ {\matrix{ {{\beta ^2} + 1} & \beta \cr \beta & 1 \cr } } \right]$$</p> <p>By (1) we get</p> <p>$$\left[ {\matrix{ {{\beta ^2} + 2\beta } + 2 & {\alpha + 1} \cr {\alpha + 3\beta + 1} & 2 \cr } } \right] = \left[ {\matrix{ 2 & 2 \cr 2 & 2 \cr } } \right]$$</p> <p>$\therefore$ $\alpha = 1\,\,\beta = 0\,\, \Rightarrow {\alpha _1} = 1$</p> <p>Similarly if ${A^2} + AB + BA = 0$ then</p> <p>$$\left( {{A^2} = \left[ {\matrix{ 1 & { - 1} \cr 2 & \alpha \cr } } \right]\left[ {\matrix{ 1 & { - 1} \cr 2 & \alpha \cr } } \right] = \left[ {\matrix{ { - 1} & { - 1 - \alpha } \cr {2 + 2\alpha } & {{\alpha ^2} - 2} \cr } } \right]} \right)$$</p> <p>$$\left[ {\matrix{ {2\beta } & {\alpha - \beta + 1 - 1 - \alpha } \cr {\alpha + 2\beta + 1 + 2 + 2\alpha } & {{\alpha ^2} - 2 + 1} \cr } } \right] = \left[ {\matrix{ 0 & 0 \cr 0 & 0 \cr } } \right]$$</p> <p>$\Rightarrow \beta = 0$ and $\alpha = - 1\,\, \Rightarrow {\alpha _2} = - 1$</p> <p>$\therefore$ $|{\alpha _1} - {\alpha _2}| = |2| = 2.$</p>