$$\begin{gathered} Let A=\left[\begin{array}{cc}1& -1\\ 2& \alpha \end{array}\right] and B= \left[\begin{array}{cc}\beta & 1\\ 1& 0\end{array}\right],\alpha ,\beta \in R. \\ Let {\alpha }_1 be the value of \alpha which satisfies \\ {\left(A+B\right)}^2=A^2+\left[\begin{array}{cc}2& 2\\ 2& 2\end{array}\right] and {\alpha }_2 be the value of \alpha \\ which satisfies {\left(A+B\right)}^2=B^2. Then \left|{\alpha }_1-{\alpha }_2\right| is \\ equal to \_\_\_\_\_\_\_\_\_\_\_\_\_. \end{gathered}$$

$$\begin{gathered} A+B=\left[\begin{array}{cc}\beta +1& 0\\ 3& \alpha \end{array}\right] \\ \Rightarrow {\left(A+B\right)}^2=\left[\begin{array}{cc}\beta +1& 0\\ 3& \alpha \end{array}\right]\left[\begin{array}{cc}\beta +1& 0\\ 3& \alpha \end{array}\right] \\ =\left[\begin{array}{cc}{\left(\beta +1\right)}^2& 0\\ 3\left(\beta +1\right)+3\alpha & {\alpha }^2\end{array}\right], \\ {A}^2=\left[\begin{array}{cc}1& -1\\ 2& \alpha \end{array}\right]\left[\begin{array}{cc}1& -1\\ 2& \alpha \end{array}\right] \\ =\left[\begin{array}{cc}-1& -1-\alpha \\ 2+2\alpha & {\alpha }^2-2\end{array}\right] \\ \therefore \left[\begin{array}{cc}1& -\alpha +1\\ 2\alpha +4& {\alpha }^2\end{array}\right]=\left[\begin{array}{cc}{\left(\beta +1\right)}^2& 0\\ 3\left(\alpha +\beta +1\right)& {\alpha }^2\end{array}\right] \\ \underset{ }{\boxed{\alpha =1}\Rightarrow {\alpha }_1=1,} \\ {B}^2=\left[\begin{array}{cc}\beta & 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}\beta & 1\\ 1& 0\end{array}\right] \\ =\left[\begin{array}{cc}{\beta }^2+1& \beta \\ \beta & 1\end{array}\right]=\left[\begin{array}{cc}{\left(\beta +1\right)}^2& 0\\ 3\left(\beta +1\right)+3\alpha & {\alpha }^2\end{array}\right] \\ \therefore \beta =0, \alpha =-1={\alpha }_2, \\ \boxed{\left|{\alpha }_1-{\alpha }_2\right|=\left|1-\left(-1\right)\right|=2} \end{gathered}$$