Let $$A = \begin{pmatrix} 1 & -1 \\ 2 & \alpha \end{pmatrix}$$ and $$B = \begin{pmatrix} \beta & 1 \\ 1 & 0 \end{pmatrix}$$, $$\alpha, \beta \in \mathbb{R}$$. Let $$\alpha_1$$ be the value of $$\alpha$$ which satisfies $$(A + B)^2 = A^2 + \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}$$ and $$\alpha_2$$ be the value of $$\alpha$$ which satisfies $$(A + B)^2 = B^2$$. Then $$|\alpha_1 - \alpha_2|$$ is equal to

We have $$A = \begin{pmatrix} 1 & -1 \\ 2 & \alpha \end{pmatrix}$$, $$B = \begin{pmatrix} \beta & 1 \\ 1 & 0 \end{pmatrix}$$.

Part 1: Finding $$\alpha_1$$ from $$(A+B)^2 = A^2 + \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}$$

Expand $$(A+B)^2 = A^2 + AB + BA + B^2$$.

So the condition becomes $$AB + BA + B^2 = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}$$.

$$AB = \begin{pmatrix} 1 & -1 \\ 2 & \alpha \end{pmatrix}\begin{pmatrix} \beta & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} \beta - 1 & 1 \\ 2\beta + \alpha & 2 \end{pmatrix}$$

$$BA = \begin{pmatrix} \beta & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 2 & \alpha \end{pmatrix} = \begin{pmatrix} \beta + 2 & -\beta + \alpha \\ 1 & -1 \end{pmatrix}$$

$$B^2 = \begin{pmatrix} \beta & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} \beta & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} \beta^2 + 1 & \beta \\ \beta & 1 \end{pmatrix}$$

$$AB + BA + B^2 = \begin{pmatrix} (\beta-1) + (\beta+2) + (\beta^2+1) & 1 + (-\beta+\alpha) + \beta \\ (2\beta+\alpha) + 1 + \beta & 2 + (-1) + 1 \end{pmatrix}$$

$$= \begin{pmatrix} \beta^2 + 2\beta + 2 & \alpha + 1 \\ 3\beta + \alpha + 1 & 2 \end{pmatrix}$$

From position $$(1,2)$$: $$\alpha + 1 = 2 \implies \alpha_1 = 1$$.

From position $$(1,1)$$: $$\beta^2 + 2\beta + 2 = 2 \implies \beta^2 + 2\beta = 0 \implies \beta(\beta + 2) = 0$$.

So $$\beta = 0$$ or $$\beta = -2$$.

From position $$(2,1)$$: $$3\beta + \alpha_1 + 1 = 2 \implies 3\beta + 2 = 2 \implies \beta = 0$$.

Therefore $$\alpha_1 = 1$$ (with $$\beta = 0$$).

Part 2: Finding $$\alpha_2$$ from $$(A+B)^2 = B^2$$

$$(A+B)^2 = B^2$$ means $$A^2 + AB + BA = O$$ (the zero matrix).

$$A^2 = \begin{pmatrix} 1 & -1 \\ 2 & \alpha \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 2 & \alpha \end{pmatrix} = \begin{pmatrix} 1-2 & -1-\alpha \\ 2+2\alpha & -2+\alpha^2 \end{pmatrix} = \begin{pmatrix} -1 & -1-\alpha \\ 2+2\alpha & \alpha^2-2 \end{pmatrix}$$

$$= \begin{pmatrix} -1 + (\beta-1) + (\beta+2) & (-1-\alpha) + 1 + (-\beta+\alpha) \\ (2+2\alpha) + (2\beta+\alpha) + 1 & (\alpha^2-2) + 2 + (-1) \end{pmatrix}$$

$$= \begin{pmatrix} 2\beta & -\beta \\ 3\alpha + 2\beta + 3 & \alpha^2 - 1 \end{pmatrix}$$

From $$(1,1)$$: $$2\beta = 0 \implies \beta = 0$$.

From $$(1,2)$$: $$-\beta = 0 \implies \beta = 0$$. (Consistent.)

From $$(2,2)$$: $$\alpha^2 - 1 = 0 \implies \alpha = 1$$ or $$\alpha = -1$$.

From $$(2,1)$$: $$3\alpha + 0 + 3 = 0 \implies \alpha = -1$$.

Therefore $$\alpha_2 = -1$$.

$$|\alpha_1 - \alpha_2| = |1 - (-1)| = 2$$.

The answer is 2.