Let A be a 2 $$\times$$ 2 matrix with real entries such that $$A' = \alpha A + 1$$, where $$\alpha \in \mathbb{R} - \{-1, 1\}$$. If det$$(A^2 - A) = 4$$, the sum of all possible values of $$\alpha$$ is equal to

Given $$A' = \alpha A + I$$ where A is a 2×2 real matrix.

Let $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$, then $$A' = \begin{pmatrix} a & c \\ b & d \end{pmatrix}$$.

From $$A' = \alpha A + I$$:

$$a = \alpha a + 1 \Rightarrow a = \frac{1}{1-\alpha}$$

$$c = \alpha b$$ and $$b = \alpha c$$

$$d = \alpha d + 1 \Rightarrow d = \frac{1}{1-\alpha}$$

From $$b = \alpha c$$ and $$c = \alpha b$$: $$b = \alpha^2 b$$, so $$b(1-\alpha^2) = 0$$. Since $$\alpha \neq \pm 1$$, we get $$b = 0$$ and $$c = 0$$.

Therefore $$A = kI$$ where $$k = \frac{1}{1-\alpha}$$.

$$A^2 - A = k^2I - kI = (k^2 - k)I$$

$$\det(A^2 - A) = (k^2 - k)^2 = 4$$

$$k^2 - k = \pm 2$$

Case 1: $$k^2 - k - 2 = 0 \Rightarrow (k-2)(k+1) = 0$$

$$k = 2 \Rightarrow \alpha = \frac{1}{2}$$ or $$k = -1 \Rightarrow \alpha = 2$$

Case 2: $$k^2 - k + 2 = 0$$

Discriminant $$= 1 - 8 = -7 < 0$$. No real solutions.

Sum of all possible values of $$\alpha = \frac{1}{2} + 2 = \frac{5}{2}$$