Let $$X = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$, $$Y = \alpha I + \beta X + \gamma X^2$$ and $$Z = \alpha^2 I - \alpha\beta X + (\beta^2 - \alpha\gamma)X^2, \alpha, \beta, \gamma \in \mathbb{R}$$.
If $$Y^{-1} = \begin{bmatrix} \frac{1}{5} & \frac{-2}{5} & \frac{1}{5} \\ 0 & \frac{1}{5} & \frac{-2}{5} \\ 0 & 0 & \frac{1}{5} \end{bmatrix}$$, then $$(\alpha - \beta + \gamma)^2$$ is equal to ______

First, note that $$X = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$, $$Y = \alpha I + \beta X + \gamma X^2$$, and $$Y^{-1} = \begin{bmatrix} 1/5 & -2/5 & 1/5 \\ 0 & 1/5 & -2/5 \\ 0 & 0 & 1/5 \end{bmatrix}$$, and we seek $$(\alpha - \beta + \gamma)^2$$.

Next, compute $$X^2$$: $$X^2 = X \cdot X = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ Since $$X^3 = 0$$ (the zero matrix), higher powers vanish.

Now, write $$Y$$ explicitly as $$Y = \alpha\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + \beta\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} + \gamma\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} \alpha & \beta & \gamma \\ 0 & \alpha & \beta \\ 0 & 0 & \alpha \end{bmatrix}.$$

Since for an upper triangular matrix with constant diagonal $$\alpha$$ the inverse has diagonal entries $$1/\alpha$$, and given that $$Y^{-1}$$ has diagonal entries $$1/5$$, it follows that $$\alpha = 5$$.

Next, the inverse of $$Y$$ can be expressed as $$Y^{-1} = \frac{1}{\alpha}I - \frac{\beta}{\alpha^2}X + \frac{\beta^2 - \alpha\gamma}{\alpha^3}X^2,$$ so its $$(1,2)$$ entry is $$-\frac{\beta}{\alpha^2}$$. Equating this to the given value $$-2/5$$ leads to $$-\frac{\beta}{\alpha^2} = -\frac{2}{5} \implies \beta = \frac{2\alpha^2}{5} = 10.$$

Substituting $$\alpha = 5$$ and $$\beta = 10$$ into the formula for the $$(1,3)$$ entry, $$\frac{\beta^2 - \alpha\gamma}{\alpha^3}$$, and setting it equal to the given $$1/5$$ yields $$\frac{100 - 5\gamma}{125} = \frac{1}{5} \implies 100 - 5\gamma = 25 \implies \gamma = 15.$$

Therefore, $$\alpha - \beta + \gamma = 5 - 10 + 15 = 10$$ and $$(\alpha - \beta + \gamma)^2 = 10^2 = 100,$$ which is the required answer.