Let $$A$$ be a $$2 \times 2$$ symmetric matrix such that $$A\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \end{bmatrix}$$ and the determinant of $$A$$ be $$1$$. If $$A^{-1} = \alpha A + \beta I$$, where $$I$$ is an identity matrix of order $$2 \times 2$$, then $$\alpha + \beta$$ equals _____

Let $$A = \begin{bmatrix}a&b\\b&d\end{bmatrix}$$ (symmetric). From $$A\begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}3\\7\end{bmatrix}$$: $$a+b=3$$ and $$b+d=7$$.

From $$\det(A) = ad - b^2 = 1$$ and $$a = 3-b, d = 7-b$$: $$(3-b)(7-b) - b^2 = 21-10b = 1$$, so $$b=2, a=1, d=5$$.

$$A = \begin{bmatrix}1&2\\2&5\end{bmatrix}$$, $$A^{-1} = \begin{bmatrix}5&-2\\-2&1\end{bmatrix}$$.

From $$A^{-1} = \alpha A + \beta I$$: comparing entries gives $$2\alpha = -2$$ (so $$\alpha = -1$$) and $$\alpha + \beta = 5$$ (so $$\beta = 6$$).

$$\alpha + \beta = -1 + 6 = \boxed{5}$$.