Matrix $$A = \left| {\matrix{ x & 3 & 2 \cr 1 & y & 4 \cr 2 & 2 & z \cr } } \right|$$, if xyz = 60 and 8x + 4y + 3z = 20, then A(adj A) is equal to

An ordered pair ($$\alpha$$, $$\beta$$) for which the system of linear $$(1 + \alpha )x + \beta y + z = 2$$, $$\alpha x + (1 + \beta )y + z = 3$$, $$\alpha x + \beta y + 2z = 2$$ has a unique solution.

Consider matrix $$A = \left[ {\matrix{ 2 & 1 \cr 1 & 2 \cr } } \right]$$, if $${A^{ - 1}} = \alpha I + \beta A$$, where $$\alpha$$, $$\beta$$ $$ \notin $$ R, then ($$\alpha$$ + $$\beta$$) is equal to (where A^$$-$$1 denotes the inverse of matrix A)