If $${A^2} - A + 1 = 0$$, then the inverse of $$A$$ is :

The system of equations

$$\matrix{ {\alpha \,x + y + z = \alpha - 1} \cr {x + \alpha y + z = \alpha - 1} \cr {x + y + \alpha \,z = \alpha - 1} \cr } $$

has no solutions, if $$\alpha $$ is :

If $${a_1},{a_2},{a_3},........,{a_n},.....$$ are in G.P., then the determinant $$$\Delta = \left| {\matrix{ {\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr {\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr {\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr } } \right|$$$
is equal to :

If $${a^2} + {b^2} + {c^2} = - 2$$ and

f$$\left( x \right) = \left| {\matrix{ {1 + {a^2}x} & {\left( {1 + {b^2}} \right)x} & {\left( {1 + {c^2}} \right)x} \cr {\left( {1 + {a^2}} \right)x} & {1 + {b^2}x} & {\left( {1 + {c^2}} \right)x} \cr {\left( {1 + {a^2}} \right)x} & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr } } \right|,$$

then f$$(x)$$ is a polynomial of degree :