Let $$A = \left( {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right)$$ and $$B = \left( {\matrix{ a & 0 \cr 0 & b \cr } } \right),a,b \in N.$$ Then

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 :