If $$p = \left[ {\matrix{ 1 & \alpha & 3 \cr 1 & 3 & 3 \cr 2 & 4 & 4 \cr } } \right]$$ is the adjoint of the $$3 \times 3$$ matrix A and det A = 4, then $$\alpha$$ is equal to

If $$A = \left( {\matrix{ 1 & 1 \cr 0 & i \cr } } \right)$$ and $${A^{2018}} = \left( {\matrix{ a & b \cr c & d \cr } } \right)$$, then $$(a + d)$$ equals

The solution of $$\det (A - \lambda {I_2}) = 0$$ be 4 and 8 and $$A = \left( {\matrix{ 2 & 2 \cr x & y \cr } } \right)$$. Then

(I₂ is identity matrix of order 2)