a) For all $p \in P, p=ar, r \in R.$ Since, $P$ is prime and $P\neq (a) \Rightarrow r\in P.$ For all $x \in aP, x=ap$ Since $P$ is an ideal $\Rightarrow ap \in P $
b) $\forall i$ $1\leq i\leq n$ $ a_i=ap_i=a(a_1r_{i1}+...+a_nr_{in}) \Rightarrow 0=a_1(-ar_{i1})+...+a_i(1-ar_{ii})+a_n(-r_{in})$
Hence, the $r_{ij}$'s are the entries on the matrix.
c) This is the part that I'm stuck at. I can't seem to find the required element. I know it satisfies $(1-ab)*a_i=0$. But I can't proceed any further.
d) Follows easily from (c).
Any hints or solutions?
It seems only (c) has some difficulty for you. denote $A^*$ be the adjoint matrix for $A$.Since $AP=0$ we get $A^*AP=0$.that is $det(A)IP=0$ where $I$ is the unit matrix.remark $det(A)=1-ab$ for some $b$.Hence you can get the result.