To check conflict serializability, we construct the precedence graph for each schedule:
For Schedule S1:
Conflict on X: r3(X) is followed by w1(X)⟹T3→T1
Conflict on Y: r2(Y) is followed by w3(Y)⟹T2→T3
Conflict on Z: w2(Z) is followed by r1(Z) and w1(Z)⟹T2→T1
The precedence graph has edges T2→T3→T1 and T2→T1, which is acyclic. Therefore, S1 is conflict-serializable.
For Schedule S2:
Conflict on X: r3(X) is followed by w1(X)⟹T3→T1
Conflict on Y: r2(Y) is followed by w3(Y)⟹T2→T3
Conflict on Z:
r1(Z) appears before w2(Z)⟹T1→T2
w2(Z) appears before w1(Z)⟹T2→T1
This creates a cycle (T1→T2→T1). Therefore, S2 is not conflict-serializable.
Hence, only S1 is conflict-serializable (A).