Correct option is (A) p = 2
Condition to have no solution is
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\)
\(\therefore\) \(\frac3{2p-1}=\frac1{p-1}\neq\frac{-1}{-(2p+1)}\)
\(\Rightarrow\) \(\frac3{2p-1}=\frac1{p-1}\)
\(\Rightarrow\) 3 (p - 1) = 2p - 1
\(\Rightarrow\) 3p - 3 = 2p - 1
\(\Rightarrow\) 3p - 2p = 3 - 1
\(\Rightarrow\) p = 2
Hence, for p = 2, the given system of equations has no solution.
