Correct option is (D) Does not exists
Given system of equations is
kx – y = 2
\(\Rightarrow\) kx - y - 2 = 0 _________(1)
6x – 2y – 3 = 0 _________(2)
By comparing given system with standard form of system, we get
\(a_1=k,b_1=-1,c_1=-2\)
and \(a_2=6,b_2=-2,c_2=-3\)
\(\therefore\) \(\frac{a_1}{a_2}=\frac k6,\)
\(\frac{b_1}{b_2}=\frac{-1}{-2}=\frac12\)
\(\frac{c_1}{c_2}=\frac{-2}{-3}=\frac23\)
\(\because\) \(\frac12\neq\frac23\)
\(\therefore\) \(\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\)
But condition for infinitely many solutions is
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) which is not possible.
Hence, for any value of k, the given system never give infinitely many solutions.
Hence, the value of k does not exists.
