Question

For the following system of equation determine the value of k for which the given system has no solution.
$2x-ky+3=0$
$3x+2y-1=0$

Answer 1

The given system of equations is
$2x-ky+3=0$
$3x+2y-1=0$

This is of the form $a_{1}x+b_{1}y+c_1=0$
$a_{2}x+b_{2}y+c_2=0$

where, $a_1=2,b_1=-k,c_2=3$

and $a_2=3,b_2=2,c_2=-1$

For no solution, we must have
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

We have, $\frac{a_1}{a_2}=\frac{2}{3}$ and $\frac{c_1}{c_2}=\frac{3}{-1}$

Clearly, $\frac{a_1}{a_2}\ne \frac{c_1}{c_2}$

so, the given system of equations, will have no solution, if
$\frac{a_1}{a_2}=\frac{b_1}{b_2}$ i.e., $\frac{2}{3}=\frac{-k}{2}\Rightarrow k=\frac{-4}{3}$

Hence, the given system of equations will have no solution, if $k=\frac{-4}{3}$.