Question

Find the value of k for which the following system of linear equations has no solutions:​kx + 3y = 3, 12x + ky = 6

Answer 1

$$\begin{gathered} \mathrm{kx}+3\mathrm{y}=3 \\ 12\mathrm{x}+\mathrm{ky}=6 \\ \mathrm{For}\mathrm{no}\mathrm{solution}, \\ \frac{\mathrm{k}}{12}=\frac{3}{\mathrm{k}}\ne \frac{3}{6} \\ {\mathrm{k}}^2=36 \\ \mathrm{k}=\pm 6 \\ \mathrm{As}\frac{3}{\mathrm{k}}\ne \frac{3}{6} \\ \mathrm{So},\mathrm{k}\ne 6 \\ \mathrm{Therefore},\mathrm{k}=-6\mathrm{ANS}... \end{gathered}$$