Find the value of k for which the system of equations $\mathrm{kx}+3 \mathrm{y}=3,12 \mathrm{x}+\mathrm{ky}=6$ has no solution.

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Find the value of k for which the system of equations $\mathrm{kx}+3 \mathrm{y}=3,12 \mathrm{x}+\mathrm{ky}=6$ has no solution.

Solution:

Given system of equations:
$\begin{array}{l} \mathrm{kx}+3 \mathrm{y}=3 \\ \mathrm{kx}+3 \mathrm{y}-3=0\dots \dots(i) \\ 12 \mathrm{x}+\mathrm{ky}=6 \\ 12 \mathrm{x}+\mathrm{ky}-6=0\dots \dots(ii) \end{array}$
The given equations are of the following form:
$a_{1} x+b_{1} y+c_{1}=0, a_{2} x+b_{2} y+c_{2}=0$
where, $a_{1}=\mathrm{k}, \mathrm{b}_{1}=3, \mathrm{c}_{1}=-3$ and $\mathrm{a}_{2}=12, \mathrm{~b}_{2}=\mathrm{k}, \mathrm{c}_{2}=-6$
For the given system has no solution, we must have:
$\begin{array}{l} \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ \text { i.e., } \frac{k}{12}=\frac{3}{k} \neq \frac{-3}{-6} \\ \frac{k}{12}=\frac{3}{k} \text { and } \frac{3}{k} \neq \frac{1}{2} \\ \Rightarrow \mathrm{k}^{2}=36 \text { and } \mathrm{k} \neq 6 \\ \Rightarrow \mathrm{k}=\pm 6 \text { and } \mathrm{k} \neq 6 \end{array}$
As a result, the given system of equations has no solution when $\mathrm{k}$ is equal to $-6$ .