For what value of α, the system of equations will have no solution?

GIVEN:

$\alpha x+3 y=\alpha-3$

$12 x+\alpha y=\alpha$

To find: To determine for what value of k the system of equation has no solution 

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

For no solution

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

Here,

$\frac{\alpha}{12}=\frac{3}{\alpha} \neq \frac{\alpha-3}{\alpha}$

Consider the following for α

$\frac{\alpha}{12}=\frac{3}{\alpha}$

$\alpha^{2}=12 \times 3$

$\alpha^{2}=36$

$\alpha=\pm 6$

Now consider the following

$\frac{3}{\alpha} \neq \frac{\alpha-3}{\alpha}$

$3 \alpha \neq \alpha(\alpha-3)$

$3 \alpha \neq \alpha^{2}-3 \alpha$

$6 \alpha \neq \alpha^{2}$

$\alpha \neq 6$

Hence the common value of α is − 6

Hence for α = -6 the system of equation has no solution