Question

For what value of $\alpha$, the system of equations
$\alpha x+3y=\alpha -3$
$12x+\alpha y=\alpha$
will have no solution

Answer 1

Given,

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

$12x+\alpha y=\alpha$

Condition for equations to have no solutions,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

$\therefore \frac{\alpha }{12}=\frac{3}{\alpha }\ne \frac{\alpha -3}{\alpha }$

$\Rightarrow \frac{\alpha }{12}=\frac{3}{\alpha }\Rightarrow \alpha =\pm 6$

$\Rightarrow \frac{3}{\alpha }\ne \frac{\alpha -3}{\alpha }\Rightarrow \alpha \ne 6$

the given equations have no solutions when$\alpha =-6$