Question

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

αx + 3y = α − 3
12x + αy = α

Answer 1

GIVEN:
$$\begin{gathered} \alpha x+3y=\alpha -3 \\ 12x+\alpha y=\alpha \end{gathered}$$
To find: To determine for what value of k the system of equation has no solution

We know that the system of equations

For no solution

Here,

$\frac{\alpha }{12}=\frac{3}{\alpha }\ne \frac{\alpha -3}{\alpha }$
Consider the following for α

$$\begin{gathered} \frac{\alpha }{12}=\frac{3}{\alpha } \\ {\alpha }^2=12\times 3 \\ {\alpha }^2=36 \\ \alpha =\pm 6 \end{gathered}$$

Now consider the following

$$\begin{gathered} \frac{3}{\alpha }\ne \frac{\alpha -3}{\alpha } \\ 3\alpha \ne \alpha (\alpha -3) \\ 3\alpha \ne {\alpha }^2-3\alpha \\ 6\alpha \ne {\alpha }^2 \\ \alpha \ne 6 \end{gathered}$$
Hence the common value of α is − 6

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