Question

Find all values of lambda such that ,the following system has a none-zero solutions.

2x+y= lambda x

4x-y =lambda y

Answer 1

From the first equation: $y = (\lambda-2)x$ . Substitute this value into the second equation we get

$4x -(\lambda-2)x = \lambda(\lambda-2)x$ , as $x \ne 0$ $6-\lambda = \lambda^2-2\lambda$ , or

$\lambda^2-\lambda - 6 =0$ . This equation has two solutions: $\lambda =-2$ and $\lambda =3$