Question

giving out brainliest !!
Let k be a real constant. Note that (x,y) = (0,0) is always a solution to the system-

x+3y = kx
3x+y = ky

Find all values of k for which (0,0) is not the only solution.

Answer 1

Answer:

$k=4 \ , \ k=-2$

Step-by-step explanation:

Trivial Solutions

We are dealing with a type of systems of equations with the following structure:

$\displaystyle \left \{ {{ax+by=0} \atop {cx+dy=0}} \right.$

There is a trivial solution where x=y=0, but the interest of the problem is to find the conditions for a given system to have other solutions than the trivial. The idea is to transform the compatible determinate system to a compatible indeterminate system that accepts infinitely many solutions. This can be achieved by computing the determinant of the system

$\Delta=\left|\begin{array}{cc}a&b\\c&d\end{array}\right|$

If this determinant is zero, the system is compatible indeterminate.

Let's analyze the given system:

$\displaystyle \left \{ {{x+3y=kx} \atop {3x+y=ky}} \right.$

Rearranging

$\displaystyle \left \{ {{(1-k)x+3y=0} \atop {3x+(1-k)y=0}} \right.$

Computing the determinant and equating to 0

$\Delta=\left|\begin{array}{cc}1-k&3\\3&1-k\end{array}\right|=0$

Expanding the determinant

(1-k)^2-9=0

Rearranging

(1-k)^2=9

Taking the square root (with both possible signs)

1-k=\pm 3

Solving for k

k=1\pm 3

The two possible values of k are

$\boxed{k=4 \ , \ k=-2}$