# System of linear algebraic equations

• Oct 18th 2013, 11:18 AM
FilipVz
System of linear algebraic equations
Hi,

can somebody help me with the problem:

x+y=kx
-x+y=ky

For what values of k will the equatioins have real nontrivial solutions.

Thanks.
• Oct 18th 2013, 11:47 AM
HallsofIvy
Re: System of linear algebraic equations
That is the same (1- k)x+ y= 0 and -x+ (1- k)y= 0. One obvious solution is x=y= 0 but that is the "trivial solution".

You would get that, for example, if you solved the first equation for y: y= (k- 1)x. Putting that into the second equation, we get (-1+ k(k-1))x= k(k-1)x so that, if x is not 0, we can divide on both sides to get (-1+ k(k-1))= k(k-1).
• Oct 18th 2013, 11:01 PM
FilipVz
Re: System of linear algebraic equations
So the equasions does not have a real nontrivial solution?
• Oct 18th 2013, 11:40 PM
votan
Re: System of linear algebraic equations
Quote:

Originally Posted by FilipVz
So the equasions does not have a real nontrivial solution?

You have to solve the equation given to you by HallSofIvy. You may have one real value for k, two distinct real values, or not real value.
• Oct 19th 2013, 12:56 AM
BobP
Re: System of linear algebraic equations
Solving a pair of equations like this is equivalent, graphically, to finding the intersection of a pair of straight lines.

The first equation can be written as $\displaystyle y=(k-1)x,$ which is a straight line passing through the origin.

The second equation can be written as $\displaystyle y = -\frac{x}{(k-1)},$ which is also a straight line passing through the origin.

So they intersect at the origin.

The only way that they can have other points in common is if the lines have the same slope, that is , if

$\displaystyle k-1 = \frac{-1}{(k-1)}.$

That would imply that $\displaystyle (k-1)^{2}=-1$ which clearly cannot happen.
• Oct 19th 2013, 07:33 AM
HallsofIvy
Re: System of linear algebraic equations
That's much nicer than my suggestion!

My point was that the two equations will have more than one (and so an infinite number) solution only of (-1+ k(k-1))= k(k-1).
Multiplying that out, -1+ k^2- k= k^2- k. Subtracting k^2- k from both sides leaves -1= 0 which is never true and so that equation is not satisfied for any value of k.