Thread: System of linear algebraic equations

1. 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.

2. 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).

3. Re: System of linear algebraic equations

So the equasions does not have a real nontrivial solution?

4. Re: System of linear algebraic equations

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.

5. 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 $y=(k-1)x,$ which is a straight line passing through the origin.

The second equation can be written as $y = -\frac{x}{(k-1)},$ 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

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

That would imply that $(k-1)^{2}=-1$ which clearly cannot happen.

6. 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.