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December 26, 2014

December 26, 2014

Posted by **Alice** on Sunday, May 17, 2009 at 5:29am.

x + ky = 0

kx + 3y = 0

Calculate the two values of k that lead to non-trivial solutions to these

equations and express y in terms of x for the two values.

I thought that for the solutions to be no trivial the determinant of the coefficients had to equal zero. From this I got k=+/- sqrt 3 but I am not sure of this is right.

- Math -
**drwls**, Sunday, May 17, 2009 at 7:39amRewrite as

y = (-1/k)x and

y = (-k/3)x

There are an infinite number of solutions if the lines coincide, which happens when

1/k = k/3

k = +/-sqrt3

In this case,

y = x/sqrt3 or -x/sqrt3

Otherwise, there are no solutions at all, other than x=y=0. That may be what they call the "trivial" solution. Nevertheless, it is still a solution.

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