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x + 2y - 3z  = 0
3x + 2y + z  = 0
5x + 6y - 5z = 0

I was thinking about solving this using Gaussian Elimination, but the result is going to be x = y = z = 0. So could anyone give me a hint how to go about this issue, please?

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1 Answer 1

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The determinant of your matrix vanishes. Thus its rank is less than $3$. By inspection, it is also greater than $1$, so it must be $2$. Thus, when you apply Gaussian elimination, you should end up with one row of zeros at the end, which allows you to choose $z=\lambda$ arbitrarily and then express $y$ and $x$ in terms of $\lambda$ using the non-zero rows. That gives you the equation of a line through the origin containing the solutions to the system of equations.

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I see. I got the last row of 0s. So my matrix would be x + 2y - 3z = 0; -4y + 10z = 0; z = λ. And then I express x and y and that's it? Thanks :) –  Sorin Cioban May 15 '11 at 12:57
    
@Sorin: Generally (taking into account other questions), I think you'd benefit from putting more effort into being precise and using terms precisely. What you wrote is a set of equations, not a matrix. (In my answer, I somewhat informally referred to "your matrix", but I was referring to the matrix corresponding to the set of equations you'd written, not to the equations.) Further, $z=\lambda$ doesn't belong with the other two; it's not part of the system of equations (the last equation in the transformed system being $0=0$), but the introduction of a parameter to parametrize the solutions. –  joriki May 15 '11 at 13:03

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