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I needed to find for which values of $\lambda$ the matrix is singular.

$ \begin{bmatrix} 1-\lambda & 0 & 3 \\ 1 & 1-\lambda & 0 \\ 0 & 2 & -\lambda \\ \end{bmatrix} $

What I did : Compared the determinant of the matrix to zero and ended up with this: $-\lambda (1-2\lambda+\lambda^2)+6=0 $ (tried 2 different ways of calculating the determinant but ended up with the same expression). How do I solve it from here? Thanks a lot!

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    Going past your issue with the error, you would have a cubic equation to solve and can resort to one of the methods here: http://en.wikipedia.org/wiki/Cubic_function or use a computer algebra system (Mathematica, Maxima, MapleV) or Matlab or WolframAlpha.2012-12-05

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Assuming your expression for the determinant is correct, Wolfram Alpha says the approximate roots are $2.53766$ and $-.268828 \pm 1.51397 i$.