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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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    yes it's supposed to be -\lambda. I feel silly... really sorry. first time posting here. thanks for your patience.2012-10-01
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    No need to feel silly, probably I have had more troubles with minus signs than you. The little correction turns the thing from an unpleasant cubic to a tame one.2012-10-01
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    You are trying to find the eigenvalues of a certain matrix. Can you type in the matrix separately? Typos are less likely if you do that.2012-10-01
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    If you are now able to solve the problem, you're encouraged to post an answer and, later, accept it. This keeps the question from popping up over and over again.2012-10-02
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    talked with the publisher of this question and he admitted of having a typo... the (2,2) number is supposed to be $1+\lambda$ ... will post my answer soon.2012-10-04
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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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