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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. –  user1685224 Oct 1 '12 at 15:32
    
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. –  André Nicolas Oct 1 '12 at 15:40
    
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. –  André Nicolas Oct 1 '12 at 15:47
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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. –  Gerry Myerson Oct 2 '12 at 0:33
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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: en.wikipedia.org/wiki/Cubic_function or use a computer algebra system (Mathematica, Maxima, MapleV) or Matlab or WolframAlpha. –  Amzoti Dec 5 '12 at 13:07

1 Answer 1

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

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