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I just wanted to check if anyone can see a simpler way to solve this. Because I am not looking forward to using the cubic formula to solve it!
$$ det(\lambda-AI) = \left| \begin{array}{ccc} \lambda + a & -b & -b \\ c & \lambda + d & 0 \\ 0 & d & \lambda \end{array} \right| = 0$$

requires us to solve

$$ \lambda^3 + (a+d)\lambda^2 + (ad+bc)\lambda -bcd = 0 $$

(I can find no rational roots. As, by the rational zero theorem, the rational roots would be $\pm$ a factor of bcd. The only factors which will cancel out bcd are $\lambda = \pm\sqrt{bc}$ or $\lambda = -d$ neither of which result in $ det(\lambda-AI) = 0$.)

Is the next step the cubic formula? :(

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  • $\begingroup$ Are you sure the signs in the matrix are correct? If the determinant were either $\lambda^3 + (a+d)\lambda^2 + (ad + bc)\lambda + bcd$ or $\lambda^3 + (a+d)\lambda^2 + (ad - bc)\lambda - bcd$, then $\lambda = -d$ would be a solution. $\endgroup$
    – Pedro M.
    Jan 28, 2015 at 17:51
  • $\begingroup$ I know! It's sooo infuriating, but i've been over the signs several times... This is what happens when you build your own model rather than doing beautiful bookwork $\endgroup$
    – user1792403
    Jan 28, 2015 at 17:51
  • $\begingroup$ I do not see the relevance of the Rational Roots Theorem. For relevance one would need the cubic to have integer coefficients. And depending on the values of $b,c,d$ the number $bcd$ may have many integer factors other than the obvious ones, $\endgroup$
    – André Nicolas
    Jan 28, 2015 at 17:54
  • $\begingroup$ So the original matrix you have is really $$A = \begin{pmatrix} -a & b & b\\ -c & -d & 0\\ 0 & -d & 0 \end{pmatrix}?$$ $\endgroup$
    – Pedro M.
    Jan 28, 2015 at 17:54
  • $\begingroup$ Oh I see, interesting point. But I'm hoping to do a theoretical rather than numerical description, so the only factors of bcd are indeed b,c,d. $\endgroup$
    – user1792403
    Jan 28, 2015 at 17:56

1 Answer 1

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Just to clarify for those who may see this question in the future, the actual matrix was

$$A = \begin{pmatrix} -a & -b & -b\\ c & -d & 0\\ 0 & d & 0 \end{pmatrix}$$

so that the polynomial to solve was $$\lambda^3 + (a+d)\lambda^2 + (ad + bc)\lambda + bcd,$$

which has an obvious root $\lambda = -d$, and hence factors out.

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