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Suppose the matrix A = \begin{bmatrix} 40&-29&-11\\ -18&30&-12\\ 26&24&-50\end{bmatrix} has a certain complex number $\lambda \neq 0$ as an eigenvalue. Which of the following numbers must also be an eigenvalue of A? The options are $\lambda+20$, $\lambda-20$,$20-\lambda$,$-20-\lambda$.

I started to find the eigenvalues of this matrix by the usual procedure. My eigenvalues were 0,58.45, -38.45. My characteristic equation was $\lambda^{3}-20\lambda^{2}-2248\lambda = 0$.

Can somebody tell me how to proceed? Thanks in advance...!!!

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    $\begingroup$ Sum of eigen values is equal to the trace of the matrix. Here trace = $40+30-50 = 20$ and hence the other eigen value must be $20 -\lambda$ Note that since the determinant value is zero, one of its eigen values is 0. $\endgroup$ – user348749 Oct 25 '16 at 9:22
  • $\begingroup$ @Muralidharan Got it. Thank you. $\endgroup$ – Sahana Oct 25 '16 at 9:26

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