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Prove or give a counterexample: For $A \in \mathbb{R}^{3\times 3}$, $\det(A + I) = \det(A) + \det(I)$ if $\mbox{tr}(A) = -\mbox{tr}(\mbox{adj}(A))$. Here, $\mbox{adj}(A)$ is the classical adjoint (the transpose of the cofactor matrix).

Thanks in advance!

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Hint: evaluate the characteristic polynomial of $A$ (that is $\det(\lambda I - A)$) at $\lambda = -1$. What are the coefficients to the various powers of $\lambda$?

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  • $\begingroup$ Thanks! The solution I had at hand used eigenvalues and assumed that the matrix $A$ is invertible (which we don't necessarily know?) I was way over thinking it! It was just a painful computation... $\endgroup$ – Suugaku Nov 20 '12 at 17:23

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