let $A$ be a 3 by 3 symmetric matrix over $\mathbb{R}$ such that $A^3+A^2-A-E=0$, then $|A+2E|=$?

Question

Since $A^3+A^2-A-E=(A-E)(A+E)^2=0$, so both 1 and -1 may be eigenvalues of $A$, what is the third eigenvalue of $A$, is it -1? On the other hand $A=P^T\Lambda P$, so $\Lambda^3+\Lambda^2-\Lambda-E=0$, hence $(\Lambda-E)(\Lambda+E)^2=0$, therefore $\Lambda=E$ or $-E$, so $|A+2E|=27$ or $1$?

Answer 1

Diagonalize $A$ and we get $B=P^{-1} A P$, where $B=diag(a_1, a_2, a_3)$ and $a_i = 1$ or $a_i = -1$. Then $$\left\lvert A+2E \right\rvert = \left\lvert P^{-1} \right\rvert \left\lvert A+2E \right\rvert \left\lvert P \right\rvert = \left\lvert P^{-1}AP+2P^{-1}EP\right\rvert = \left\lvert B+2E \right\rvert$$ and you can calculate it. The answer is 1, 3, 9, 27.

Answer 2

As $A=P^T\Lambda P$, so $\Lambda^3+\Lambda^2-\Lambda-E=0$, hence $(\Lambda-E)(\Lambda+E)^2=0$, therefore $\Lambda=E$ or $-E$, now we have $|A+2E|=27$ or $1$, where $P$ is an orthogonal matrix over $\mathbb{R}$.

Note. This answer is wrong, As $\Lambda$ may not be neither $E$ nor $-E$.