Trace and determinant of a 3×3 matrix via eigenvalues

Trace. $\textrm{tr}(A)=2+(-1)+4=5$.

Determinant. $\det(A)=2\cdot(-1)\cdot4=-8$.

Characteristic polynomial.
$$\chi_A(\lambda)=(\lambda-2)(\lambda+1)(\lambda-4).$$
Expanding:
$$(\lambda-2)(\lambda+1)=\lambda^2-\lambda-2,$$
$$(\lambda^2-\lambda-2)(\lambda-4)=\lambda^3-5\lambda^2+2\lambda+8.$$
So $\chi_A(\lambda)=\lambda^3-5\lambda^2+2\lambda+8$.

Check. Coefficient of $\lambda^2$ is $-\textrm{tr}(A)=-5$, constant term is $\det(A)=8$. Correct.