How to calculate the determinant $\det(A+xI)$?

Question

Let $A$ be an $n\times n$ symmetry matrix with the diagonal elements of $A$ are $0$ (the diagonal elements of $A$ are can also be any constant ). Let $I$ be an $n\times n$ identity matrix, and Let $x$ be a constant. How to calculate $\det (A+xI)$? We wish $\det (A+xI)$ be express by the combinations of $\det(A)$.

Answer 1

That's impossible (unless $n=1$), as the determinant of $A+xI$ depends on the eigenvalues of $A$ rather than the determinant of $A$. Consider, e.g. $A_1=\pmatrix{1\\ &4}$ and $A_2=2I_2$. They have the same determinants, but $\det(A_1+xI)=(x+1)(x+4)\ne(x+2)^2=\det(A_2+xI)$.

Answer 2

If $A$ is diagonalizable:

$$A=U^{-1}DU$$

And $$\det(A)=\prod \lambda_i$$ Where $\lambda_i$ is the eigenvalues

Thus $$\det(A+xI)=\det\big(U^{-1}(D+xI)U\big)=\det(D+xI)=\prod (\lambda_i+x)$$

Answer 3

for $n=3$, $$\det(A+xI)=x^3-tr(A)x^2+(A_11+A_22+A_33)x+\det A$$