Check if $\det(I + S) = 1 + \operatorname{trace}(S)$ holds ?

I saw the following statement in my homework and we are asked to make use of the statement:

If $S$ is a symmetric matrix then

$$\det(I + S ) = 1 + \operatorname{trace}(S).$$

However, I am not sure if this really holds. Please give me a hand here please

This statement is false : take $S = I$ for example. It would give us $2^n = 1 + n$.

What is true is that the expansion of the characteristic polynomials is given by traces of powers of the matrix $A$; explicitly, the characteristic polynomial $\chi_A(T)=\det(T\cdot{\rm Id}-A)$ is given by

$$T^n-{\rm tr}(A)T^{n-1}+\frac{{\rm tr}(A)^2-{\rm tr}(A^2)}{2}T^{n-2}-\frac{{\rm tr}(A)^3-3{\rm tr}(A){\rm tr}(A^2)+2{\rm tr}(A^3)}{6}T^{n-3}+\cdots $$

with the final term being $(-1)^n\det A$. This does not depend on $A$ being symmetric or anything else and so $\det(1+A)=(-1)^n\chi_A(-1)\approx (-1)^n[(-1)^n-{\rm tr}(A)(-1)^{n-1}]=1+{\rm tr}(A)$ is in some sense a valid first-order approximation (we set $T=-1$ and truncated our expansion at just two terms).

This can also be seen more simply as in Alex's answer: if $\{\lambda_i\}$ are the eigenvalues of $A$ with multiplicity, then $I+A$ has eigenvalues $\{\lambda_i+1\}$ with multiplicity, and $$\det(1+A)=\prod_{i=1}^n(1+\lambda_i)=1+\left(\sum_{i=1}^n\lambda_i\right)+{\rm higher~order~terms}\approx 1+{\rm tr}(A).$$

I have been assume $A$ is an $n\times n$ matrix throughout, but the underlying field is arbitrary.

The $\chi_A(T)$ expansion is possible thanks to the fundamental theorem of symmetric polynomials, and in particular may be recursively computed by hand using the NG formulas (see this section).

This is false. By the (finite-dimensional) spectral theorem, every symmetric matrix is diagonalizable, so we can write $S=QDQ^{-1}$. Thus we have $$\det(I+S)=\det(Q^{-1}(I+S)Q)=\det(I+D)=\prod\limits_{i=1}^n (1+d_i)\\\ne 1+\sum\limits_{i=1}^n d_i=1+\mathrm{trace}(D)=1+\mathrm{trace}(S).$$ Perhaps the formula $\prod\limits_{i=1}^n (1+d_i)$ may be of use, however.

If the matrix is also real, it suffice to check it over diagonal matrices. But it doesn't hold in dimension $\geqslant 2$, because if $S:=\pmatrix{1&0\\0& a}$, $\det(I+S)=2(a+1)$ and $1+\operatorname{Tr}(S)=2+a$