We use the idea suggested by Alekk.

Let $A_{i,j}$ the entries of $A$. Then in the first case, the entries of $M:=A+G$ are Gaussian independent random variables, that is, $M_{i,j}\sim N(A_{i,j},1)$. Denote $N$ the set of elements of $\Bbb R^{n^2}$ such that the matrix of generic term $x_{i,j}$ is not invertible. This set has null Lebesgue measure in $\Bbb R^{n^2}$ as it's the zeros of a polynomial. By independence, the family $(M_{i,j},i,j\in [n])$ is Gaussian, so
$$p:=P(A+G\mbox{ is not invertible})=P((M_{i,j})_{i,j=1}^n\in S).$$

As the law of $(M_{i,j},i,j\in [n])$ is absolutely continuous with respect to Lebesgue measure in $\Bbb R^{n^2}$, we conclude that $p=0$.

In the second case, write $\det(A+G)$ as a polynomial of the $G_{i,j},i\leqslant j$, and use the fact that $(G_{i,j},1\leqslant i\leqslant j\leqslant n)$ is Gaussian to conclude in the same way as in the first case.

- The result doesn't depend on the choice of the
*deterministic* matrix $A$.
- We don't need i.i.d.ness, just the fact that $(G_{i,j},i,j\in [n])$ is Gaussian in the first case, $(G_{i,j},i\leqslant j,i,j\in [n])$ in the second.
- We can have a more general result when the law of $(A_{i,j}+G_{i,j})_{i,j\in [n]}$ is absolutely continuous with respect to Lebesgue measure on $\Bbb R^{n^2}$.

nondegenerateGaussian measure on the vector space of $n\times n$ matrices. As indicated somewhere else on this site, such a measure is absolutely continuous w.r.t. the Lebesgue measure. Since the Lebesgue measure of the hypersurface $\det A=0$ is zero, so will be its Gaussian measure. Thus with probability $0$ a a random matrix (of the first type you indicated) is not invertible. – Liviu Nicolaescu Oct 26 '12 at 10:02