Is $GL(n;\mathbb{C})$ algebraic or not? The set of $n\times n$ matrices can be identified with $\mathbb{C}^{n^2}$.
1) Consider the subset $V$ of the affine space $\mathbb{A}^{n^2+1}$ (note plus one) given by
$$V:=\{(x_{ij},t): \det(x_{ij})\cdot t-1=0\}$$
This is an algebraic subset, indeed $V=V(f)$ where $f=\det(x_{ij})\cdot t-1\in\mathbb{C}[x_{ij},t]$. Then consider the map $\phi:GL(n;\mathbb{C})\longrightarrow V$ sending the matrix $(a_{ij})$ to the $(n^2+1)$-tuple $(a_{ij},\frac{1}{\det(a_{ij})})$. This is a bijection, making $GL(n;\mathbb{C})$ also algebraic.
2) $GL(n;\mathbb{C})$ is the complement of the algebraic variety in $\mathbb{C}^{n^2}$ defined by the vanishing of the determinant polynomial, and so is open in the Euclidean topology. Since every affine algebraic variety in $\mathbb{C}^n$ is closed in the Euclidean topology, we can conclude that the general linear group is not an algebraic variety.
So what should I argue from the above reasoning?
1) $GL(n;\mathbb{C})$ is algebraic
2) $GL(n;\mathbb{C})$ is not algebraic
3) It is a common fact that some subsets can be algebraic in one affine space and not algebraic in an affine space of different dimension, so both 1) and 2) can be correct.

The general linear group is definitely an affine algebraic variety. It is a closed subvariety of $\mathbb{A}^{n^2+1}$, but also an open subvariety of $\mathbb{A}^{n^2}$. This is not a contradiction. I suggest that you look at the case $n=1$ more closely and learn the abstract coordinate-free definition of varieties.
