A basic question about algebraic groups on inversion being a morphism

Question

Let $G = GL_n(k)$ where $k$ is an algebraically closed field. I know that the inversion $i$ is a morphism but I was wondering how I can show this.

By definition I have to show that 1) $i: G \rightarrow G$ is continuous 2) for all open $V \subseteq G$ for all $f \in O_G(V)$, we have $f \circ i \in O_G(i^{-1} (V) )$.

I know that $GL_n(k)$ is an algebraic group so $i$ is a morphism but I am struggling to provide the details for it. Details would be appreciated. Thank you.

Answer 1

By using cofactor matrices, you can determine an expression of $i(A)$ as a rational function of the coefficient of $A$.

Continuity of rational functions between affine algebraic sets