= Solution
Embed $\operatorname{GL}_n(F)$ by $A\mapsto(A,0)$. The multiplication and inverse formulas show that this is an injective <group homomorphism>, so its image $H$ is a <subgroup>. The translations form $N=\{(I,c):c\in F^n\}$, with $(I,c)(I,d)=(I,c+d)$ and inverse $(I,-c)$. Hence $N$ is isomorphic to the additive <group> of $F^n$.
For an arbitrary affine transformation,
$$
\boxed{(A,b)(I,c)(A,b)^{-1}=(I,Ac).}
$$
<Conjugation> therefore preserves $N$, proving that $N$ is normal in $G$.
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