Solution
= Solution
The map $A\mapsto(A,0)$ is an <injective> <group homomorphism>, so its image is the required copy of $\operatorname{GL}_n(F)$. Translations are the <subgroup> $N=\{(I,b):b\in F^n\}$, isomorphic to the additive <group> of $F^n$. For any affine map,
$$
\boxed{(A,b)(I,t)(A,b)^{-1}=(I,At).}
$$
Thus <conjugation> preserves translations, proving $N\trianglelefteq G$.