Prove or disprove that $GL_2(\mathbb{Z})$ is a P.I.D.
So, ideals of $GL_2(\mathbb{Z})$ are precisely of the form $GL_2(I)$ where $I \triangleleft \mathbb{Z}$, and since $\mathbb{Z}$ is a P.I.D. we have $I = n\mathbb{Z}$ for some $n \in \mathbb{N}$
It seems like it would be awfully hard for there to be one matrix that generates all the invertible matrices with coefficients in $n\mathbb{Z}$, which is what an ideal of $GL_2(\mathbb{Z})$ looks like, but I'm having a hard time showing this. Can somebody help me out?
As everyone says in the comments: $\mathrm{GL}_2(\mathbb{Z})$ is not a ring --- it's a nonabelian group under multiplication. PIDs are commutative rings, so these are very different objects. Unless you have defined some addition/multiplication on the set $\mathrm{GL}_2(\mathbb{Z})$ to make it a commutative ring, the question makes no sense. Even entrywise operations would not make $\mathrm{GL}_2(\mathbb{Z})$ into a ring.
Either way: it is an abuse of terminology. To most mathematicians, the symbol $\mathrm{GL}_2(\mathbb{Z})$ refers not just to the set of invertible matrices, but the resulting group under multiplication. To separate this set from its group operation is somewhat offensive to most people, as you can see from the comments.
But I think this is the most charitable interpretation of your question: "is every ideal of $M_2(\mathbb{Z})$ principal?"
This question makes sense and the answer is "yes". Every ideal of $M_n(\mathbb{Z})$ is of the form $M_n(I)$ for some ideal $I$ of $\mathbb{Z}$, and all such $I$ are principal.