I feel like I might have made a mistake on this question, and would appreciate some feedback from someone more experienced than me.
If $G$ acts on $S$, we write $S^G = \{s \in S: gs = s \, \forall \, g \in G\}$ for the invariant ring. Our 2 x 2 matrices $P_{ij}$ act on $\mathbb{C}[X,Y]$ by $X \longmapsto P_{11}X + P_{12}Y$ and $Y \longmapsto P_{21}X + P_{22}Y$, and extending these mappings to $X^i$ and $Y^i$. So, I want to calculate the invariant rings $(i): \mathbb{C}[X,Y]^{GL_2}$ and $(ii): \mathbb{C}[X,Y]^{SL_2}$. My arguments went like this:
$(i)$ Suppose we have some element of $\mathbb{C}[X,Y]^{GL_2}$, say $z = \sum \limits_{i,j \geq 0} A_{i j}X^i Y^j$. Take the matrix $\lambda I$, where $I$ is the identity matrix and $\lambda$ is large. Then $z$ must satisfy $\sum \limits_{i,j \geq 0} A_{i j}X^i Y^j = \sum \limits_{i,j \geq 0} A_{i j}\lambda^{i+j} X^i Y^j$; so clearly $A_{ij}$ must be zero unless $i=j=0$; thus our invariant ring is just $\mathbb{C}$.
$(ii)$ Now our matrices have to have determinant 1. Take the matrix with $A_{11} =A_{22} = 0$ and $A_{12} = \alpha$, $A_{21}=-1/\alpha$ (sorry I can't draw matrices in latex!) Then we must have $A_{ji} = (-1)^{j} \alpha^{i-j}A_{ij}$; so for $i \neq j$, clearly this can't be true for all $\alpha$: thus $A_{ij} = 0$ unless $i=j$. So any element of the invariant ring must be of the form $\sum_i A_i (XY)^i$. But then taking any upper triangular matrix with diagonal elements 1, don't we deduce again that the invariant ring is just $\mathbb{C}$?
The reason I am confused is that these were 2 consecutive problems in the same worksheet, and I didn't expect them to have the same invariant ring. If someone could point out where I've gone wrong or confirm my solutions, I'd be very grateful.