Determining which elements of a matrix group form a one-parameter subgroup We have just learned about one-parameter subgroups in my Algebra class and I am not sure if I am approaching the following proof in the right way.

Problem Statement: Let $G$ be a group of real matrices of the form $\begin{bmatrix} x && y \\ 0 && 1 \end{bmatrix}$ with $x>0$.

a) Determine the matrices $A$ such that $e^{tA}$ is a one-parameter group in $G$.
b) Compute $e^{tA}$ explicitly for the matrices in (a).
c) Make a drawing showing some one-parameter subgroups in the xy-plane.


So for a) I must show that $e^{(s+t)A}=e^{sA}e^{tA}$ in order to be a one-parameter subgroup, correct? So I thought to compute the first few terms of the expansion for each of these, to determine which matrices worked, but this may be too complicated.
I have
$$e^{(t+s)A}=I+(s+t)A+\frac{(s^2+2st+t^2)}{2}A^2+...$$
$$e^{sA}e^{tA}=I+(s+t)A+\frac{(s^2+2st+t^2)}{2}A^2+\frac{(s^2t+st^2)}{2}A^3+\frac{(s^2t^2)}{4}A^4+...$$
(I mutlplied the first 3 terms of $e^{sA}$ by the first 3 terms of $e^{tA}$, is that what I should be doing?)
So should I be choosing matrices such that $A^3,A^4=\mathbf{0}$? I am not thinking that my strategy is correct, because $$A^n=\begin{bmatrix} x^n && x^ny+x^{n-1}y+...+xy+y \\ 0 && 1 \end{bmatrix}$$ and I believe I would be left with very few matrices which form a one-parameter subgroup if $A^3,A^4=\mathbf{0}$ must hold.
Any advice on how I should be approaching a)? I should be able to figure out b) once I finish a). Then for c), once I compute $e^{tA}$ explicitly, should I be drawing points on the plane after I determine the appropriate values for $x,y$? I am also not quite sure how to approach this part.
Any strategies/suggestions are appreciated!
 A: I guess you have a typo and you are looking for the matrices $A \in M_2(\mathbb{R})$ for which $e^{tA}$ is a one-parameter subgroup in $G$. Note that for any matrix $A \in M_2(\mathbb{R})$, the family $e^{tA}$ is a one-parameter subgroup of $\mathrm{GL}_2(\mathbb{R})$ (we always have $e^{(t+s)A} = e^{tA}e^{sA}$ as can be checked by expanding the power series). The question is for which matrices $A \in M_2(\mathbb{R})$, the image $e^{tA}$ lands inside the group $G$.
Assume that $A \in M_2(\mathbb{R})$ such that for $\alpha_A(t) := e^{tA}$ we have $\alpha(t) \in G$ for all $t \in \mathbb{R}$. Writing
$$ \alpha_A(t) = \begin{pmatrix} x(t) & y(t) \\ 0 & 1 \end{pmatrix} $$
we have
$$ \dot{\alpha}_A(0) = A = \begin{pmatrix} \dot{x}(0) & \dot{y}(0) \\ 0 & 0 \end{pmatrix} $$
so this suggests that for matrices of the form
$$ A = \begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix} $$
the one-parameter subgroups $\alpha_A(t)$ will land inside $G$. You can check this directly by computing $\alpha_A(t) = e^{tA}$ (there will be two cases - where $a \neq 0$ and when $a = 0$ (in which case $A$ is nilpotent)).
