Let $G$ be $ \left( \begin{matrix} a & b \\ 0 & 1 \end{matrix} \right)$, which is a group endowed with matrix multiplication (This has been already shown). Suppose that we identify $G$ with the right-half plane in $\mathbb{R}^2$ by associating the point $(a,b)$ to the matrix $ \left( \begin{matrix} a & b \\ 0 & 1 \end{matrix} \right)$.
Show that the formula
$$\mu(A)=\int\int_A \frac{1}{a^2}dadb$$
defines a left Haar measure on $G$, and that the formula $$\mu(A)=\int\int_A \frac{1}{a}dadb$$
defines a right Haar measure on $G$
$\textbf{This is my approach}$
First, let us consider the left Haar measure. A left translation by $u=(x,y)$ is given by: $$ \left( \begin{matrix} x & y \\ 0 & 1 \end{matrix} \right) \left( \begin{matrix} a & b \\ 0 & 1 \end{matrix} \right)= \left( \begin{matrix} xa & xb+y \\ 0 & 1 \end{matrix} \right)$$
Then $da$ becomes $xda$, $db$ becomes $xdb$, and $a$ becomes $xa$. Therefore $$\mu(uA)=\int\int_{uA}\dfrac{1}{a^2}dadb=\int\int_A\dfrac{1}{(ax)^{2}}xdaxdb=\int\int_A\dfrac{1}{a^2}dadb=\mu(A) $$
Now, let us consider the right Haar measure. A right translation by $u=(x,y)$ is given by: $$ \left( \begin{matrix} a & b \\ 0 & 1 \end{matrix} \right) \left( \begin{matrix} x & y \\ 0 & 1 \end{matrix} \right)= \left( \begin{matrix} xa & ya+b \\ 0 & 1 \end{matrix} \right)$$
Then $da$ becomes $xda$, $db$ becomes $db$, and $a$ becomes $xa$. Therefore
$$\mu(Au)=\int\int_{Au}\dfrac{1}{a}dadb=\int\int_A\dfrac{1}{ax}xdadb=\int\int_A\dfrac{1}{a}dadb=\mu(A)$$
Is this enough to answer the question? and more importantly, is this correct? I'd appreciate if anyone can check my procedure.