Modular function for certain upper triangular matrices (Anton Deitmar Automorphic Forms Exercise 3.4) I am doing the following exercise from Anton Deitmar's Automorphic Forms

Let $B$ be the group of all real matrices of the form $\begin{bmatrix}1 & x\\
0 & y\end{bmatrix}$ with $y\neq 0$. Show that the modular function is $\Delta(\begin{bmatrix}1 & x\\
0 & y\end{bmatrix})=\vert y\vert$

For context, we shall let $G=\text{GL}(n,\mathbb{R})$, then if $\mu$ is the Haar measure on $G$, then we have that given $g=\begin{bmatrix}1 & x\\
0 & y\end{bmatrix}$, we have that $\mu_g$ is the measure $\mu_g(A)=\mu(Ag)$. Then by the uniqueness up to scaling of the Haar measure we have that $\mu_g(A)$ is a scalar multiple of $\mu(A)$, and we have that the factor is $\Delta(g)$, so $\mu_g(A)=\Delta(g)\mu(A)$. From my understanding the Haar measure on $G$ is $\frac{dxdydzdw}{\vert xw-yz\vert^2}$ where we have $\begin{bmatrix}x & y\\
z & w\end{bmatrix}\in G$. My thought is to take some arbitary matrix say
$\begin{bmatrix}a & b\\
c & d\end{bmatrix}\in G$, then multiply on the right by $g$, so we see that
$$
\begin{bmatrix}a & b\\
c & d\end{bmatrix}\begin{bmatrix}1 & x\\
0 & y\end{bmatrix}=\begin{bmatrix}a & ax+by\\
c & cx+dy\end{bmatrix}
$$
Thus, taking the determinant after multiplying by $g$, we find that we get $y(ad-bc)$, so I feel as though $\Delta(g)=\frac{1}{\vert y\vert^2}$, but this is not the case as that is not what the problem tells me to find. I am just learning about Haar measure, so I would appreciate any help.
Note: Due to comment below, I think that the question must be asking about the Haar measure on the group $B$ rather than $\text{GL}_2(\mathbb{R})$, and in that case I do not know what the Haar measure on this group is, and I am lost.
 A: You'll probably need to develop a bit of intuition for Haar measures more step-wise.
First, the group $N=\{\pmatrix{1&x\cr 0&1}:x\in\mathbb R\}$ is isomorphic (not only algebraically, but topologically, etc.) to $\mathbb R$ itself, so (invoking uniqueness of Haar measure), the Haar integral on it is (up to scalar multiples)
$$
\int_N f\pmatrix{1&x\cr 0&1}\;dn \;=\; \int_{\mathbb R} f\pmatrix{1&x\cr0&1}\;dx
$$
with usual Haar measure on $\mathbb R$.
Preliminary to the second point: on $\mathbb R^\times$, with multiplication, ${dx\over |x|}$ is readily seen to be a Haar measure.
Second, the group $H=\{\pmatrix{1&0\cr 0&y}:y\in \mathbb R^\times\}$ is in every regard isomorphic to $\mathbb R^\times$, so has Haar measure ${dy\over |y|}$.
Your group $B=\{\pmatrix{1&x\cr 0&y}\}$ is a semi-direct product of $N$ and $H$, with $H$ normalizing $N$, so $dx\,{dy\over |y|}$ is a Haar measure on it... either left or right, which I can never remember. :) But you can check which it is, and the discrepancy in the opposite case is the computation of the modular function on it. :)
