[proofplan] We reduce an arbitrary integer matrix of determinant $n$ by left multiplication with matrices in $SL_2(\mathbb{Z})$, which is exactly the operation of applying determinant-preserving integer row operations. First, we use Bézout coefficients for the lower row to make the lower-left entry zero and the lower-right entry positive. The determinant condition then forces the upper-left entry to be a positive divisor $a$ of $n$, with lower-right entry $d=n/a$. Finally, an elementary shear changes the upper-right entry by multiples of $d$, so we choose its unique residue modulo $d$ in the interval $0 \le b < d$; a direct matrix comparison proves uniqueness. [/proofplan] [step:Kill the lower-left entry by a unimodular row operation on the first column] Let \begin{align*} A = \begin{pmatrix} p & q \\ r & s \end{pmatrix} \in \Delta_n. \end{align*} Thus $p,q,r,s \in \mathbb{Z}$ and \begin{align*} ps - qr = n. \end{align*} Since $n \in \mathbb{N}$, the first column $(p,r)$ is not $(0,0)$; otherwise $\det A=0$. Let $a := \gcd(p,r)$, chosen positive. Then $a$ divides both $p$ and $r$, and there exist $x,y \in \mathbb{Z}$ such that \begin{align*} xp + yr = a. \end{align*} Define \begin{align*} \gamma := \begin{pmatrix} x & y \\ -r/a & p/a \end{pmatrix} \in M_2(\mathbb{Z}). \end{align*} The entries are integral because $a$ divides both $p$ and $r$. Its determinant is \begin{align*} \det \gamma = x\frac{p}{a} - y\left(-\frac{r}{a}\right) = \frac{xp+yr}{a} = 1, \end{align*} so $\gamma \in SL_2(\mathbb{Z})=\Gamma$. Multiplying on the left by $\gamma$ gives \begin{align*} \gamma A &= \begin{pmatrix} x & y \\ -r/a & p/a \end{pmatrix} \begin{pmatrix} p & q \\ r & s \end{pmatrix} \\ &= \begin{pmatrix} xp+yr & xq+ys \\ -rp/a+pr/a & -rq/a+ps/a \end{pmatrix} \\ &= \begin{pmatrix} a & xq+ys \\ 0 & (ps-qr)/a \end{pmatrix} \\ &= \begin{pmatrix} a & b_0 \\ 0 & d_1 \end{pmatrix}, \end{align*} where $b_0 := xq+ys \in \mathbb{Z}$ and \begin{align*} d_1 := \frac{n}{a} \in \mathbb{Z}. \end{align*} The integer $d_1$ is positive because $n>0$ and $a>0$. Therefore the left coset of $A$ contains a triangular matrix of the form \begin{align*} \begin{pmatrix} a & b_0 \\ 0 & d_1 \end{pmatrix}, \qquad a>0, \qquad d_1>0. \end{align*} [guided] The goal is to remove the lower-left entry using only left multiplication by an element of $SL_2(\mathbb{Z})$. Left multiplication performs integer row operations, so the lower-left entry of the product is obtained by applying a new lower row to the first column of $A$. For that reason the relevant pair is the first column $(p,r)$, not the lower row $(r,s)$. Let \begin{align*} A = \begin{pmatrix} p & q \\ r & s \end{pmatrix} \in \Delta_n. \end{align*} By the definition of $\Delta_n$, the entries satisfy $p,q,r,s \in \mathbb{Z}$ and \begin{align*} ps - qr = n. \end{align*} Because $n \in \mathbb{N}$, we have $n>0$. Hence the first column $(p,r)$ cannot be $(0,0)$, since then the determinant $ps-qr$ would be $0$. Set $a := \gcd(p,r)$ with $a>0$. Bézout's identity gives integers $x,y \in \mathbb{Z}$ such that \begin{align*} xp + yr = a. \end{align*} We now build a matrix whose first row produces this gcd and whose second row is orthogonal to the first column. Define \begin{align*} \gamma := \begin{pmatrix} x & y \\ -r/a & p/a \end{pmatrix} \in M_2(\mathbb{Z}). \end{align*} The entries $-r/a$ and $p/a$ are integers because $a$ divides both $p$ and $r$. Its determinant is \begin{align*} \det \gamma = x\frac{p}{a} - y\left(-\frac{r}{a}\right) = \frac{xp+yr}{a} = 1. \end{align*} Thus $\gamma \in SL_2(\mathbb{Z})=\Gamma$, so multiplying by $\gamma$ on the left keeps us in the same left coset of $\Gamma \backslash \Delta_n$. Now compute the product explicitly: \begin{align*} \gamma A &= \begin{pmatrix} x & y \\ -r/a & p/a \end{pmatrix} \begin{pmatrix} p & q \\ r & s \end{pmatrix} \\ &= \begin{pmatrix} xp+yr & xq+ys \\ -rp/a+pr/a & -rq/a+ps/a \end{pmatrix}. \end{align*} The lower-left entry is zero because \begin{align*} -rp/a+pr/a=0. \end{align*} The upper-left entry is $a$ by the Bézout identity, and the lower-right entry is determined by the determinant of $A$: \begin{align*} -rq/a+ps/a = \frac{ps-qr}{a} = \frac{n}{a}. \end{align*} Therefore \begin{align*} \gamma A = \begin{pmatrix} a & xq+ys \\ 0 & n/a \end{pmatrix}. \end{align*} Define $b_0 := xq+ys \in \mathbb{Z}$ and \begin{align*} d_1 := \frac{n}{a} \in \mathbb{Z}. \end{align*} The integer $d_1$ is positive because $n>0$ and $a>0$. Hence the left coset of $A$ contains a matrix \begin{align*} \begin{pmatrix} a & b_0 \\ 0 & d_1 \end{pmatrix} \end{align*} with $a>0$ and $d_1>0$, which is the triangular form needed before reducing the upper-right entry modulo the lower-right entry. [/guided] [/step] [step:Use the determinant condition to identify the diagonal entries] From the previous step, the left coset of $A$ contains a matrix \begin{align*} T_0 := \begin{pmatrix} a & b_0 \\ 0 & d \end{pmatrix} \end{align*} with $a,b_0 \in \mathbb{Z}$ and $d \in \mathbb{N}$. Since $T_0 \in \Delta_n$, \begin{align*} n=\det T_0=ad. \end{align*} Because $n>0$ and $d>0$, it follows that \begin{align*} a=\frac{n}{d}\in \mathbb{N}. \end{align*} Thus $d$ is a positive divisor of $n$, and the diagonal entries satisfy $ad=n$ with $a,d \in \mathbb{N}$. [/step] [step:Choose the upper-right entry as the unique residue modulo $d$] Let $b_0 \in \mathbb{Z}$ be the upper-right entry of $T_0$. By Euclidean division, there exist unique integers $m,b \in \mathbb{Z}$ such that \begin{align*} b_0 = md+b, \qquad 0 \le b < d. \end{align*} Define the shear matrix \begin{align*} \sigma_m := \begin{pmatrix} 1 & -m \\ 0 & 1 \end{pmatrix} \in SL_2(\mathbb{Z}). \end{align*} Then \begin{align*} \sigma_m T_0 = \begin{pmatrix} 1 & -m \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b_0 \\ 0 & d \end{pmatrix} = \begin{pmatrix} a & b_0-md \\ 0 & d \end{pmatrix} = \begin{pmatrix} a & b \\ 0 & d \end{pmatrix}. \end{align*} Since $\sigma_m \in \Gamma$, this matrix lies in the same left coset as $A$. Therefore every left coset has a representative of the stated form. [/step] [step:Compare two triangular representatives in the same left coset] Suppose two displayed representatives lie in the same left coset. Thus there exist $a,d,a',d' \in \mathbb{N}$ and $b,b' \in \mathbb{Z}$ with \begin{align*} ad=n, \qquad a'd'=n, \qquad 0 \le b < d, \qquad 0 \le b' < d', \end{align*} and there exists \begin{align*} \gamma = \begin{pmatrix} \alpha & \beta \\ \gamma_1 & \delta \end{pmatrix} \in SL_2(\mathbb{Z}) \end{align*} such that \begin{align*} \gamma \begin{pmatrix} a & b \\ 0 & d \end{pmatrix} = \begin{pmatrix} a' & b' \\ 0 & d' \end{pmatrix}. \end{align*} Multiplying the matrices gives \begin{align*} \begin{pmatrix} \alpha a & \alpha b+\beta d \\ \gamma_1 a & \gamma_1 b+\delta d \end{pmatrix} = \begin{pmatrix} a' & b' \\ 0 & d' \end{pmatrix}. \end{align*} Since $a>0$, the equality $\gamma_1 a=0$ implies $\gamma_1=0$. The determinant condition $\det \gamma=1$ then gives \begin{align*} \alpha\delta=1, \end{align*} so $\alpha=\delta=1$ or $\alpha=\delta=-1$. The lower-right entry satisfies \begin{align*} d'=\delta d. \end{align*} Because $d,d'>0$, we must have $\delta=1$, hence also $\alpha=1$. Therefore \begin{align*} a'=a, \qquad d'=d, \qquad b'=b+\beta d. \end{align*} Since $0 \le b < d$ and $0 \le b' < d$, the equality $b'=b+\beta d$ forces $\beta=0$ and $b'=b$. Hence the two representatives are identical. [/step] [step:Conclude that the displayed triangular matrices form the complete representative system] The existence steps show that every coset in $\Gamma \backslash \Delta_n$ contains at least one matrix \begin{align*} \begin{pmatrix} a & b \\ 0 & d \end{pmatrix} \end{align*} with $a,d \in \mathbb{N}$, $ad=n$, and $0 \le b<d$. The comparison step shows that no two distinct matrices of this form lie in the same left coset. Therefore these triangular matrices form a unique set of representatives for $\Gamma \backslash \Delta_n$, as claimed. [/step]