If $(k,l) = 1$, show that $(b_2k-b_1l, a_1l-a_2k) = 1$ for $a_1b_2-a_2b_1=1$ Problem:

If $(k,l) = 1$, show that $(b_2k-b_1l, a_1l-a_2k) = 1$ for $a_1b_2-a_2b_1=1$
  Note: ($a_1, a_2,b_1,b_2,k,l \in \mathbb{Z}$)

Also note that the actual (bigger) problem is:

If $m = a_1x+b_1y$ (Eq1) and $n=a_2x+b_2y$ (Eq2) and $a_1b_2-a_2b_1=1$, prove that $(m,n)=(x,y)$.

I have proved that $(m,n) = d | x \text{ and } d|y$ I am left to prove that $d = (x,y)$, that's why I asked above question.
I had made $m = dk$ and $n = dl$ ($\therefore (k,l)=1$) and after substitution in Eq1 and Eq2 for $x$ and $y$ values I got $d(b_2k-b_1l)=x$ and $d(a_1l-a_2k) = y$. If I can show  $(b_2k-b_1l, a_1l-a_2k) = 1$ then definitely $d = (x,y)$
Method:
Suppose $(b_2k-b_1l, a_1l-a_2k) = d > 1$. Then, $d | b_2k, d | b_1l, d|a_1l,d|a_2k$ must be true.
This implies $d|a_2b_1kl$ and $d|a_1b_2kl$ must be true. Let, $dp = a_2b_1kl \implies a_2b_1 = \frac{dp}{kl}$ and $dq=a_1b_2kl\implies a_1b_2 = \frac{dq}{kl}$
So, $$a_1b_2-a_2b_1=1 \implies \frac{d(q-p)}{kl}=1 \implies kl =d(q-p)$$
I am stuck here. How to prove a contradiction so that $d = 1$?
 A: The 2nd statement in this paragraph is wrong.

Suppose $(b_2k-b_1l, a_1l-a_2k) = d > 1$. Then, $d | b_2k, d | b_1l, d|a_1l,d|a_2k$ must be true.

Rather it should be as follows: 
$d|b_2k-b_1l$ and $d|a_1l-a_2k$. 
Now, $$d|b_2k-b_1l \Rightarrow d|a_1b_2k-a_1b_1l$$ and 
$$d|a_1l-a_2k \Rightarrow d|a_1b_1l-a_2b_1k$$
Hence $d|(a_1b_2k-a_1b_1l)+(a_1b_1l-a_2b_1k) \Rightarrow d|(a_1b_2-a_2b_1)k \Rightarrow d|k$ 
Similarly, you can show that $d|l$. 
So $d|k$ and $d|l$ .  
But by assumption $(k,l)=1$ 
Hence contradiction $\Rightarrow d=1$ .
A: If you learned a bit about determinants, the equation involving $a_1b_2-a_2b_1$ might remind you of them. So let us try whether linear algebra might be useful here.
Knowing that $(k,l)=1$ means that there exist some integers $u$ and $v$ such that $uk+vl=1$.
We also know that $a_1b_2-a_2b_1=1$.
This means that these determinants are both equal to one:
$$
\begin{vmatrix}
  a_1 & a_2 \\
  b_1 & b_2
\end{vmatrix}=
\begin{vmatrix}
  l & u \\
 -k & v
\end{vmatrix}=1
$$
Since we know that determinant of product of two matrices is product of determinants 
The product is
$$\begin{pmatrix}
  a_1 & a_2 \\
  b_1 & b_2
\end{pmatrix}
\begin{pmatrix}
  l & u \\
 -k & v
\end{pmatrix}=
\begin{pmatrix}
  a_1l-a_2k & x \\
  b_1l-b_2k & y 
\end{pmatrix}
$$
where $x,y\in\mathbb Z$. (Namely $x=a_1u+a_2v$, $y=b_1u+b_2v$.)
If we expand the last determinant, we get
$$x(b_1l-b_2k)+y(a_1l-a_2k)=1$$
which means that $(b_1l-b_2k)$ and $(a_1l-a_2k)$ are coprime.
