$(a_{1},a_{2})$ and $(b_{1},b_{2})\in \mathbb{Z}^{2}$ with certain property Can we find $(a_{1},a_{2})$ and $(b_{1},b_{2})\in \mathbb{Z}^{2}$ such that $$a_{1}a_{2} - 6\cdot b_{1}b_{2} = 2$$ and $$a_{1}b_{2}+a_{2}b_{1} = 1$$ 
$\textbf{Note.}$ $a_{1},a_{2},b_{1},b_{2} \neq 0$.

The given system can be considered as a linear system with parameters $a_1$ and $b_1$, and thus solved in the following way
$$a_2= 2\dfrac{a_1 + 3b_1}{a_1^2 + 6 b_1^2}, \ \ \ b_2= \dfrac{a_1 - 2b_1}{a_1^2 + 6b_1^2}$$
These rational numbers have to be integers.
In particular $a_2-2b_2=\dfrac{10b_1}{a_1^2 + 6 b_1^2}$ should be an integer.
In other words, there should exist $k \in \mathbb{Z}$ such that:
$$k(a_1^2 + 6 b_1^2)=10 b_1 \ \ \ (1)$$
As $a_1^2>0$, the absolute value of the LHS is $>6|k|b_1^2$. Thus, if there is a solution, it is with $|k|=1$ and $|b_1|=1$ (otherwise the abs. value of the LHS of (1) would be $>12|b_1|$, which is not compatible with the RHS).
Equation (1) then becomes 
$$a_1^2 +6 = +10 \ \ \ \ (2)$$
or $a_1^2 +6 = -10$ (this last case being without solution).
A consequence of (2) is that $|a_1|=2$.
Plugging these values of $(a_1,b_1)=(2,1)$ or $(-2,1)$, or $(2,-1)$ or $(-2,-1)$ into the expressions of $x$ and $y$ above give no valid solution (either $b_2=0$, or $x$ not integer, or $y=0$, resp. ).
As a conclusion, there are no solutions for which $a_1a_2b_1b_2 \neq 0$.
