<s>[INST] Show that the given linear system has this specific solution if $\det \neq 0$ 
Show that the linear system  $\begin{pmatrix} a_{11} & a_{12}\\ 
a_{21} & a_{22}  \end{pmatrix} \begin{pmatrix} x_1\\  x_2
\end{pmatrix}= \begin{pmatrix} b_1\\  b_2 \end{pmatrix} \,\,\,\,\,\,
\text{ with }\,\,\,\, A= \begin{pmatrix} a_{11} & a_{12}\\  a_{21} &
a_{22}  \end{pmatrix}$
has the unique solution $\begin{pmatrix} x_1\\  x_2 \end{pmatrix}$
  with $x_1 = \frac{\det \begin{pmatrix} b_1 & a_{12}\\  b_2 & a_{22}
\end{pmatrix}}{\det A} \,\,\,\,\,\,; \,\,\,\,\,  x_2 = \frac{\det
\begin{pmatrix} a_{11} & b_1\\  a_{21} & b_2 \end{pmatrix}}{\det A}$,
  if $\det A \neq 0.$

I have an idea and I think it's almost correct but something is missing or wrong..
Let $\vec{x} = \begin{pmatrix}
x_1\\ 
x_2
\end{pmatrix}$ and let $\vec{b} = \begin{pmatrix}
b_1\\ 
b_2
\end{pmatrix}$. For the linear system $A\vec{x} = \vec{b}$ there exists the unique solution $\begin{pmatrix}
x_1\\ 
x_2
\end{pmatrix}$, if $\det A \neq 0$ because we can use the inverse of matrix $A$:
$$A\vec{x}=\vec{b} \Leftrightarrow A^{-1}A\vec{x} = A^{-1} \vec{b} \Leftrightarrow A^{-1} \vec{b} = \vec{x}$$
Know that $A^{-1} = \frac{1}{\det A} \begin{bmatrix}
a_{22} & -a_{12}\\ 
-a_{21} & a_{11}
\end{bmatrix}$
Then we have $A^{-1} \vec{b} = \vec{x} \Leftrightarrow \frac{1}{\det A} \begin{bmatrix}
a_{22} & -a_{12}\\ 
-a_{21} & a_{11}
\end{bmatrix}
\begin{bmatrix}
b_1\\ 
b_2
\end{bmatrix} =
\begin{bmatrix}
x_1\\ 
x_2
\end{bmatrix}$
But how is this supposed to be equal to the solution given in the task? 
I did a mistake somewhere? :(
 [/INST] You are almost there; simply write out what the two determinants 
$$\det \begin{pmatrix} b_1 & a_{12}\\  b_2 & a_{22}
\end{pmatrix}\qquad\text{ and }\qquad\det
\begin{pmatrix} a_{11} & b_1\\  a_{21} & b_2 \end{pmatrix}$$
are. You will find that these are precisely the coefficients of the vector
$$\begin{bmatrix}
a_{22} & -a_{12}\\ 
-a_{21} & a_{11}
\end{bmatrix}
\begin{bmatrix}
b_1\\ 
b_2
\end{bmatrix}.$$
</s>