Why does the determinant $D$, have to be $0$ for equation to have a solution? Suppose $2\times2$ equation:
$$
  \begin{cases}
   a_1x + b_1y = c_1  \\
   a_2x + b_2y = c_2 
  \end{cases} $$  
We can make determinants:
$$D=\begin{vmatrix}{a_1}&{b_1}\\{a_2}&{b_2}\end{vmatrix}$$
$$A=\begin{vmatrix}{a_1}&{c_1}\\{a_2}&{c_2}\end{vmatrix}$$
$$B=\begin{vmatrix}{c_1}&{b_1}\\{c_2}&{b_2}\end{vmatrix}$$
Solution to the $2\times2$ equation exists if $D = 0$ 
Why?
 A: Some intuitions.
From some point of view, your equation is $DX=B$, hence, if it is possible, $D^{-1}DX=X=D^{-1}B$ and  Cauchy's theorem about determinants  implies, that $\det D\neq 0$, because $D^{-1}D=I$.
A: No no, you got this the wrong way. If $ \det D \ne 0$ which is the "general" case then the system has a unique solution . If $\det D = 0$ which is the "specific" case further investigation is required and the system has either no solution or infinitely many solutions.
A: Consider this linear system: 
\begin{equation}
\begin{matrix}
a_1x+b_1y&={c_1}\\ a_2x + b_2y&= {c_2}\end{matrix}
\end{equation}
which, in matrix format is
\begin{equation}
\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} {c_1} \\ {c_2} \end{bmatrix}.
\end{equation}
Solutions to which can be found with Cramer's rule, as
\begin{equation}
x = \begin{vmatrix} {{c_1}} & b_1 \\ c_2 & b_2 \end{vmatrix}/\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}  = { c_1b_2 - b_1c_2 \over a_1b_2 - b_1a_2}
\end{equation}
Which shows that if the determinant is zero, there are no solutions. So in general this is not true.
