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:

$$D=\begin{vmatrix}{a_1}&amp;{b_1}\\{a_2}&amp;{b_2}\end{vmatrix}$$

$$A=\begin{vmatrix}{a_1}&amp;{c_1}\\{a_2}&amp;{c_2}\end{vmatrix}$$

$$B=\begin{vmatrix}{c_1}&amp;{b_1}\\{c_2}&amp;{b_2}\end{vmatrix}$$

Solution to the $2\times2$ equation exists if $D = 0$ 

Why?
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 $D^{-1}D=I$.
Consider this linear system: 
\begin{equation}
\begin{matrix}
a_1x+b_1y&amp;={c_1}\\ a_2x + b_2y&amp;= {c_2}\end{matrix}
\end{equation}
which, in matrix format is
\begin{equation}
\begin{bmatrix} a_1 &amp; b_1 \\ a_2 &amp; 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}} &amp; b_1 \\ c_2 &amp; b_2 \end{vmatrix}/\begin{vmatrix} a_1 &amp; b_1 \\ a_2 &amp; 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.