Why does the determinant $D$, have to be $0$ for equation to have a solution?

Question

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?

Answer 1

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 $D^{-1}D=I$.

Answer 2

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.

Answer 3

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.