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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?

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  • $\begingroup$ This is not true; consider any case with $a_1 = b_1 = 0$, $c_1 \neq 0$. Then, $D = 0$ but the system admits no solution because its first equation doesn't. $\endgroup$ Commented Dec 16, 2014 at 13:30
  • $\begingroup$ I think your question needs work before we can answer... $\endgroup$
    – JohnD
    Commented Dec 16, 2014 at 13:34

3 Answers 3

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

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  • $\begingroup$ But If $B=0$ the system has infinitely solutions. $\endgroup$ Commented Dec 16, 2014 at 4:14
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    $\begingroup$ This is incorrect imo: you are assuming $\;D^{-1}\;$ exists to begin with in your first line: obviously that then it must be that $\;\det D\neq 0\;$ . The fact is that a non-homogeneous square system still can have solution even if $\;\det D=0\;$ . $\endgroup$
    – Timbuc
    Commented Dec 16, 2014 at 4:45
  • $\begingroup$ @Timbuc In fact, the question is about the Cramer method. Hence my "from some point of view" and the rest. $\endgroup$ Commented Dec 16, 2014 at 12:17
  • $\begingroup$ @PrzemysławScherwentke, the answer's still incorrect. True, Cramer's Rule assumes the determinanto isn't zero, yet the OP didn't mention this and he asks why there's a solution if it is zero, which apparently shows some deep misunderstanding in what he does. Anyway, in your answer, again, you assume (a priori, without any further condition, which doesn't exist) that the matrix is invertible, which doesn't address the OP's question, imo. $\endgroup$
    – Timbuc
    Commented Dec 16, 2014 at 12:23
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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.

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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.

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    $\begingroup$ This is not accurate. It is possible for there to be solutions when the determinant is zero. Consider the case where all $a_i$,$b_i$, and $c_i$ are zero. Then any $x$ and $y$ are a solution. $\endgroup$ Commented Jan 24, 2015 at 11:00
  • $\begingroup$ Ok, so I'll make the case for non-zero coefficients. My point is now valid. $\endgroup$
    – Autolatry
    Commented Jan 26, 2015 at 8:58
  • $\begingroup$ Not so. If all $a_i$,$b_i$, and $c_i$ are $1$ then $x=y=\frac{1}{2}$ is a solution, but the determinant is zero. The condition for existence of a solution is that $[c_1,c_2]$ is in the column space of the matrix. If the determinant is non-zero, then the column space is everything, so you're guaranteed a solution, but you can still have a solution if the determinant is zero. $\endgroup$ Commented Jan 26, 2015 at 23:37

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