1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Travis Willse Dec 16 '14 at 13:30
  • $\begingroup$ I think your question needs work before we can answer... $\endgroup$ – JohnD Dec 16 '14 at 13:34
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ But If $B=0$ the system has infinitely solutions. $\endgroup$ – AsdrubalBeltran Dec 16 '14 at 4:14
  • 1
    $\begingroup$ 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 Dec 16 '14 at 4:45
  • $\begingroup$ @Timbuc In fact, the question is about the Cramer method. Hence my "from some point of view" and the rest. $\endgroup$ – Przemysław Scherwentke Dec 16 '14 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 Dec 16 '14 at 12:23
0
$\begingroup$

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.

$\endgroup$
-2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Callus Jan 24 '15 at 11:00
  • $\begingroup$ Ok, so I'll make the case for non-zero coefficients. My point is now valid. $\endgroup$ – Autolatry Jan 26 '15 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. $\endgroup$ – Callus Jan 26 '15 at 23:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.