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This question already has an answer here:

Given a system of linear equations:

$$(S):\begin{cases} a_1x+b_1y+c_1z=d_1\\ a_2x+b_2y+c_2z=d_2\\ a_3x+b_3y+c_3z=d_3 \end{cases}$$

And we use matrix to represent:

$$\left( \begin{array}{ccc|c} a_1 & b_1 & c_1 & x\\ a_2 & b_2 & c_2 & y\\ a_3 & b_3 & c_3 & z\\ \end{array}\right)= \begin{pmatrix} d_1\\ d_2\\ d_3\\ \end{pmatrix} $$Let $$M=\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \\ \end{pmatrix}$$

$$\begin{pmatrix} x\\ y\\ z\\ \end{pmatrix}=M^{-1}\begin{pmatrix} d_1\\ d_2\\ d_3\\ \end{pmatrix}$$

For (S) to have a unique solution, $det M\ne0$, and no solution for $det M=0$

But I am wondering why $det M=0$ also implies it has infinity many solution.

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marked as duplicate by Dietrich Burde, Pedro Tamaroff Oct 13 '16 at 15:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Why should we have "no solutions for $\det(M)=0$ ? Just take $M=0$ and $d_1=d_2=d_3=0$, then every triple $(x,y,z)$ is a solution. $\endgroup$ – Dietrich Burde Oct 13 '16 at 15:37
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Zero determinant means that one of the equations can be expressed as linear combination of the others (at least, as far as the left hand side is concerned). However, it depends on what happens with the right hand side in this linear combination: If the values coincide, there is simply a superfluous equation and in essence you have less equations than variables, leading to infinitely many solutions. But if the values differ, you arrive at a contradiction and have no solution.

Example 1:

$$\begin {matrix}2x_1&-x_2&+5x_3&=&15\\ 3x_1&+2x_2&-x_3&=&\hphantom04\\ -x_1&-3x_2&+6x_3&=&11\end{matrix} $$ Here, the third equation is the first minus the second, both on the left and on the right. Hence the third equation is redundant and we simply need to solve $$\begin {matrix}2x_1&-x_2&+5x_3&=&15\\ 3x_1&+2x_2&-x_3&=&\hphantom04\end{matrix} $$

Example 2:

$$\begin {matrix}2x_1&-x_2&+5x_3&=&15\\ 3x_1&+2x_2&-x_3&=&\hphantom04\\ -x_1&-3x_2&+6x_3&=&10\end{matrix} $$ This time we again have third = first minus second on the left, but on the right we have $15-4\ne 10$, hence a contradiction and no solution

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  • $\begingroup$ Most Amazing Answer :) $\endgroup$ – Jon Garrick Mar 29 '17 at 3:34

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