# Why the discriminant of a system of linear equation=0 implies it has infinity many solution? [duplicate]

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.

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

• 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. – Dietrich Burde Oct 13 '16 at 15:37

$$\begin {matrix}2x_1&-x_2&+5x_3&=&15\\ 3x_1&+2x_2&-x_3&=&\hphantom04\\ -x_1&-3x_2&+6x_3&=&11\end{matrix}$$ 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 $$\begin {matrix}2x_1&-x_2&+5x_3&=&15\\ 3x_1&+2x_2&-x_3&=&\hphantom04\end{matrix}$$
$$\begin {matrix}2x_1&-x_2&+5x_3&=&15\\ 3x_1&+2x_2&-x_3&=&\hphantom04\\ -x_1&-3x_2&+6x_3&=&10\end{matrix}$$ 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