# Thread: matrix of 3 lines intersecting at 1 point

1. ## matrix of 3 lines intersecting at 1 point

Given 3 lines
$k: a_1x+b_1y+c_1=0$
$l: a_2x+b_2y+c_2=0$
$m: a_3x+b_3y+c_3=0$

Prove that in order to make $k,l,m$ intersect at one point, must satisfy this condition
det $\left[ \begin{matrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{matrix} \right] = 0$

2. Originally Posted by GTK X Hunter
Given 3 lines
$k: a_1x+b_1y+c_1=0$
$l: a_2x+b_2y+c_2=0$
$m: a_3x+b_3y+c_3=0$

Prove that in order to make $k,l,m$ intersect at one point, must satisfy this condition
det $\left[ \begin{matrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{matrix} \right] = 0$
I presume that you know that a point, (x,y), where all three lines intersect is a solution to that system of equations and that such a system of equations is equivalent to the matrix equation $\begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$, while requiring that z= 1. If that matrix has an inverse, we could multiply both sides by its inverse and get the unique solution x= 0, y= 0, z= 0.

We can only have a solution with z= 1 if the solution is not unique. That means that the matrix must not have an inverse and so its determinant must be 0.