[Math] How do collinear points on a matrix affect its rank

Question

Consider the matrix

\begin{matrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end{matrix}

what effect does $({x_1},{y_1})$,$({x_2},{y_2})$,$({x_3},{y_3})$ being collinear have on the rank of the above matrix ?

Answer 1

$$(x_i\,,\,y_i)\;,\,\,i=1,2,3\;,\;\;\text{are collinear}\;\;\iff \frac{y_3-y_2}{x_3-x_2}=\frac{y_3-y_1}{x_3-x_1}=\frac{y_2-y_1}{x_2-x_1}\iff$$

$$\iff (x_2-x_1)(y_3-y_2)=(y_2-y_1)(x_3-x_2)\\\;\;\;\;\;\;\;\;\;\;(x_3-x_1)(y_2-y_1)=(x_2-x_1)(y_3-y_1)\\\;\;\;\;\;\;\;\;\;\;(x_3-x_2)(y_2-y_1)=(x_2-x_1)(y_3-y_2)$$

If one of the denominators vanishes (and thus all of them), then $\,x_1=x_2=x_3\,$ , and we have two columns linearly dependent and the determinant in then zero, otherwise and using the above equality:

$$\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix}=x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=$$

$$=(x_3-x_2)(y_2-y_1)+x_2\left[(y_3-y_1)-(y_3-y_2)\right]+x_3(y_1-y_2)=$$

$$=0$$

so again zero. Do it slowly...!