Question

If the points ($x_1$, $y_1$), ($x_2$, $y_2$) and ($x_3$, $y_3$) are collinear, then the rank of the matrix
$\left[\begin{array}{ccc}{x}_{{}_1}& y_{{}_1}& 1\\ x_{{}_2}& y_{{}_2}& 1\\ x_{{}_3}& y_{{}_3}& 1\end{array}\right]$will always be less than

• A. 3
• B. 2
• C. 1
• D. None of these

Answer 1

Answer: (B)

The given matrix is $\left[\begin{array}{ccc}{x}_{{}_1}& y_{{}_1}& 1\\ x_{{}_2}& y_{{}_2}& 1\\ x_{{}_3}& y_{{}_3}& 1\end{array}\right]$
using $R_{{}_2}\to R_{{}_2}-R_{{}_1},R_{{}_3}\to R_{{}_3}-R_{{}_1}$
$\Delta =\left[\begin{array}{ccc}{x}_{{}_1}& y_{{}_1}& 1\\ x_{{}_2}-x_{{}_1}& y_{{}_2}-y_{{}_1}& 0\\ x_{{}_3}-x_{{}_1}& y_{{}_3}-y_{{}_1}& 0\end{array}\right]=0$
$(\because$ points are collinear i.e., area of triangle =0 )
$\Rightarrow \left|\begin{array}{cc}{x}_{{}_2}-x_{{}_1}& y_{{}_2}-y_{{}_1}\\ x_{{}_3}-x_{{}_1}& y_{{}_3}-y_{{}_1}\end{array}\right|=0$
So, the rank of matrix is always less than $2$.