Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am writing a bilinear interpolation method.

This method can be abstract by solve the equation A*x = b, A is a 4x4 matrix below: $A=\begin{pmatrix} 1 &x_1 &y_1 &x_1y_1\\ 1 & x_2 & y_2 & x_2y_2\\ 1 & x_3 & y_3 & x_3y_3\\ 1 & x_4 & y_4 & x_4y_4\end{pmatrix}$

Here, $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and $(x_4, y_4)$ are four points containing the dst interpolation point.

My problem is when $\det(A) = 0$ (then $x \neq A^{-1}b$), what is the quadrangle looks like?

share|cite|improve this question

The determinant is trivially zero, if the four points are collinear (if $y=mx+n$ then the third row is $m$ times the second plus $n$ times the first column). It is also zero if to points coincide (two equal rows). More generally, $\det A=0$ only if all points are on a curve of the form $$\tag1a+bx+cy+dxy=0$$ with not all of $a,b,c,d$ equal to zero. If $d=0$, we re-obtain the result above about collinerar points; if $d\ne0$, then $(1)$ can be rewritten as $$ \left(x+\frac cd\right)\left(y+\frac bd\right)=\frac{bc-ad}{d^2},$$ that is a hyperbola (or two axe-parallel lines)

share|cite|improve this answer

Consider the rows of your matrix without the leading $1$. This gives four vectors in $\mathbb{R}^3$, namely $v_1 = (x_1, y_1, x_1 y_1)$, $v_2 = (x_2, y_2, x_2 y_2)$, $v_3 = (x_3, y_3, x_3 y_3)$, and $v_4 = (x_4, y_4, x_4 y_4)$. These vectors are affinely dependent if and only if the determinant of your matrix $A$ is $0$. Put differently, they are all contained in a plane in $\mathbb{R}^3$ if and only if $\det(A) = 0$. Since such a plane cuts the $xy$-coordinate plane in a line (except for the trivial case), all points must be collinear.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.