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I'd like to find the determinant of following matrix

$$ \begin{pmatrix} {x_1}^2 & x_1y_1 & {y_1}^2 & x_1 & y_1 \\ {x_2}^2 & x_2y_2 & {y_2}^2 & x_2 & y_2 \\ {x_3}^2 & x_3y_3 & {y_3}^2 & x_3 & y_3 \\ {x_4}^2 & x_4y_4 & {y_4}^2 & x_4 & y_4 \\ {x_5}^2 & x_5y_5 & {y_5}^2 & x_5 & y_5 \\ \end{pmatrix} $$

where $ x_i \not=x_j $, $ y_i \not= y_j $ for $ i \not= j $
($ i,j=1,2,3,4,5$)

And I'd like to verify the determinant is not zero.

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I was wondering this while studying

Vandermonde matrix, conic section - five points ... and so on.

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By the way,

could recommend any program or website which calculate determinant of matrix?

Actually I touched "wolframalpha.com", but wolfram can't recognize my input. (maybe I did mistake...)

$$ $$ Thank you for your attention to this matter.

$$ $$


The conic section equation will be of the form

$$ Ax^2 + Bxy+Cy^2 +D x+Ey +F= 0 --- (*) $$with A, B, C not all zero

If the points $(x_1, y_1), (x_2, y_2), \cdots, (x_5, y_5)$ satisfy (*) , then

$$ \begin{pmatrix} {x_1}^2 & x_1y_1 & {y_1}^2 & x_1 & y_1 \\ {x_2}^2 & x_2y_2 & {y_2}^2 & x_2 & y_2 \\ {x_3}^2 & x_3y_3 & {y_3}^2 & x_3 & y_3 \\ {x_4}^2 & x_4y_4 & {y_4}^2 & x_4 & y_4 \\ {x_5}^2 & x_5y_5 & {y_5}^2 & x_5 & y_5 \\ \end{pmatrix}\begin{pmatrix} A \\ B \\ C \\ D \\ E \\ \end{pmatrix} = \begin{pmatrix} -F \\ -F \\ -F \\ -F \\ -F \\ \end{pmatrix} $$

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BUT in this time

I know the fact(statement) "five points decide conic section UNIQUELY"

So, I thought the determinant of matrix will be nonzero

(and I asked you guys that 'convinced me the determinant is nonzero'...)

BUT the determinant is zero...

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Where I did mistake?

PLEASE help me. Thanks a lot.

share|improve this question
Matlab's symbolic toolbox... –  draks ... Jul 17 at 14:23
wolfram can recognize the matrix as the following {{x_1^2%2Cx_1y_1%2Cy_1^2%2Cx_1%2Cy_1}%2C{x_2^2%2Cx_2y_2%2Cy_2^2%2Cx_2%2Cy_2}%2C{‌​x_3^2%2Cx_3y_3%2Cy_3^2%2Cx_3%2Cy_3}%2C{x_4^2%2Cx_4y_4%2Cy_4^2%2Cx_4%2Cy_4}%2C{x_5‌​^2%2Cx_5y_5%2Cy_5^2%2Cx_5%2Cy_5}} –  Eul Can Jul 17 at 14:25
A $5\times 5$ determinant can be computed using any symbolic CAS. Maple, MATLAB Sym, Mathematica etc. or, quite cumbersome, using Leibniz formula. –  AlexR Jul 17 at 14:25
$\det(M)=0$ @ W|A... –  draks ... Jul 17 at 14:27
Determinant is non trivial iff the rank is full. Therefore i would try to show linear independence of the rows or columns. –  Dan Jul 17 at 14:28

2 Answers 2

For $x_i = y_i$ this is pretty wrong. Determinant is zero iff rank is not full. But if $(x_1,\cdots,x_5)=(y_1,\cdots,y_5)$ this is already wrong.

share|improve this answer
no it's not. only $x_i \neq x_j$ –  Dan Jul 17 at 14:33
sorry, I'd misread it –  Eul Can Jul 17 at 14:33
I also realized it at second view. –  Dan Jul 17 at 14:34

If the determinant is zero then the columns are related, which means there exist $a,b,c,d,e$ such that $$ ax_i^2 + bx_iy_i+cy_i^2 +d x_i+ey_i = 0,\ i=1..5.$$ Then $P_i(x_i,y_i)$ are on the same conic.

Therefore the determinant can be zero, and is zero if and only if the points $P_i$ lie on the same conic in the plane.

share|improve this answer
Is $ax_i^2 + bx_iy_i+cy_i^2 +d x_i+ey_i +f= 0$ right, ins't it? –  user143993 Jul 17 at 15:25
No, there is no $f$ here. The equation still defines a conic. –  Beni Bogosel Jul 17 at 15:27
I'd like to check this, [the columns are related] if and only if [the column vectors are independendent] –  user143993 Jul 17 at 15:27
@user143993: The statement you wrote is not right. The determinant is zero if and only if the columns (or the lines) are related. You can prove it like this: if the determinant is not zero, then any relation between the columns implies that the coefficients are zero (because the system has the unique solution of zeros). Therefore the columns are not related. Conversely, if the determinant is zero, then the system doesn't have a unique solution, i.e. has a non zero solution, i.e. you can find a relation between the column vectors. –  Beni Bogosel Jul 17 at 15:29
Ummm, Is $ax_i^2 + bx_iy_i+cy_i^2 +d x_i+ey_i +f= 0$ general conic section equation ?? –  user143993 Jul 17 at 15:29

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