# How to find the determinant of this matrix

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.



I was wondering this while studying

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



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

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



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...



Where I did mistake?

• Matlab's symbolic toolbox... Jul 17, 2014 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}}
– Jam
Jul 17, 2014 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. Jul 17, 2014 at 14:25
• Jul 17, 2014 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. Jul 17, 2014 at 14:28

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.

• no it's not. only $x_i \neq x_j$ Jul 17, 2014 at 14:33
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.
• Is $ax_i^2 + bx_iy_i+cy_i^2 +d x_i+ey_i +f= 0$ right, ins't it? Jul 17, 2014 at 15:25
• No, there is no $f$ here. The equation still defines a conic. Jul 17, 2014 at 15:27
• Ummm, Is $ax_i^2 + bx_iy_i+cy_i^2 +d x_i+ey_i +f= 0$ general conic section equation ?? Jul 17, 2014 at 15:29