Determinant of a Vandermonde matrix

• Apr 15th 2008, 08:19 PM
pakman
Determinant of a Vandermonde matrix
Consider the 3x3 Vandermonde matrix:

V=[1 x1 x1^2; 1 x2 x2^2; 1 x3 x3^2]

Show that the det(V) = (x2-x1)(x3-x1)(x3-x2)

Sorry for the format. I did R2-R1 and R3-R1 then found the det to be (x2-x1)(x3^2-x1)-(x2^2-x1^2)(x3-x1)

Is this just an algebraic nightmare or did I do something wrong? Thank you!
• Apr 15th 2008, 08:28 PM
mr fantastic
Quote:

Originally Posted by pakman
Consider the 3x3 Vandermonde matrix:

V=[1 x1 x1^2; 1 x2 x2^2; 1 x3 x3^2]

Show that the det(V) = (x2-x1)(x3-x1)(x3-x2)

Sorry for the format. I did R2-R1 and R3-R1 then found the det to be (x2-x1)(x3^2-x1)-(x2^2-x1^2)(x3-x1)

Is this just an algebraic nightmare or did I do something wrong? Thank you!

You've done something wrong. The answer is $-(x_1 - x_3) (x_1 - x_2) (x_2 - x_3)$.

I hope you took the easy road and expanded along the first column.
• Apr 15th 2008, 08:32 PM
pakman
Quote:

Originally Posted by mr fantastic
You've done something wrong. The answer is $-(x_1 - x_3) (x_1 - x_2) (x_2 - x_3)$.

I hope you took the easy road and expanded along the first column.

That's what I did. I subtracted row 1 from both row 2 and 3 to get 0's in a21 and a31. But then it gets a bit messy from there (Crying)
• Apr 15th 2008, 08:54 PM
mr fantastic
Quote:

Originally Posted by pakman
Consider the 3x3 Vandermonde matrix:

V=[1 x1 x1^2; 1 x2 x2^2; 1 x3 x3^2]

Show that the det(V) = (x2-x1)(x3-x1)(x3-x2)

Sorry for the format. I did R2-R1 and R3-R1 then found the det to be (x2-x1)(x3^2-x1)-(x2^2-x1^2)(x3-x1)

Is this just an algebraic nightmare or did I do something wrong? Thank you!

You should have (x2 - x1)(x3^2 - x1^2) - (x2^2 - x1^2)(x3 - x1) ....

= (x2 - x1)(x3 - x1)(x3 + x1) - (x2 - x1)(x2 + x1)(x3 - x1)

= (x2 - x1)(x3 - x1) [(x3 + x1) - (x2 + x1)] = ......
• Apr 15th 2008, 09:19 PM
pakman
Ahh okay, stupid algebraic mistake. Thanks!