A Vandermonde matrix of order n has the form

1 1 1 . . . 1 x_{1}x_{2}x_{3}. . . x_{n}x_{1}^{2}x_{2}^{2}x_{3}^{2}. . . x_{n}^{2}. . . . . . . . . . . . . . . . . . . . . x_{1}^{n-1}x_{2}^{n-1}x_{3}^{n-1}. . . x_{n}^{n-1}

Although the transpose of this may be used as the definition; and it may also be referred to as an *alternant* matrix. Whilst solving an equation with an nXn Vandermonde matrix requires O(n^{2}) operations, it turns out that the determinant is very easy to calculate, using the formula:

_____ | | | | (x_{j}-x_{i}) | | 1≤i<j≤n

For example, the matrix

1 1 1 x y z xOr its transpose, has determinant (y-x)(z-x)(z-y). This could be proved by multiplying out this expression and checking that it gives the same result as a row expansion of the above matrix, but a more elegant solution (which also illustrates why the general nXn result holds) is to use elementary column operations to obtain a lower-triangular matrix, for which the determinant is simply the product of the diagonal.^{2}y^{2}z^{2}

|1 1 1 | |1 0 0 | |x y z | = |x y-x z-x | |xBy subtracting column 1 from each of columns 2 and 3. Then factorising the bottom row we get^{2}y^{2}z^{2}| |x^{2}y^{2}-x^{2}z^{2}-x^{2}|

|1 0 0 | |x y-x z-x | |xSo we can pull out factors of (y-x) and (z-x) by the multilinearity of the determinant function:^{2}(y-x)(y+x) (z-x)(z+x) |

|1 0 0 | (y-x)(z-x) |x 1 1 | |xNow we perform a further column operation, subtracting the second column from the third, which cancels the 1 and x:^{2}(y+x) (z+x) |

|1 0 0 | (y-x)(z-x) |x 1 0 | |xAgain pull a factor forward:^{2}(y+x) (z-y) |

|1 0 0 | (y-x)(z-x)(z-y) |x 1 0 | |xObserve that this is a triangular matrix, so its determinant is the product of the diagonal =1*1*1 =1. So we are left with simply the factors (y-x)(z-x)(z-y) as desired.^{2}(y+x) 1 |