Question

Can you determine the determinant below ? (Would have put it here, but system wouldn't take it -- determinant just a tiny bit too big.) ...

`| ( (x-1)^5, (x-1)^4, (x-1)^3, (x-1)^2, (x-1), 1 ), ( (x-2)^5, (x-2)^4, (x-2)^3, (x-2)^2, (x-2), 1 ), ( (x-3)^5, (x-3)^4, (x-3)^3, (x-3)^2,(x-3), 1 ), ( (x-4)^5,(x-4)^4,(x-4)^3, (x-4)^2,(x-4), 1 ), ( (x-5)^5,(x-5)^4,(x-5)^3, (x-5)^2,(x-5), 1 ), ( (x-6)^5,(x-6)^4,(x-6)^3, (x-6)^2,(x-6), 1 ) |`\ = ?

Answer 1

`34560`

Explanation:

We seek the value of the determinant, `D`, where:

# D = | ( (x-1)^5, (x-1)^4, (x-1)^3, (x-1)^2, (x-1), 1 ), ( (x-2)^5, (x-2)^4, (x-2)^3, (x-2)^2, (x-2), 1 ), ( (x-3)^5, (x-3)^4, (x-3)^3, (x-3)^2, (x-3), 1 ), ( (x-4)^5, (x-4)^4, (x-4)^3, (x-4)^2, (x-4), 1 ), ( (x-5)^5, (x-5)^4, (x-5)^3, (x-5)^2, (x-5), 1 ), ( (x-6)^5, (x-6)^4, (x-6)^3, (x-6)^2, (x-6), 1 ) | #

For Simplicity, write:

`x_1=(x-1)`, `x_2=(x-2)`, `x_3=(x-3)`,
`x_4=(x-4)`, `x_5=(x-5)`, `x_6=(x-6)`,

Then we can write the given determinant as:

# D = | ( x_1""^5, x_1""^4, x_1""^3, x_1""^2, x_1"", 1 ), ( x_2""^5, x_2""^4, x_2""^3, x_2""^2, x_2"", 1 ), ( x_3""^5, x_3""^4, x_3""^3, x_3""^2, x_3"", 1 ), ( x_4""^5, x_4""^4, x_4""^3, x_4""^2, x_4"", 1 ), ( x_5""^5, x_5""^4, x_5""^3, x_5""^2, x_5"", 1 ), ( x_6""^5, x_6""^4, x_6""^3, x_6""^2, x_6"", 1 ) | #

Which is a Vandermonde matrix of order `6`. As such we can write the determinant as a product of the factors of the various permutations:

`D = prod_(1 le i lt j le n) (x_i-x_j)`

`\ \ \ = (x_1-x_2)(x_1-x_3)(x_1-x_4)(x_1-x_5)(x_1-x_6) *`
`\ \ \ \ \ \ \ \ (x_2-x_3)(x_2-x_4)(x_2-x_5)(x_2-x_6) *`
`\ \ \ \ \ \ \ \ (x_3-x_4)(x_3-x_5)(x_3-x_6) * (x_4-x_5)(x_4-x_6) * (x_5-x_6)`

`\ \ \ = (1)(2)(3)(4)(5) *`
`\ \ \ \ \ \ \ \ (1)(2)(3)(4) *`
`\ \ \ \ \ \ \ \ (1)(2)(3) * (1)(2) * (1)`

`\ \ \ = (5!)(4!)(3!)(2!)`

`\ \ \ = 120 * 24 * 6 * 2`

`\ \ \ = 34560`