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

1 Answer
Steve M
Feb 9, 2018

#288#

Explanation:

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

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

For Simplicity, write:

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

Then we can write the given determinant as:

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

Which is a Vandermonde matrix of order #5#. 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_2-x_3)(x_2-x_4)(x_2-x_5) * #
# \ \ \ \ \ \ \ \ (x_3-x_4)(x_3-x_5) * (x_4-x_5) #

# \ \ \ = (1)(2)(3)(4) * (1)(2)(3) * (1)(2) * (1) #

# \ \ \ = (4!)(3!)(2!) #

# \ \ \ = 24 * 6 * 2 #

# \ \ \ = 288 #

Related questions
Impact of this question
1225 views around the world
You can reuse this answer
Creative Commons License