#|
( (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 )
| = # ?

Question

#| 
  ( (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 )
 | = # ?

Evelyn Agard · Accepted Answer

#288#

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 ) | #

Write: for Simplicity

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

The supplied determinant can then be written as follows:

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 #

Evelyn Agard · Answer

undefined The given expression appears to represent a 5x5 matrix with entries derived from polynomial expressions involving $ x $. Each row consists of powers of $ x $ subtracted by constants from 1 to 5, and each column represents different powers of these expressions, from $ (x-a)^4 $ to $ 1 $, where $ a $ ranges from 1 to 5.

To simplify the matrix, we can expand each polynomial expression:

For the first row:
$$ (x-1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1 $$
$$ (x-1)^3 = x^3 - 3x^2 + 3x - 1 $$
$$ (x-1)^2 = x^2 - 2x + 1 $$
$$ (x-1) = x - 1 $$
$$ 1 = 1 $$

We apply the same expansion process to the remaining rows, replacing $ x $ with $ x-a $ for each row from 2 to 5.

For the second row:
$$ (x-2)^4 = x^4 - 8x^3 + 24x^2 - 32x + 16 $$
$$ (x-2)^3 = x^3 - 6x^2 + 12x - 8 $$
$$ (x-2)^2 = x^2 - 4x + 4 $$
$$ (x-2) = x - 2 $$
$$ 1 = 1 $$

For the third row:
$$ (x-3)^4 = x^4 - 12x^3 + 54x^2 - 108x + 81 $$
$$ (x-3)^3 = x^3 - 9x^2 + 27x - 27 $$
$$ (x-3)^2 = x^2 - 6x + 9 $$
$$ (x-3) = x - 3 $$
$$ 1 = 1 $$

For the fourth row:
$$ (x-4)^4 = x^4 - 16x^3 + 96x^2 - 256x + 256 $$
$$ (x-4)^3 = x^3 - 12x^2 + 48x - 64 $$
$$ (x-4)^2 = x^2 - 8x + 16 $$
$$ (x-4) = x - 4 $$
$$ 1 = 1 $$

For the fifth row:
$$ (x-5)^4 = x^4 - 20x^3 + 150x^2 - 500x + 625 $$
$$ (x-5)^3 = x^3 - 15x^2 + 75x - 125 $$
$$ (x-5)^2 = x^2 - 10x + 25 $$
$$ (x-5) = x - 5 $$
$$ 1 = 1 $$

Thus, the expanded matrix is:

$$
\begin{pmatrix}
x^4 - 4x^3 + 6x^2 - 4x + 1 & x^3 - 3x^2 + 3x - 1 & x^2 - 2x + 1 & x - 1 & 1 \
x^4 - 8x^3 + 24x^2 - 32x + 16 & x^3 - 6x^2 + 12x - 8 & x^2 - 4x + 4 & x - 2 & 1 \
x^4 - 12x^3 + 54x^2 - 108x + 81 & x^3 - 9x^2 + 27x - 27 & x^2 - 6x + 9 & x - 3 & 1 \
x^4 - 16x^3 + 96x^2 - 256x + 256 & x^3 - 12x^2 + 48x - 64 & x^2 - 8x + 16 & x - 4 & 1 \
x^4 - 20x^3 + 150x^2 - 500x + 625 & x^3 - 15x^2 + 75x - 125 & x^2 - 10x + 25 & x - 5 & 1 \
\end{pmatrix}
$$

#| ( (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 ) | = # ?

D = |

D = |