# | ( (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 ) | =  ?

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:

${x}_{1} = \left(x - 1\right)$, ${x}_{2} = \left(x - 2\right)$, ${x}_{3} = \left(x - 3\right)$,
${x}_{4} = \left(x - 4\right)$, ${x}_{5} = \left(x - 5\right)$

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:

$D = {\prod}_{1 \le i < j \le n} \left({x}_{i} - {x}_{j}\right)$

$\setminus \setminus \setminus = \left({x}_{1} - {x}_{2}\right) \left({x}_{1} - {x}_{3}\right) \left({x}_{1} - {x}_{4}\right) \left({x}_{1} - {x}_{5}\right) \cdot$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left({x}_{2} - {x}_{3}\right) \left({x}_{2} - {x}_{4}\right) \left({x}_{2} - {x}_{5}\right) \cdot$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left({x}_{3} - {x}_{4}\right) \left({x}_{3} - {x}_{5}\right) \cdot \left({x}_{4} - {x}_{5}\right)$

$\setminus \setminus \setminus = \left(1\right) \left(2\right) \left(3\right) \left(4\right) \cdot \left(1\right) \left(2\right) \left(3\right) \cdot \left(1\right) \left(2\right) \cdot \left(1\right)$

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

$\setminus \setminus \setminus = 24 \cdot 6 \cdot 2$

$\setminus \setminus \setminus = 288$