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

## $| \left({\left(x - 1\right)}^{5} , {\left(x - 1\right)}^{4} , {\left(x - 1\right)}^{3} , {\left(x - 1\right)}^{2} , \left(x - 1\right) , 1\right) , \left({\left(x - 2\right)}^{5} , {\left(x - 2\right)}^{4} , {\left(x - 2\right)}^{3} , {\left(x - 2\right)}^{2} , \left(x - 2\right) , 1\right) , \left({\left(x - 3\right)}^{5} , {\left(x - 3\right)}^{4} , {\left(x - 3\right)}^{3} , {\left(x - 3\right)}^{2} , \left(x - 3\right) , 1\right) , \left({\left(x - 4\right)}^{5} , {\left(x - 4\right)}^{4} , {\left(x - 4\right)}^{3} , {\left(x - 4\right)}^{2} , \left(x - 4\right) , 1\right) , \left({\left(x - 5\right)}^{5} , {\left(x - 5\right)}^{4} , {\left(x - 5\right)}^{3} , {\left(x - 5\right)}^{2} , \left(x - 5\right) , 1\right) , \left({\left(x - 6\right)}^{5} , {\left(x - 6\right)}^{4} , {\left(x - 6\right)}^{3} , {\left(x - 6\right)}^{2} , \left(x - 6\right) , 1\right) |$\ = ?

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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Steve M Share
Feb 9, 2018

$34560$

#### 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:

${x}_{1} = \left(x - 1\right)$, ${x}_{2} = \left(x - 2\right)$, ${x}_{3} = \left(x - 3\right)$,
${x}_{4} = \left(x - 4\right)$, ${x}_{5} = \left(x - 5\right)$, ${x}_{6} = \left(x - 6\right)$,

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:

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

$\setminus \setminus \setminus = \left({x}_{1} - {x}_{2}\right) \left({x}_{1} - {x}_{3}\right) \left({x}_{1} - {x}_{4}\right) \left({x}_{1} - {x}_{5}\right) \left({x}_{1} - {x}_{6}\right) \cdot$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left({x}_{2} - {x}_{3}\right) \left({x}_{2} - {x}_{4}\right) \left({x}_{2} - {x}_{5}\right) \left({x}_{2} - {x}_{6}\right) \cdot$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left({x}_{3} - {x}_{4}\right) \left({x}_{3} - {x}_{5}\right) \left({x}_{3} - {x}_{6}\right) \cdot \left({x}_{4} - {x}_{5}\right) \left({x}_{4} - {x}_{6}\right) \cdot \left({x}_{5} - {x}_{6}\right)$

$\setminus \setminus \setminus = \left(1\right) \left(2\right) \left(3\right) \left(4\right) \left(5\right) \cdot$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(1\right) \left(2\right) \left(3\right) \left(4\right) \cdot$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(1\right) \left(2\right) \left(3\right) \cdot \left(1\right) \left(2\right) \cdot \left(1\right)$

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

$\setminus \setminus \setminus = 120 \cdot 24 \cdot 6 \cdot 2$

$\setminus \setminus \setminus = 34560$

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