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Computing the determinant of a Vandermonde-like matrix

This is a problem from a students competition but it looks quite straightforward to solve.

The task is to compute the determinant of the following matrix.

A=[123...n122232...n2...............1n2n3n...nn]

What I noticed is that it looks similar to Vandermonde's matrix even though it's not quite that. 

Let's pull out from its determinant a common multiple k from the k-th column for each k=1,2,3,...,n.

We get 

det(A)=n!|102030...n0112131...n1...............1n12n13n1...nn1|

The matrix whose determinant we got in the RHS of the last equality is the transpose of the Vandermonde matrix (with αs=s, for s=1,2,...,n). Hence its determinant equals the determinant of the Vandermonde matrix itself (since we know that det(A)=det(AT)).

So (leaving aside the n! multiplier) the determinant we are left with now is Vandermonde's determinant for the numbers 1,2,3,...,n. We know its value is equal to 

1i<jn(ji)

So the result we get now is 

det(A)=n!1i<jn(ji)

This is the same as 

n!(n1)1(n2)2(n3)3 ... 3n32n21n1

Finally it is not difficult to realize that this expression is equal to 

n! (n1)! (n2)! ... 3! 2! 1!

So this is our final answer here for the determinant of A.


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