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## Homework Statement

Let a_1,...,a_n be Real. Then define a 2n x 2n matrix A as follows.

The following are the first 4 and the last 2 columns;

note that the (2k) column equals (a_k)(derivative of the (2k-1) column

A[,1] = a_1^(2n),a_1^(2n-1),a_1^(2n-2),...,a_1^(2), a_1

A[,2] = (2n)a_1^(2n),(2n-1)a_1^(2n-1),(2n-2)a_1^(2n-2),...,2a_1^(

2), a_1

A[,3] = a_2^(2n),a_2^(2n-1),a_2^(2n-2),...,a_2^(2), a_2

A[,4] = (2n)a_2^(2n),(2n-1)a_2^(2n-1),(2n-2)a_2^(2n-2),...,2a_2^(2), a_2

.

.

.

A[,2n-1] = a_n^(2n),a_n^(2n-1),a_n^(2n-2),...,a_n^(2), a_n

A[,2n] = (2n)a_n^(2n),(2n-1)a_n^(2n-1),(2n-2)a_n^(2n-2),...,2a_n^(2), a_n

Calulate det(A)

## Homework Equations

I'd like to need to know if there's a special name for this kind of matrix? (It seems to be a special case, which can be generalized to any order derivative)

Also, I outlined my idea below, and I'm curious if there's an easier way of solving this problem

## The Attempt at a Solution

I partition A into 4 square matrices, say P,Q,R, and S. Then proceed inductively, with the top-left partition P at step k+1 being equal to the whole A matrix from the previous kth step. This way at each step the dimensions are as follows

top-left P: (2k-2)x(2k-2)

top-right Q: (2k-2)x2

bottom-left R: 2x(2k-2)

bottom-right S: 2x2

Next, at each step, I use det(A)=det(P)det(S-RP^(-1)Q)

This is where I'm getting confused. Can I somehow use the dependence between the columns without going through all the gory calculations?

Appreciate any help

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