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I want to prove that the following 2 matrices have the same determinant (which they do and it is equal to 288) Apologies for quality: [Matrix one][1] [Matrix two][2]

1 2

I know that the first matrix is a Vandermonde matrix, and it is easy to calculate its determinant. The second matrix looks similar, but I can't find the connection. Any ideas would be helpful.

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    $\begingroup$ Matrix 2 is obtained by doing some column operations namely C2-C1, C3-C1,.... These preserve the determinant. $\endgroup$
    – daruma
    May 5 at 0:35
  • $\begingroup$ How difficult is it to compute the determinants with software? $\endgroup$
    – CroCo
    May 5 at 2:10

1 Answer 1

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Let $A$ be the first matrix, and $B$ the second matrix , then

$ B = A C $

where

$ C = \begin{bmatrix} 1 && -1 && 0 && 0 && 0 \\0 && 1 && -1 && 0 && 0 \\ 0 && 0 && 1 && -1 && 0 \\ 0 && 0 && 0 && 1 && -1 \\ 0 && 0 && 0 && 0 && 1 \end{bmatrix} $

From the well-known property of determinants we know that

$ \det(B)= \det(A) \det(C) $

Since $C$ is upper-triangular, then its determinant is the product of its diagonal elements, which is $1$, hence

$ \det(B) = \det(A) $

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  • $\begingroup$ Thank you, this helps a lot! $\endgroup$
    – Uncle True
    May 5 at 1:02
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    $\begingroup$ Can you explain how you arrived at $B=AC$? $\endgroup$
    – CroCo
    May 5 at 2:11
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    $\begingroup$ Just by examining the matrices $A$ and $B$, you can see that the first column of $B$ is the same as $A$, and the $k$-th column of $B$ is the $k$-th column of $A$ minus the $(k-1)$-th column of $A$, for $ k = 2,3,4,5 $. $\endgroup$
    – Hosam H
    May 5 at 5:31

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