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\begin{vmatrix}5- 3 \cdot 2^4 & 2^1 & 2^2 & 2^3 \\ 10 - 3 \cdot 3^4 & 3^1 & 3^2 & 3^3 \\ 2 -3 (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\ 37-3(-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix}

This is a determinant that came up while I was doing a problem.. it has almost the structure of the Vandermonde determinant but I can't see if there are easy simplifications. Any help/ hint is appreciated.

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It is easy if you see $5=2^2+1$.

$\begin{vmatrix}5- 3 \cdot 2^4 & 2^1 & 2^2 & 2^3 \\ 10 - 3 \cdot 3^4 & 3^1 & 3^2 & 3^3 \\ 2 -3 (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\ 37-3(-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} = $

$\begin{vmatrix} 2^4 & 2^1 & 2^2 & 2^3 \\ 3^4 & 3^1 & 3^2 & 3^3 \\ (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\ (-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix}\cdot (-3)+$

$\begin{vmatrix}2^2 & 2^1 & 2^2 & 2^3 \\ 3^2 & 3^1 & 3^2 & 3^3 \\ (-1)^2 & (-1)^1 &(-1)^2 & (-1)^3 \\ (-6)^2 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} +$

$\begin{vmatrix}1 & 2^1 & 2^2 & 2^3 \\ 1 & 3^1 & 3^2 & 3^3 \\ 1 & (-1)^1 &(-1)^2 & (-1)^3 \\ 1 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} $

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  • $\begingroup$ Nice, this is clever. $\endgroup$ – Morgan Rodgers Oct 11 '20 at 11:00
  • $\begingroup$ unbelievable solution.. isn't the second one after simplification zero? $\endgroup$ – Buraian Oct 11 '20 at 11:06
  • $\begingroup$ You are right, the matrix has same columns. $\endgroup$ – Zhang Oct 11 '20 at 11:20
  • $\begingroup$ You maybe interested in where it came from :-) here $\endgroup$ – Buraian Oct 11 '20 at 11:24
  • $\begingroup$ Thanks. I'm glad I helped. $\endgroup$ – Zhang Oct 11 '20 at 11:48
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Do an expansion down the first column, then you have subdeterminants to calculate that ARE all essentially Vandermonde.

What I mean:

$$\begin{vmatrix}5- 3 \cdot 2^4 & 2^1 & 2^2 & 2^3 \\ 10 - 3 \cdot 3^4 & 3^1 & 3^2 & 3^3 \\ 2 -3 (-1)^4 & (-1)^1 &(-1)^2 & (-1)^3 \\ 37-3(-6)^4 & (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} = (5-3\cdot2^{4}) \begin{vmatrix} 3^1 & 3^2 & 3^3 \\ (-1)^1 &(-1)^2 & (-1)^3 \\ (-6)^1 & (-6)^2 & (-6)^3 \\ \end{vmatrix} + \ldots \\ = (5-3\cdot2^{4})\cdot3\cdot(-1)\cdot(-6) \begin{vmatrix} 1 & 3 & 3^2 \\ 1 &(-1) & (-1)^2 \\ 1 & (-6) & (-6)^2 \\ \end{vmatrix} + \ldots$$

(Also, since this is a $4\times 4$ matrix it should be easy to compute the determinant without any special tricks).

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