# Almost Vandermonde determinant

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

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.

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

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

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

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

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

• Nice, this is clever. – Morgan Rodgers Oct 11 '20 at 11:00
• unbelievable solution.. isn't the second one after simplification zero? – Buraian Oct 11 '20 at 11:06
• You are right, the matrix has same columns. – Zhang Oct 11 '20 at 11:20
• You maybe interested in where it came from :-) here – Buraian Oct 11 '20 at 11:24
• Thanks. I'm glad I helped. – Zhang Oct 11 '20 at 11:48

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).