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I want to calculate the determinant of the matrix

$M=\begin{pmatrix} 1 & 1 & 1\\ x & y & z\\ x^2 & y^2 & z^2 \end{pmatrix}$

where $x,y,z\in\mathbb{F}_4=\{0,1,a,b\}$ (the field of four elements). $x,y,z$ could be distinct elements of $\mathbb{F}_4$ or not.

All I know is that for any element $\mathbb{F}_4$, say $t$, then $t+t=0$, that $b=a^2$ and that $+$ and $\cdot$ are commutative in $\mathbb{F}_4$. Because $t+t=0$, then $t=-t$ for all $t\in\mathbb{F}_4$. I have found that:

$\det{M}=1\cdot(y\cdot z^2-y^2\cdot z)-1\cdot(x\cdot z^2-z\cdot x^2)+1\cdot(x\cdot y^2-y\cdot x^2)$

Using $t=-t$ and the fact that $1\cdot t=t$ for all $t\in\mathbb{F}_4$, then we have:

$\det{M}=yz^2+zy^2+xz^2+zx^2+xy^2+yx^2$

Is there a way of simplifying things further?

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    $\begingroup$ Look up the formula for Vandermonde determinant, and then compute it over $\mathbb{F}_4$. $\endgroup$ – Dietrich Burde Oct 5 '16 at 18:49
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Hint: The formula for the Vandermonde determinant gives $$ \det(M)=(x+y)(x+z)(y+z) $$ over any field of characteristic $2$. Now use the addition and multiplication table for $\mathbb{F}_4=\{0,1,a,b\}$ (if you google MSE, then you will find it here, or at other MSE-questions).

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  • $\begingroup$ Alternatively WA it. $\endgroup$ – Git Gud Oct 5 '16 at 19:28
  • $\begingroup$ Yes, WA is very useful, indeed. $\endgroup$ – Dietrich Burde Oct 5 '16 at 19:45

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