# What are actually proportional rows(columns) in a determinant?

So, I posted the question about determinant equality and thoght that two columns are proportional. But they are not. Please explain me this on the next exaple (I thought the first and the second column are proportional):

$$\begin{matrix} 1 & 1 & 1 \\ a & a^2 & a^3 \\ a^2 & a^4 & a^6 \\ \end{matrix}$$

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## 2 Answers

For them to be proportional you would need a $b$ such that $1 = b \cdot 1$, so that $b = 1$, and thus $a^2 = b a = a$ (and $a^4 = b a^{2} = a^{2}$), so that $a = 0, 1$ (provided we are working over a field).

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So $a, a^2, a^3$ is propotional with $a^3, a^3, a^4$? – A6SE Mar 23 '13 at 18:24
@A6Tech, if the second list is $a^2, a^3, a^4$, yes. – Andreas Caranti Mar 23 '13 at 18:28
Yeah, $a^2$, i accidently made a mistake. Thanks a lot :D – A6SE Mar 23 '13 at 18:30
@A6Tech, that was my guess, misprints happen, I just wanted to make sure I was giving you a correct answer. – Andreas Caranti Mar 23 '13 at 18:31

The matrix you have been given is actually a special case of a Vandermonde matrix.

The property is not that the columns or rows are proportional (in which case the determinant would have to be $0$), but that the columns form geometric sequences $x^0 , x^1 , x^2$.

In general, the determinant of the $3 \times 3$ Vandermonde matrix $$\left( \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array} \right)$$ is $( b-a ) ( c-a ) ( c-b )$.

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