Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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):

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

share|cite|improve this question
up vote 1 down vote accepted

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).

share|cite|improve this answer
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 $( b-a ) ( c-a ) ( c-b )$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.