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

I am studying for a test, and I am completely lost at this problem.

Let $f(t)=detV$, with $x_1, x_2, x_3$ all distinct. Explain why $f(t)$ is a cubic polynomial, show that the coefficient of $t^3$ is nonzero, and find 3 points on the graph of $f$.

Compared to the other questions in my textbook, this one is a curveball, and I can't find others like it.

share|cite|improve this question
What does $V$ have to do with $x_1, x_2, x_3$? – Qiaochu Yuan Oct 17 '12 at 7:54
It doesn't say. That's all the book gives me. I assume it represents a square matrix. – Grace C Oct 17 '12 at 7:57
Then the problem is badly underspecified. – Qiaochu Yuan Oct 17 '12 at 7:58
What's $V$? ${}{}$ – Rudy the Reindeer Oct 17 '12 at 8:12
-1 Sometimes it is necessary to know how and what to read when trying to solve a problem. This question is a rather poorly written, presented and a very incomplete one, in particular since by Martini's answer the whole info is in the same page. Anyone at this level shouldn't ever ask a question that she/he can't basically understand. – DonAntonio Oct 17 '12 at 11:54
up vote 2 down vote accepted

If the matrix is the one on the last answer, then the condition of $x_1$, $x_2$ and $x_3$ being different implies that the minors $V_1^n$ are non-zero (check by computation). Particularly, $V_1^4 \neq 0$, which will be the coefficient of $t^3$ (up to sign change): $$\mbox{coeff}(t^3) = (-1)^{1+4}( x_2x_3^2 + x_3x_1^2 + x_1x_2^2 - x_2x_1^2 - x_3x_2^2 - x_1x_3^2). $$

share|cite|improve this answer

On the bottom of the page where this exercise is from (D. Lay, Linear Algebra and its applications, p. 186) it is written:

Exercises 9 and 10 concern determinants of the following Vandermonde matrices: \[ V(t) = \begin{pmatrix} 1 & t & t^2 & t^3 \\ 1 & x_1 & x_1^2 & x_1^3\\ 1 & x_2 & x_2^2 & x_2^3 \\ 1 & x_3 & x_3^2 & x_3^3 \end{pmatrix} \]

(Remark: and another one which is not important for the exercise above). The exercise above is No. 10:

10. Let $f(t) = \det V$, with $x_1$, $x_2$, $x_3$ all distinct. Explain why $f(t)$ is a cubic polynomial, show that the coefficient of $t^3$ is nonzero, and find 3 points on the graph of $f$.

I dont't know if this qualifies as an answer, but it's too long for a comment.

share|cite|improve this answer
Bravo for the research! – Marc van Leeuwen Oct 17 '12 at 9:17
Did you recognize the question, use google very carefully, or are you a wizard? – Aaron Oct 17 '12 at 9:49
I had luck using google ... – martini Oct 17 '12 at 9:54

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.