# Explain why a determinant function is a cubic polynomial

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.

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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

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): $$\mbox{coeff}(t^3) = (-1)^{1+4}( x_2x_3^2 + x_3x_1^2 + x_1x_2^2 - x_2x_1^2 - x_3x_2^2 - x_1x_3^2).$$

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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.

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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