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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 $x_1, x_2, x_3$?2012-10-17
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    It doesn't say. That's all the book gives me. I assume it represents a square matrix.2012-10-17
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    Then the problem is badly underspecified.2012-10-17
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    What's $V$? ${}{}$2012-10-17
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    -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.2012-10-17
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    What's $t$?????2012-10-19

2 Answers 2

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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!2012-10-17
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    Did you recognize the question, use google very carefully, or are you a wizard?2012-10-17
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    I had luck using google ...2012-10-17