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Show that $$\left|\begin{matrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{matrix}\right|=(a-b)(b-c)(c-a)$$

Gerry Myerson
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manish kumar
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    You need to seriously edit your question to include (a) Mathjax formatting, (b) a little information as to the context of the problem and (c) your efforts in attempting this problem. –  May 26 '16 at 09:20
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    @manishkumar Welcome to MSE. You need to learn how to format math. There is a useful guide here http://math.stackexchange.com/help/notation In the meantime I have edited your question. Let me know if that is not what you want. – almagest May 26 '16 at 09:23
  • Expand the determinant through the first column and notice that the $(1,1)$ minor $M_{1, 1}=(1)^1.1.(bc^{2}-b^{2}c)$ then find $M_{2, 1}$ and $M_{3, 1}$. –  May 26 '16 at 09:25
  • The determinant is zero (since two rows are equal) when $a=b$, $b=c$ or $c = a$. The determinant is also a polynomial of degree $2$ in each of $a,b,c$. This can be used to find the determinant up to a constant without explicitly computing it. The constant can be found by considering the diagonal contribution $1\cdot b\cdot c^2$. – Winther May 26 '16 at 09:36
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    This is a Vandermonde Matrix. – sTertooy May 26 '16 at 10:00

3 Answers3

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$$\left|\begin{matrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{matrix}\right| =\left|\begin{matrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{matrix}\right| =\left|\begin{matrix}b-a&b^2-a^2\\c-a&c^2-a^2\end{matrix}\right|=(b-a)(c-a)\left|\begin{matrix}1&b+a\\1&c+a\end{matrix}\right|=(b-a)(c-a)(c-b).$$

Alternatively, paving the way for a generalization, $$\left|\begin{matrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{matrix}\right| =\left|\begin{matrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{matrix}\right| =\left|\begin{matrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{matrix}\right|=(b-a)(c-a)\left|\begin{matrix}1&a&a^2\\0&1&b+a\\0&1&c+a\end{matrix}\right|=(b-a)(c-a)\left|\begin{matrix}1&a&a^2\\0&1&b+a\\0&0&c-b\end{matrix}\right|=(b-a)(c-a)(c-b)\left|\begin{matrix}1&a&a^2\\0&1&b+a\\0&0&1\end{matrix}\right|.$$

  • The most elegant approach, +1. –  May 26 '16 at 09:56
  • @Bacon Not going into comparisons but I already mentioned this approach in my answer and I do believe my approach is better. – User Not Found May 26 '16 at 16:31
  • @ArghyaChakraborty It seems you did; however I disagree with you insofar as I think Yves' answer is more elegant than yours and why I upvoted his/her answer. –  May 27 '16 at 07:32
  • @bacon: I would rate as elegant an answer that is easy to understand, requires little computation and generalizes to higher order in an obvious way. Based on these criteria, my best choice is elsewhere on this page :) –  May 27 '16 at 07:43
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    @YvesDaoust I love your arrogance! :-) –  May 27 '16 at 07:47
  • @Bacon: you are welcome. –  May 27 '16 at 08:04
  • @YvesDaoust "generalizes to higher order in an obvious way" this answer wins..."elegant an answer that is easy to understand, requires little computation" your answer below wins. – User Not Found May 27 '16 at 13:34
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Observe

  1. The expression must be a homogeneous polynomial of degree at most $3$ as all terms involved are made of factors of degree $0+1+2$.

  2. The expression cancels when two of the parameters are equal.

Then the expression must be of the form $$K(a-b)(b-c)(c-a)$$ for some constant $K$.

As the main diagonal generates the term $+bc^2$, by identification we must have $K=1$.

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Yes this determinant can be expanded row-wise or row operations $R_2=R_2-R_1$ and $R_3=R_3-R_1$ and then expanded...but there's a better way.

Clearly setting $a=b$ or $b=c$ or $a=c$ gives determinant as a multiple of $(a-b)(b-c)(c-a)$

Also the fact that determinant can at most have power 2 over $a$ , $b$ or $c$ gives that it is a constant times $(a-b)(b-c)(c-a)$.

And the constant can be easily determined by setting $a=-1,b=0,c=1$ and evaluating the determinant and the expression.

Hope this helps.

User Not Found
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