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Qestion: $\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=K(a-b)(b-c)(c-a)$, solve for $K$

Answer: $K=(ab+bc+ca)$

My attempt: $$\begin{align}\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}&=\begin{vmatrix}1&0&0\\a^2&b^2-a^2&c^2-a^2\\a^3&b^3-a^3&c^3-a^3\end{vmatrix}\\&=\begin{vmatrix}b^2-a^2&c^2-a^2\\b^3-a^3&c^3-a^3\end{vmatrix}\\&=(b-a)(c-a)\begin{vmatrix}b+a& c+a\\b^2+ba+a^2&c^2+ca+a^2\end{vmatrix}\\&=(c-b)(a-b)(b+a)(c+a)\begin{vmatrix}1&1\\(b+a)-ba&(c+a)-ca\end{vmatrix}\\\end{align}$$

I don't know what should I do next, maybe I've made a mistake but I didn't notice it

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    $\begingroup$ How did you take $(b+a)$ and $(c+a)$ out of the determinant in the second last step? $\endgroup$ – tatan Mar 17 at 17:11
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Consider the matrix \begin{bmatrix} 1& 1 &1 &1 \\ X & a&b&c \\X^2 & a^2 & b^2 & c^2 \\ X^3 & a^3 & b^3 & c^3 \end{bmatrix}

Its determinant is a Vandermonde determinant, equal to $$(a-X)(b-X)(c-X)(b-a)(c-a)(c-b)$$

But developping with respect to the first column, you see that the determinant you are looking for is just equal to the coefficient (with a sign minus) of the $X$ term in this polynomial, which is $$(-bc-ac-ab)(b-a)(c-a)(c-b)$$

Finally you get that the $K$ you are looking for is equal to $$K = ab + ac + bc$$

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The mistake is here

$$\begin{vmatrix}b+a& c+a\\b^2+ba+a^2&c^2+ca+a^2\end{vmatrix}=(c-b)(a-b)(b+a)(c+a)\begin{vmatrix}1&1\\(b+a)-ba&(c+a)-ca\end{vmatrix}$$

These aren't equal to each other. If you want to continue row reduction methods, you should continue with

$$\begin{vmatrix}b+a& c+a\\b^2+ba+a^2&c^2+ca+a^2\end{vmatrix} = \begin{vmatrix}b-c& c+a\\(b-c) a + b^2 - c^2 &c^2+ca+a^2\end{vmatrix} = (b-c)\begin{vmatrix}1& c+a\\a+b+c &c^2+ca+a^2\end{vmatrix}$$

And you should be able to take it from there.

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Note that we can rewrite the determinant as a sum: $$\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{vmatrix}=\begin{vmatrix}b^2&c^2\\b^3&c^3\end{vmatrix} - \begin{vmatrix}a^2&c^2\\a^3&c^3\end{vmatrix} + \begin{vmatrix}a^2&b^2\\a^3&b^3\end{vmatrix} = (b^2c^3 - b^3c^2) - (a^2c^3 - a^3c^2) + (a^2b^3 - a^3b^2)$$

Now, we can solve for $K$ by simply solving $K = \frac{(b^2c^3 - b^3c^2) - (a^2c^3 - a^3c^2) + (a^2b^3 - a^3b^2)}{(a - b)(b - c)(c - a)}$. If you need another hint just comment and I will add more.

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  • $\begingroup$ No, this is not a Vandermonde determinant, because there misses the line $(a \quad b \quad c)$... However, one can link this determinant with a Vandermonde : see my answer. $\endgroup$ – TheSilverDoe Mar 17 at 17:47
  • $\begingroup$ Ah I didnt notice that, thanks for the heads up :) $\endgroup$ – Victoria M Mar 17 at 18:01

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