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Prove without expanding: \begin{equation} \begin{vmatrix}bc&a^2&a^2\\b^2&ac&b^2\\c^2&c^2 & ab\end{vmatrix} = \begin{vmatrix}ac&bc&ab\\bc&ab&ac\\ab&ac&bc\end{vmatrix} \end{equation}


  • Tried to multiply by 'abc' in all rows then take common factors.
  • Tried to expand determinant into two determinants.
  • I used the determinant properties shown here: http://www.vitutor.com/alg/determinants/properties_determinants.html
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    • $\begingroup$ Sorry, maybe I missed something, but why don't you just take the determinant of both matrices and show they give the same result? $\endgroup$ – mattos May 27 '15 at 11:23
    • $\begingroup$ @Mattos because The problem stated that I must use determinant properties, sorry I forgot to mention that $\endgroup$ – Mario May 27 '15 at 11:25
    • $\begingroup$ Oh ok. Well see here, equation $(32)$, I think it may help using the 6 statements above it. $\endgroup$ – mattos May 27 '15 at 11:29
    • $\begingroup$ Have you tried the property that $\det AB = \det A\det B$ Perhaps you can multiply both sides by some matrix to make them equal $\endgroup$ – Dan Robertson May 27 '15 at 11:29
    • $\begingroup$ @DanRobertson That's the first thing I've tried, looked reasonable but didn't work $\endgroup$ – Mario May 27 '15 at 11:35
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    We check that the property holds when one of $a,b,c$ is zero. Now assume $a,b,c$ are all non-zero.

    We divide the columns by $bc$, $ac$, and $ab$, respectively, and multiply the rows by $bc$, $ac$, and $ab$ respectively. This gives

    $$\begin{vmatrix}bc&a^2&a^2\\b^2&ac&b^2\\c^2&c^2 & ab\end{vmatrix} = \begin{vmatrix}bc&ab&ac\\ab&ac&bc\\ac&bc&ab\end{vmatrix}$$ then we cycle the rows to obtain $$=\begin{vmatrix}ac&bc&ab\\bc&ab&ac\\ab&ac&bc\end{vmatrix}.$$

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