# Determinant proof using its properties

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

• 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
• Sorry, maybe I missed something, but why don't you just take the determinant of both matrices and show they give the same result? – mattos May 27 '15 at 11:23
• @Mattos because The problem stated that I must use determinant properties, sorry I forgot to mention that – Mario May 27 '15 at 11:25
• Oh ok. Well see here, equation $(32)$, I think it may help using the 6 statements above it. – mattos May 27 '15 at 11:29
• 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 – Dan Robertson May 27 '15 at 11:29
• @DanRobertson That's the first thing I've tried, looked reasonable but didn't work – Mario May 27 '15 at 11:35

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 $$=\begin{vmatrix}ac&bc&ab\\bc&ab&ac\\ab&ac&bc\end{vmatrix}.$$