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# Without expanding the determinant, prove that $\begin{vmatrix} a &a^2 &bc \\ b &b^2 &ca \\ c &c^2 &ab \end{vmatrix}\; = \; \begin{vmatrix} 1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3 \end{vmatrix}$

Toolbox:
• If each element of a row(or a column) of a determinant is multiplied by k.
• By this property we can take out any common factor from any one row or any column of a given determinant.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns. rows.
• 2. Mutiplication of the elements of any row or column by a non-zero number
• 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.

$\begin{vmatrix}a & a^2 & bc\\b & b^2 & ca\\c & c^2 & ab\end{vmatrix}=\begin{vmatrix}a & a^2 & a^3\\b & b^2 & b^3\\c & c^2 & c^3\end{vmatrix}$

Expanding the determinant .

Now let us consider the LHS

Let us multiply the $R_1$ by a,$R_2$ by b and $R_3$ by c.

So,$\begin{vmatrix}a & a^2 & bc\\b & b^2 & ca\\c & c^2 & ab\end{vmatrix}=\frac{1}{abc}\begin{vmatrix}a^2 & a^3 & abc\\b^2 & b^3 & abc\\c^2 & c^3 & abc\end{vmatrix}$

Now we can take abc as a common factor from $C_3$

LHS=$\frac{abc}{abc}\begin{vmatrix}a^2 & a^3& 1\\b^2 & b^3 &1\\c^2& c^3 & 1\end{vmatrix}$

Let us interchange $C_1$ and $C_3$ and $C_2$ and$C_3$

(i.e)$C_1\leftrightarrow C_3$,$C_2\leftrightarrow C_3$

Therefore LHS=$\begin{vmatrix}1 & a^2 & a^3\\1 & b^2 & b^3\\1 & c^2 & c^3\end{vmatrix}=RHS.$

Hence LHS=RHS.proved.

edited Feb 28, 2013