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# Using the properties of determinants, evaluate $\begin{vmatrix} 0 & xy^2 & xz^2 \\ x^2y & 0 & yz^2 \\ x^2z & zy^2 & 0 \end{vmatrix}$

Toolbox:
• If each element of a row(or a column)of A is expressed as the sum of two or more terms,then the determinants of A can be expressed as the sum of determinants of two or more matrices.
• If each element of a row(or column)of a determinant is multiplied by a constant 'k' then |A|=k|A|.
Let $\Delta=\begin{vmatrix}0 & xy^2 &xz^2\\x^2y& 0 & yz^2\\x^2z & zy^2&0\end{vmatrix}$

Take x as a common factor from $R_1$,y from $R_2$ and z from $R_3$

We get $\Delta=xyz\begin{vmatrix}0 & y^2 &z^2\\x^2& 0 & z^2\\x^2 & y^2&0\end{vmatrix}$

Again take $x^2$ as the common factor from $C_1$,$y^2$ from $C_2$ and $z^2$ from $C_3$ we get,

$\Delta=(xyz)(x^2y^2z^2)\begin{vmatrix}0 & 1 &1\\1& 0 & 1\\1 & 1&0\end{vmatrix}$

Apply $C_2\rightarrow C_2-C_3$

$\Delta=(x^3y^3z^3)\begin{vmatrix}0 & 0 &1\\1& -1 & 1\\1 & 1&0\end{vmatrix}$

On expanding along $R_1$ we get,

$\Delta=(x^3y^3z^3)[1(1+1)]$

Therefore $\Delta=2x^3y^3z^3$

edited Mar 14, 2013