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Using the properties of determinants, prove that: $\begin{vmatrix} y^2z^2 & yz &y+ z \\ z^2x^2 & zx & z+x \\ x^2y^2 & xy & x+y \end{vmatrix}\;=\;0$

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  • If each element of a matrix A of a row(or column)of a determinant is multiplied by a constant 'k' then |A|=k|A|.
  • Let A be a square matrix,such that 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.
Let $\Delta=\begin{vmatrix}y^2z^2 & yz &y+z\\z^2x^2& zx & z+x\\x^2y^2 & xy&x+y\end{vmatrix}$
Apply $R_1\rightarrow xR_1,R_2\rightarrow yR_2,R_3\rightarrow zR_3$
$\Delta=\frac{1}{xyz}\begin{vmatrix}xy^2z^2 & xyz &xy+xz\\yz^2x^2& xyz & yz+xy\\zx^2y^2 & xyz&xz+yz\end{vmatrix}$
Let us take xyz as a common factor from $C_1$ and $C_2$
$\Delta=\frac{(xyz)^2}{xyz}\begin{vmatrix}1 & 1 &xy+xz\\1& 1 & yz+xy\\1 & 1&xz+yz\end{vmatrix}=0$
Since two rows are identical,the value of the determinant is 0.
Hence |$\Delta|=0.$


answered Mar 15, 2013 by sreemathi.v

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