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# Prove the following:$\left| {\begin{array}{*{20}{c}} {ax}&{by}&{cz} \\ {{x^2}}&{{y^2}}&{{z^2}} \\ 1&1&1 \end{array}} \right| = \left| {\begin{array}{*{20}{c}} a&b&c \\ x&y&z \\ {yz}&{zx}&{xy} \end{array}} \right|$  Hint: -Take $x,y,z$ common from column 1, column 2, and column 3 respectively.
Consider L.H.S
$\Rightarrow \left| {\begin{array}{*{20}{c}} {ax}&{by}&{cz} \\ {{x^2}}&{{y^2}}&{{z^2}} \\ 1&1&1 \end{array}} \right|$
Take $x,y,z$ common from column 1, column 2, and column 3 respectively.
$\Rightarrow \left| {\begin{array}{*{20}{c}} {ax}&{by}&{cz} \\ {{x^2}}&{{y^2}}&{{z^2}} \\ 1&1&1 \end{array}} \right| = xyz\left| {\begin{array}{*{20}{c}} a&b&c \\ x&y&z \\ {\frac{1}{x}}&{\frac{1}{y}}&{\frac{1}{z}} \end{array}} \right|$
Now multiply $xyz$ in the third row of the determinant.
$\Rightarrow xyz\left| {\begin{array}{*{20}{c}} a&b&c \\ x&y&z \\ {\frac{1}{x}}&{\frac{1}{y}}&{\frac{1}{z}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} a&b&c \\ x&y&z \\ {\frac{{xyz}}{x}}&{\frac{{xyz}}{y}}&{\frac{{xyz}}{z}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} a&b&c \\ x&y&z \\ {yz}&{zx}&{xy} \end{array}} \right| = {\text{R}}{\text{.H}}{\text{.S}}$
Hence Proved.

Note: - In such types of questions first take $x,y,z$ common from column 1, column 2, and column 3 respectively, then multiply $xyz$ in the third row of the determinant we will get the required result.

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