Question

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|

Answer 1

Hint: -Take $x,y,z$ common from column 1, column 2, and column 3 respectively.
Consider L.H.S

{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.