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Prove the following:
$\left| {\begin{array}{*{20}{c}}
  {ax}&{by}&{cz} \\
  {{x^2}}&{{y^2}}&{{z^2}} \\
  1&1&1
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  a&b&c \\
  x&y&z \\
  {yz}&{zx}&{xy}
\end{array}} \right|$

Answer
VerifiedVerified
150k+ views
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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