# 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|$

Last updated date: 22nd Mar 2023

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Answer

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308.4k+ 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.

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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