Question

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

Answer 1

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