Question

Prove that :

\[\begin{vmatrix}z & x & y \\ z^2 & x^2 & y^2 \\ z^4 & x^4 & y^4\end{vmatrix} = \begin{vmatrix}x & y & z \\ x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4\end{vmatrix} = \begin{vmatrix}x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \\ x & y & z\end{vmatrix} = xyz \left( x - y \right) \left( y - z \right) \left( z - x \right) \left( x + y + z \right) .\]

 

Answer 1

\[Let \Delta_1 = \begin{vmatrix} z & x & y\\ z^2 & x^2 & y^2 \\ z^4 & x^4 & y^4 \end{vmatrix}, \Delta_2 = \begin{vmatrix} x & y & z\\ x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \end{vmatrix}, \Delta_3 = \begin{vmatrix} x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \\ x & y & z \end{vmatrix} \text{ and }\Delta_4 = xyz\left( x - y \right)\left( y - z \right)\left( z - x \right) \left( x + y + z \right)\]
Now,
\[ \Delta_{1 =} \begin{vmatrix} z & x & y\\ z^2 & x^2 & y^2 \\ z^4 & x^4 & y^4 \end{vmatrix}\]
Using the property that if two rows ( or columns ) of a determinant are interchanged, the value of the determinant becomes negetive, we get

\[ \Rightarrow \Delta_1 = \left( - 1 \right) \begin{vmatrix} x & z & y\\ x^2 & z^2 & y^2 \\ x^4 & z^4 & y^4 \end{vmatrix} \left[ \because C_1 \leftrightarrow C_2 \right]\]
\[ = \left( - 1 \right)\left( - 1 \right)\begin{vmatrix} x & y & z\\x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \end{vmatrix} \left[ \because C_2 \leftrightarrow C_3 \right]\]

\[ = \begin{vmatrix} x & y & z\\ x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \end{vmatrix} = \Delta_2 . . . (1)\]
\[ = \left( - 1 \right) \begin{vmatrix} x^2 & y^2 & z^2 \\x & y & z\\ x^4 & y^4 & z^4 \end{vmatrix} \left[\text{ Applying }R_1 \leftrightarrow R_2 \right]\]
\[ = \left( - 1 \right) \left( - 1 \right) \begin{vmatrix} x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \\x & y^{} & z \end{vmatrix} \left[\text{ Applying }R_2 \leftrightarrow R_3 \right] \]
\[ = \begin{vmatrix} x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \\x & y^{} & z \end{vmatrix} = \Delta_3 . . . (2)\]
\[Thus, \]

\[ \Delta_1 = \Delta_2 = \Delta_3 \left[\text{ From eqs }. (1)\text{ and }(2) \right]\]
\[∆_2 = \begin{vmatrix} x & y & z\\ x^2 & y^2 & z^2 \\ x^4 & y^4 & z^4 \end{vmatrix}\] 
\[ = xyz \begin{vmatrix} 1 & 1 & 1\\x & y^{} & z\\ x^3 & y^3 & z^3 \end{vmatrix} \left[\text{ Taking out common factor x from } C_{1 ,}\text{ y from }C_2\text{ and z from }C_3 \right]\] 
\[ = xyz\begin{vmatrix} 0 & 0 & 1\\ x - y & y - z^{} & z\\ x^3 - y^3 & y^3 - z^3 & z^3 \end{vmatrix} \left[\text{ Applying }C \to C_1 \hspace{0.167em} - C_2\text{ and }C_2 \to C_2 - C_3 \right]\] 
\[ = xyz\left( x - y \right) \left( y - z \right) \begin{vmatrix} 0 & 0 & 1\\1 & 1 & z\\ x^2 + 2xy + y^2 & y^2 + 2yz + z^2 & z^3 \end{vmatrix} \left[ \because \left( a^3 - b^3 \right) = \left( a - b \right)\left( a^2 + ab + b^2 \right) \right] \left[\text{ Taking out common factor }\left( x - y \right)\text{ from }C_1\text{ and }\left( y - z \right)\text{ from }C_2 \right]\] 
\[ = xyz\left( x - y \right) \left( y - z \right)\left\{ 1 \times \begin{vmatrix} 1 & 1 \\ x^2 + xy + y^2 & y^2 + yz + z^2 \end{vmatrix} \right\} \left[\text{ Expanding along }R_1 \right]\] 
\[ = xyz\left( x - y \right) \left( y - z \right)\left\{ y^2 + yz + z^2 - x^2 - xy - y^2 \right\}\] 
\[ = xyz\left( x - y \right) \left( y - z \right)\left\{ yz - xy + z^2 - x^2 \right\}\] 
\[ = xyz\left( x - y \right) \left( y - z \right)\left\{ y\left( z - x \right) + \left( z - x \right)\left( z + x \right) \right\}\] 
\[ = xyz\left( x - y \right) \left( y - z \right)\left( z - x \right)\left( y + x + z \right)\] 
\[ = xyz\left( x - y \right) \left( y - z \right)\left( z - x \right)\left( x + y + z \right)\] 
\[ = ∆_4 \] 
\[Thus, \] 
\[ ∆_1 = ∆_2 = ∆_3 = ∆_4 \]