Question

\[If ∆ = \begin{vmatrix}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{vmatrix}, ∆_1 = \begin{vmatrix}1 & 1 & 1 \\ yz & zx & xy \\ x & y & z\end{vmatrix},\text{ then prove that }∆ + ∆_1 = 0 .\]

Answer 1

\[∆ + ∆_1 = \begin{vmatrix}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{vmatrix} + \begin{vmatrix}1 & 1 & 1 \\ yz & zx & xy \\ x & y & z\end{vmatrix}\]

\[ = \begin{vmatrix}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{vmatrix} + \begin{vmatrix}1 & yz & x \\ 1 & zx & y \\ 1 & xy & z\end{vmatrix} \left[\text{ Interchanging rows and coloumns in }∆_1 \right]\]

\[ = \begin{vmatrix}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{vmatrix} - \begin{vmatrix}1 & x & yz \\ 1 & y & zx \\ 1 & z & xy\end{vmatrix} \left[\text{ Applying }C_2 \leftrightarrow C_3\text{ in }∆_1 \right]\]

\[ = \begin{vmatrix}1 & x & x^2 \\ 0 & y - x & y^2 - x^2 \\ 0 & z - x & z^2 - x^2\end{vmatrix} - \begin{vmatrix}1 & x & yz \\ 0 & y - x & zx - yz \\ 0 & z - x & xy - yz\end{vmatrix} \left[ \text{ Applying }R_2 \to R_2 - R_1\text{ and }R_3 \to R_3 - R_1 \right]\]

\[ = \left( y - x \right)\left( z - x \right)\begin{vmatrix}1 & x & x^2 \\ 0 & 1 & y + x \\ 0 & 1 & z + x\end{vmatrix} - \left( y - x \right)\left( z - x \right)\begin{vmatrix}1 & x & yz \\ 0 & 1 & - z \\ 0 & 1 & - y\end{vmatrix} \left[\text{ Taking }\left( y - x \right) \text{ common from }R_2\text{ and }\left( z - x \right)\text{ common from }R_3 \right]\]

\[ = \left( y - x \right)\left( z - x \right)\left( z + x - y - x \right) - \left( y - x \right)\left( z - x \right)\left( - y + z \right) \left[\text{ Expanding along first column }\right]\]

\[ = \left( y - x \right)\left( z - x \right)\left( z - y \right)\left( 1 - 1 \right)\]

\[ = 0\]

\[ \therefore ∆ + ∆_1 = 0 .\]