Question

$\mathrm{If}∆=\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|,{∆}_1=\left|\begin{array}{ccc}1& 1& 1\\ yz& zx& xy\\ x& y& z\end{array}\right|,\mathrm{then}\mathrm{prove}\mathrm{that}∆+{∆}_1=0.$

Answer 1

$$\begin{gathered} ∆+{∆}_1=\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|+\left|\begin{array}{ccc}1& 1& 1\\ yz& zx& xy\\ x& y& z\end{array}\right| \\ =\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|+\left|\begin{array}{ccc}1& yz& x\\ 1& zx& y\\ 1& xy& z\end{array}\right|\left[\mathrm{Interchanging}\mathrm{rows}\mathrm{and}\mathrm{coloumns}\mathrm{in}{∆}_1\right] \\ =\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|-\left|\begin{array}{ccc}1& x& yz\\ 1& y& zx\\ 1& z& xy\end{array}\right|\left[\mathrm{Applying}{C}_2\leftrightarrow C_3\mathrm{in}{∆}_1\right] \\ =\left|\begin{array}{ccc}1& x& x^2\\ 0& y-x& y^2-x^2\\ 0& z-x& z^2-x^2\end{array}\right|-\left|\begin{array}{ccc}1& x& yz\\ 0& y-x& zx-yz\\ 0& z-x& xy-yz\end{array}\right|\left[\mathrm{Applying}{R}_2\to R_2-R_1\mathrm{and}{R}_3\to R_3-R_1\right] \\ =\left(y-x\right)\left(z-x\right)\left|\begin{array}{ccc}1& x& x^2\\ 0& 1& y+x\\ 0& 1& z+x\end{array}\right|-\left(y-x\right)\left(z-x\right)\left|\begin{array}{ccc}1& x& yz\\ 0& 1& -z\\ 0& 1& -y\end{array}\right|\left[\mathrm{Taking}\left(y-x\right)\mathrm{common}\mathrm{from}{R}_2\mathrm{and}\left(z-x\right)\mathrm{common}\mathrm{from}{R}_3\right] \\ =\left(y-x\right)\left(z-x\right)\left(z+x-y-x\right)-\left(y-x\right)\left(z-x\right)\left(-y+z\right)\left[\mathrm{Expanding}\mathrm{along}\mathrm{first}\mathrm{column}\right] \\ =\left(y-x\right)\left(z-x\right)\left(z-y\right)\left(1-1\right) \\ =0 \\ \therefore ∆+{∆}_1=0. \end{gathered}$$