Practicing Success
If x, y, z are all distinct and $\begin{vmatrix} x & x^2 & 1+x^3\\y & y^2& 1+y^3\\z & z^2 & 1+z^3\end{vmatrix}= 0 $ then the value of xyz is : |
-2 -1 -3 2 |
-1 |
The correct answer is Option (2) → -1 $\begin{vmatrix} x & x^2 & 1+x^3\\y & y^2& 1+y^3\\z & z^2 & 1+z^3\end{vmatrix}=\begin{vmatrix} x & x^2 & 1\\y & y^2& 1\\z & z^2 & 1\end{vmatrix}+\begin{vmatrix} x & x^2 & x^3\\y & y^2& y^3\\z & z^2 & z^3\end{vmatrix}=0$ $=\begin{vmatrix} x & x^2 & 1\\y & y^2& 1\\z & z^2 & 1\end{vmatrix}+xyz\begin{vmatrix} 1 & x & x^2\\1 & y& y^2\\1 & z& z^2\end{vmatrix}$ $(C_1↔C_3)$ $=\begin{vmatrix} x & x^2 & 1\\y & y^2& 1\\z & z^2 & 1\end{vmatrix}+xyz\begin{vmatrix} x^2 & x & 1\\y^2 & y& 1\\z^2 & z & 1\end{vmatrix}$ $(C_1↔C_2)$ $=\begin{vmatrix} x & x^2 & 1\\y & y^2& 1\\z & z^2 & 1\end{vmatrix}+xyz\begin{vmatrix} x & x^2 & 1\\y & y^2& 1\\z & z^2 & 1\end{vmatrix}=0$ so $xyz=-1$ |