If $x ≠ y ≠ z$ then $\begin{vmatrix}1&x&x^2\\1&y&y^2\\1&z&z^2\end{vmatrix}$ is equal to |
$x + y + z$ $x\,y\,z$ $(x-y)(y-z)(z - x)$ 0 |
$(x-y)(y-z)(z - x)$ |
The correct answer is Option (3) → $(x-y)(y-z)(z - x)$ $\begin{vmatrix}1&x&x^2\\1&y&y^2\\1&z&z^2\end{vmatrix}$ $R_2\to R_2-R_1,\ R_3\to R_3-R_1$ $=\begin{vmatrix}1&x&x^2\\0&y-x&y^2-x^2\\0&z-x&z^2-x^2\end{vmatrix}$ $=\begin{vmatrix}1&x&x^2\\0&y-x&(y-x)(y+x)\\0&z-x&(z-x)(z+x)\end{vmatrix}$ $=(y-x)(z-x)\begin{vmatrix}1&x&x^2\\0&1&y+x\\0&1&z+x\end{vmatrix}$ $C_3\to C_3-C_2$ $=(y-x)(z-x)\begin{vmatrix}1&x&0\\0&1&z-y\\0&1&z+x\end{vmatrix}$ $=(y-x)(z-x)(z-y)$ $(x-y)(y-z)(z-x)$ |