Question

If $x\ne y\ne z$ and $\left|\begin{array}{ccc}x& x^2& x^3\\ y& y^2& y^3\\ z& z^2& z^3\end{array}\right|=0$, then xyz is equal to:

• A. 1
• B. -1
• C. 0
• D. x+y+z

Answer 1

Answer: (C)

Let $A=\left|\begin{array}{ccc}x& x^2& x^3\\ y& y^2& y^3\\ z& z^2& z^3\end{array}\right|$
By taking x, y, z common from the rows $R_1,R_2$ and $R_3$ respectively. So,
$A=xyz\left|\begin{array}{ccc}1& x& x^2\\ 1& y& y^2\\ 1& z& z^2\end{array}\right|$
Operate $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$
$\Rightarrow A=xyz\left|\begin{array}{ccc}1& x& x^2\\ 0& y-x& y^2-x^2\\ 0& z-x& z^2-x^2\end{array}\right|$
Now take common y - x and z - x from the rows $R_2$ and $R_3$ respectively. Thus
$A=xyz(y-x)(z-x)\left|\begin{array}{ccc}1& x& x^2\\ 0& 1& y+x\\ 0& 1& z+x\end{array}\right|$
= xyz (y - x) (z - x) (z - y)
= xyz (x - y) (y - z) (z - x)
Given | A | = 0
So, xyz = 0 $\because x\ne y\ne z$ (given)