Question

If $\omega$ is an imaginary cube root of unity and if $\left|\begin{array}{ccc}x+{\omega }^2& \omega & 1\\ \omega & {\omega }^2& 1+x\\ 1& x+\omega & {\omega }^2\end{array}\right|=0$ then one of the value of $x$ is

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

The correct option is B

$0$

Explanation for the correct option.

Step 1: Prerequisites for the solution

Since $\omega$ is an imaginary cube root of unity then we know that

${\omega }^3=1\text{and}1+\omega +{\omega }^2=0$

Step 2: Simplification of the determinant for the value of $\mathit{x}$

To use the fact that $1+\omega +{\omega }^2=0$, we need to do the column transformation in the determinant which will be

$C_1\to C_1+C_2+C_3$

$$\begin{gathered} \therefore \left|\begin{array}{ccc}x+{\omega }^2+\omega +1& \omega & 1\\ {\omega }^2+\omega +1+x& {\omega }^2& 1+x\\ 1+x+\omega +{\omega }^2& x+\omega & {\omega }^2\end{array}\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}x+0& \omega & 1\\ 0+x& {\omega }^2& 1+x\\ 0+x& x+\omega & {\omega }^2\end{array}\right|=0 \end{gathered}$$

Now, suppose that $x=0$ then the determinant will be,

$\left|\begin{array}{ccc}0& \omega & 1\\ 0& {\omega }^2& 1\\ 0& \omega & {\omega }^2\end{array}\right|$

We know that if a determinant has any row or column with all the elements as $0$ then its value is $0$.

Hence, the correct option is (B).