Question

If $\omega$ and ${\omega }^2$ are cube roots of unity, then the value of the determinant $\left|\begin{array}{ccc}x+1& \omega & {\omega }^2\\ \omega & x+{\omega }^2& 1\\ {\omega }^2& 1& x+\omega \end{array}\right|$ is

• A. $0$
• B. $1$
• C. $x$
• D. $x^3$

Answer 1

Answer: (D)

$x^3$

$\left|\begin{array}{ccc}x+1& \omega & {\omega }^2\\ \omega & x+{\omega }^2& 1\\ {\omega }^2& 1& x+\omega \end{array}\right|$
$C_1\to C_1+C_2+C_3$
$=\left|\begin{array}{ccc}x& \omega & {\omega }^2\\ x& x+{\omega }^2& 1\\ x& 1& x+\omega \end{array}\right|$ ($\because 1+\omega +{\omega }^2=0$)
$=x\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ 1& x+{\omega }^2& 1\\ 1& 1& x+\omega \end{array}\right|$
On expanding and simplifying we get
$\Delta =x^3$