If $\omega $ is the cube root of unity, then $\left| {\begin{array}{*{20}{c}}1&\omega &{{\omega ^2}}\\\omega &{{\omega ^2}}&1\\{{\omega ^2}}&1&\omega \end{array}} \right|$=

  • A

    $1$

  • B

    $0$

  • C

    $\omega $

  • D

    ${\omega ^2}$

Similar Questions

Evaluate $\Delta=\left|\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|$

Using the property of determinants and without expanding, Prove that 

$\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$

Show that $\left|\begin{array}{ccc}a & b & c \\ a+2 x & b+2 y & c+2 z \\ x & y & z\end{array}\right|=0$

If $a, b $ and $ c$ are non zero numbers, then $\Delta = \left| {\,\begin{array}{*{20}{c}}{{b^2}{c^2}}&{bc}&{b + c}\\{{c^2}{a^2}}&{ca}&{c + a}\\{{a^2}{b^2}}&{ab}&{a + b}\end{array}\,} \right|$ is equal to

The number of real values of $x$ satisfying $\left| {\,\begin{array}{*{20}{c}} x&{3x + 2}&{2x - 1}\\{2x - 1}&{4x}&{3x + 1}\\{7x - 2}&{17x + 6}&{12x - 1}\end{array}\,} \right|$ $= 0$ is