Question

If $\omega$is a cube root of unity, then $\left| \begin{matrix} x + 1 & \omega & \omega^{2} \\ \omega & x + \omega^{2} & 1 \\ \omega^{2} & 1 & x + \omega \end{matrix} \right| =$

• A. $x^{3} + 1$
• B. $x^{3} + \omega$
• C. $x^{3} + \omega^{2}$
• D. $x^{3}$

Answer 1

Answer: (D)

$x^{3}$

Answer 2

$\Delta = \left| \begin{matrix} x + 1 & \omega & \omega^{2} \\ \omega & x + \omega^{2} & 1 \\ \omega^{2} & 1 & x + \omega \end{matrix} \right|$

=$\left| \begin{matrix} x + 1 + \omega + \omega^{2} & \omega & \omega^{2} \\ x + 1 + \omega + \omega^{2} & x + \omega^{2} & 1 \\ x + 1 + \omega + \omega^{2} & 1 & x + \omega \end{matrix} \right|$,$(C_{1} \rightarrow C_{1} + C_{2} + C_{3})$

= $x\left| \begin{matrix} 1 & \omega & \omega^{2} \\ 1 & x + \omega^{2} & 1 \\ 1 & 1 & x + \omega \end{matrix} \right|$ $\left| \begin{matrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{matrix} \right| \neq 0$

= $x\lbrack 1\{(x + \omega^{2})(x + \omega) - 1\} + \omega\{ 1 - (x + \omega)\} + \omega^{2}\{ 1 - (x + \omega^{2})\}\rbrack$

= $x(x^{2} + \omega x + \omega^{2}x + \omega^{3} - 1 + \omega - \omega x - \omega^{2} + \omega^{2} - \omega^{2}x - \omega^{4})$

= $x^{3}$ , $(\because\omega^{3} = 1)$.