Question

If $\omega \ne 1$ is a cube root of unity, then the value of $\left| \begin{matrix} 1+2{{\omega }^{100}}+{{\omega }^{200}} \\\ 1 \\\ \omega \\\\\end{matrix}\begin{matrix} {{\omega }^{2}} \\\ 1+{{\omega }^{100}}+2{{\omega }^{200}} \\\ {{\omega }^{2}} \\\\\end{matrix}\begin{matrix} 1 \\\ \omega \\\ 1+{{\omega }^{100}}+2{{\omega }^{200}} \\\\\end{matrix} \right|$ is equal to

• A. $0$
• B. $1$
• C. $\omega$
• D. ${{\omega }^{2}}$

Answer 1

Answer: (A)

$0$

Answer 2

Let $\Delta =\left| \begin{matrix} 1+2{{\omega }^{100}}+{{\omega }^{200}} & {{\omega }^{2}} \\\ 1 & 1+{{\omega }^{100}}+2{{\omega }^{200}} \\\ \omega & {{\omega }^{2}} \\\ \end{matrix} \right.$ $\left. \begin{matrix} 1 \\\ \omega \\\ 2+{{\omega }^{100}}+{{\omega }^{200}} \\\ \end{matrix} \right|$
$=\left| \begin{matrix} 1+2\omega +{{\omega }^{2}} & {{\omega }^{2}} & 1 \\\ 1 & 1+\omega +2{{\omega }^{2}} & \omega \\\ \omega & {{\omega }^{2}} & 2+\omega +{{\omega }^{2}} \\\ \end{matrix} \right|$
$=\left| \begin{matrix} \omega & {{\omega }^{2}} & 1 \\\ 1 & {{\omega }^{2}} & \omega \\\ \omega & {{\omega }^{2}} & 1 \\\ \end{matrix} \right|$
$=0$ ( $\because$ Rows ${{R}_{1}}$ and ${{R}_{3}}$ are identical)