Question

If $A=\left[\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right],B=\left[\begin{array}{ccc}\omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \\ \omega & {\omega }^2& 1\end{array}\right]$ and $C=\left[\begin{array}{c}1\\ \omega \\ {\omega }^2\end{array}\right]$ where $\omega$ is the complex cube root of $1$ ,then $(A+B)C$ is equal to

• A. $\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
• B. $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
• C. $\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$
• D. $\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

Given, $A=\left[\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right],B=\left[\begin{array}{ccc}\omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \\ \omega & {\omega }^2& 1\end{array}\right]$ and $C=\left[\begin{array}{c}1\\ \omega \\ {\omega }^2\end{array}\right]$
Now, $(A+B)=\left[\begin{array}{ccc}1+\omega & \omega +{\omega }^2& {\omega }^2+1\\ \omega +{\omega }^2& {\omega }^2+1& 1+\omega \\ {\omega }^2+\omega & 1+{\omega }^2& \omega +1\end{array}\right]$
$\Rightarrow A+B=\left[\begin{array}{ccc}-{\omega }^2& -1& -\omega \\ -1& -\omega & -{\omega }^2\\ -1& -\omega & -{\omega }^2\end{array}\right]$
Now, $(A+B)C=\left[\begin{array}{ccc}-{\omega }^2& -1& -\omega \\ -1& -\omega & -{\omega }^2\\ -1& -\omega & -{\omega }^2\end{array}\right]\left[\begin{array}{c}1\\ \omega \\ {\omega }^2\end{array}\right]$
$=\left[\begin{array}{c}-({\omega }^2+\omega +{\omega }^3)\\ -({\omega }^2+\omega +{\omega }^3)\\ {-(\omega }^2+\omega +{\omega }^3)\end{array}\right]$
$=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$