Question

If $A=\left[\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right],B=\left[\begin{array}{ccc}\omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \\ \omega & {\omega }^2& 1\end{array}\right],C=\left[\begin{array}{c}1\\ \omega \\ {\omega }^2\end{array}\right]$ then $(A+B)C$ is

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

Answer 1

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

$A+B=\left(\begin{array}{ccc}{\omega }^2& 1& \omega \\ 1& \omega & {\omega }^2\\ 1& \omega & {\omega }^2\end{array}\right)$
$\therefore A+B=-\left(\begin{array}{ccc}{\omega }^2& 1& \omega \\ 1& \omega & {\omega }^2\\ 1& \omega & {\omega }^2\end{array}\right)$
$\therefore (A+B)C=-\left(\begin{array}{ccc}{\omega }^2& 1& \omega \\ 1& \omega & {\omega }^2\\ 1& \omega & {\omega }^2\end{array}\right)\left(\begin{array}{c}1\\ \omega \\ {\omega }^2\end{array}\right)=\left(\begin{array}{c}1+\omega +{\omega }^2\\ 1+{\omega }^2+\omega \\ 1+{\omega }^2+\omega \end{array}\right)$
$=O_{3\times 1}$
Hence choice (d) is correct.