Mathematics - Complex Number Question with Solution | TestHub

$$\begin{array}{rl}& \text{ Given, }[abc{]}_{1\times 3}{\left[\begin{array}{lll}1& 9& 7\\ 8& 2& 7\\ 7& 3& 7\end{array}\right]}_{3\times 3}=\left[\begin{array}{lll}0& 0& 0\end{array}\right]\\ & \Rightarrow \left[\begin{array}{c}a+8b+7c\\ 9a+2b+3c\\ 7a+7b+7c\end{array}\right]=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]\\ & \Rightarrow a+8b+7c=0\\ & \Rightarrow 9a+2b+3c=0\\ & \Rightarrow a+b+c=0\\ & \end{array}$$ On multiplying Eq. (iii) by 2 , then subtract from Eq. (ii), we get $$7a+c=0$$ Again multiplying Eq. (iii) by 3 , then subtract from Eq. (ii), we get $$\begin{array}{ll}& 6a-b=0\\ \therefore & b=6a\text{ and }c=-7a\end{array}$$ If $a=2,b=12$ and $c=-14$ $$\begin{array}{rl}\therefore & \frac{3}{{\omega }^a}+\frac{1}{{\omega }^b}+\frac{3}{{\omega }^c}\\ \Rightarrow & \frac{3}{{\omega }^2}+\frac{1}{{\omega }^{12}}+\frac{3}{{\omega }^{-14}}=\frac{3}{{\omega }^2}+1+3{\omega }^2\\ & =3\omega +1+3{\omega }^2\\ & =1+3\left(\omega +{\omega }^2\right)\\ & =1-3=-2\end{array}$$