Question

Let $a,b\text{and}c$ be three real numbers satisfying
$\left[\begin{array}{ccc}a& b& c\end{array}\right]\left[\begin{array}{ccc}1& 9& 7\\ 8& 2& 7\\ 7& 3& 7\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\end{array}\right]$ . . . (i)

Let $\omega$ be a solution of $x^3$ - 1 = 0 with Im $\omega$ > 0. If a = 2 with b and c satisfying Eq. (i) then the value of $\frac{3}{{\omega }^a}+\frac{1}{{\omega }^b}+\frac{3}{{\omega }^c}$ is

• A. -2
• B. 2
• C. 3
• D. -3

Answer 1

The correct option is A -2

On multiplying matrices, we have
$\left[\begin{array}{ccc}a+8b+7c& 9a+2b+3c& 7a+7b+7c\end{array}\right]=\left[\begin{array}{ccc}0& 0& 0\end{array}\right]$
On solving the 3 equations, we get $b=6a\text{and}c=-7a$

If a = 2, we get , b = 12, c = -14
$\therefore$ $\frac{3}{{\omega }^a}+\frac{1}{{\omega }^b}+\frac{3}{{\omega }^c}$
$\Rightarrow \frac{3}{{\omega }^2}+\frac{1}{{\omega }^{12}}+\frac{3}{{\omega }^{-14}}=\frac{3}{{\omega }^2}+1+3{\omega }^2=3\omega +1+3{\omega }^2$
= 1 + 3 $(\omega +{\omega }^2)$= 1 - 3 = -2