Paragraph: Let $a$, b and $c$ be three real numbers satisfying $\left[\begin{array}{lll}a & b &…
Paragraph:
Let $a$, b and $c$ be three real numbers satisfying $\left[\begin{array}{lll}a & b & c\end{array}\right]\left[\begin{array}{lll}1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]$
Question:
Let $\omega$ be a solution of $x^3-1=0$ with $\operatorname{Im}(\omega)>0$. If $a=2$, with $b$ and $c$ satisfying Eq. (E), then the value of $\frac{3}{\omega^a}+\frac{1}{\omega^b}+\frac{3}{\omega^c}$
$-2$
2
3
$-3$
Solution
$
\begin{aligned}
& \text { Given, }[a b c]_{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 \quad\left[\begin{array}{c}
a+8 b+7 c \\
9 a+2 b+3 c \\
7 a+7 b+7 c
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right] \\
& \Rightarrow \quad a+8 b+7 c=0 \\
& \Rightarrow 9 a+2 b+3 c=0 \\
& \Rightarrow \quad a+b+c=0 \\
&
\end{aligned}
$
On multiplying Eq. (iii) by 2 , then subtract from Eq. (ii), we get
$
7 a+c=0
$
Again multiplying Eq. (iii) by 3 , then subtract from Eq. (ii), we get
$
\begin{array}{ll}
& 6 a-b=0 \\
\therefore \quad & b=6 a \text { and } c=-7 a
\end{array}
$
If $a=2, b=12$ and $c=-14$
$
\begin{aligned}
\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{aligned}
$